\(\dfrac{-4}{5}\cdot13\dfrac{1}{3}-\dfrac{4}{5}\cdot\left(-13\dfrac{1}{2}\right)\)Tính
H24
Những câu hỏi liên quan
a) \(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)
b) \(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)
câu này làm cx đc hoặc ko làm cx ko sao :)
\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)
`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`
`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`
`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`
`=(1-1/7).[130-4-125]/3`
`=6/7 . 1/3 = 2/7`
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`8/9+1/9 . 2/9+1/9 . 7/9`
`=8/9+1/9.(2/9+7/9)`
`=8/9+1/9 . 9/9`
`=8/9+1/9=9/9=1`
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a) \(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)
\(=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}\)
\(=-\dfrac{5}{24}\)
b) \(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)\)
\(=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}\)
\(=\dfrac{6}{7}\cdot\dfrac{1}{3}\)
\(=\dfrac{2}{7}\)
\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)
\(=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}\)
\(=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}\)
\(=1\)
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Tính giá trị biểu thức :
1. Adfrac{dfrac{2}{5}+dfrac{2}{7}-dfrac{2}{9}-dfrac{2}{11}}{dfrac{4}{5}+dfrac{4}{7}-dfrac{4}{9}-dfrac{4}{11}}
2. Bdfrac{1^2}{1cdot2}cdotdfrac{2^2}{2cdot3}cdotdfrac{3^2}{3cdot4}cdotdfrac{4^2}{4cdot5}
3. Cdfrac{2^2}{1cdot3}cdotdfrac{3^2}{2cdot4}cdotdfrac{4^2}{3cdot5}cdotdfrac{5^2}{4cdot6}cdotdfrac{5^2}{4cdot6}
4. Dleft(dfrac{4}{5}-dfrac{1}{6}right)cdotleft(dfrac{2}{3}cdotdfrac{1}{4}right)^2
5. Cho M8dfrac{2}{7}-left(3dfrac{4}{9}+4dfrac{2}{7}right) ; Nleft(10dfrac{2}...
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Tính giá trị biểu thức :
1. \(A=\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}\)
2. \(B=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
3. \(C=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4\cdot6}\cdot\dfrac{5^2}{4\cdot6}\)
4. \(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right)\cdot\left(\dfrac{2}{3}\cdot\dfrac{1}{4}\right)^2\)
5. Cho \(M=8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\) ; \(N=\left(10\dfrac{2}{9}+2\dfrac{3}{5}\right)-6\dfrac{2}{9}\). Tính \(P=M-N\)
6. \(E=10101\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3\cdot7\cdot11\cdot13\cdot37}\right)\)
7. \(F=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{256}+\dfrac{3}{64}}{1-\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
8. \(G=\text{[}\dfrac{\left(6-4\dfrac{1}{2}\right):0,03}{\left(3\dfrac{1}{20}-2,65\right)\cdot4+\dfrac{2}{5}}-\dfrac{\left(0,3-\dfrac{3}{20}\right)\cdot1\dfrac{1}{2}}{\left(1,88+2\dfrac{3}{25}\right)\cdot\dfrac{1}{80}}\text{]}:\dfrac{49}{60}\)
9. \(H=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{4\cdot5\cdot6}+...+\dfrac{1}{98\cdot99\cdot100}\)
10. \(I=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)
11. \(K=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{999}\right)\)
12. \(L=1\dfrac{1}{3}+1\dfrac{1}{8}+1\dfrac{1}{15}...\) (98 thừa số)
13. \(M=-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{3}}}}\)
14. \(N=\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}\)
15. \(P=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{5}-1\right)...\left(\dfrac{1}{2001}-1\right)\)
16. \(Q=\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2005\cdot2006}\right):\left(\dfrac{1}{1004\cdot2006}+\dfrac{1}{1005\cdot2005}+...+\dfrac{1}{2006\cdot1004}\right)\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
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Tính giá trị của các biểu thức sau 1) A1+2+2^2+...+2^{2015} 2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right) 3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89} 4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3} 5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}...
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Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
\(\left(\dfrac{4}{13}\cdot\dfrac{6}{5}+\dfrac{4}{13}\cdot\dfrac{2}{5}\right)\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
Ta có: \(\left(\dfrac{4}{13}\cdot\dfrac{6}{5}+\dfrac{4}{13}\cdot\dfrac{2}{5}\right)\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{4}{13}\cdot\left(\dfrac{6}{5}+\dfrac{2}{5}\right)\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{4}{13}\cdot\dfrac{8}{5}\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{32}{65}\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\left(2x+1\right)^2=\dfrac{10}{13}:\dfrac{32}{65}=\dfrac{10}{13}\cdot\dfrac{65}{32}=\dfrac{25}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{5}{4}\\2x+1=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{4}-1=\dfrac{1}{4}\\2x=-\dfrac{5}{4}-1=-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}:2=\dfrac{1}{8}\\x=-\dfrac{9}{4}:2=-\dfrac{9}{8}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{1}{8};-\dfrac{9}{8}\right\}\)
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Tính giá trị của các biểu thức sau1) A1+2+2^2+...+2^{2015}2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right)3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89}4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3}5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}{2005...
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Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
Tính :
a, dfrac{3cdot13-13cdot18}{15cdot40-80};
b, dfrac{18cdot34+left(-18right)cdot124}{-36cdot17+9cdotleft(-52right)};
c, dfrac{dfrac{3}{41}-dfrac{12}{47}+dfrac{27}{53}}{dfrac{4}{41}-dfrac{16}{47}+dfrac{36}{53}}+dfrac{-0,25cdotdfrac{-2}{3}-0,75:left(dfrac{-1}{2}+dfrac{2}{3}right)}{left|-1dfrac{1}{2}right|cdotleft(dfrac{-2}{3}-75%:dfrac{3}{-2}right)}.
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Tính :
a, \(\dfrac{3\cdot13-13\cdot18}{15\cdot40-80}\);
b, \(\dfrac{18\cdot34+\left(-18\right)\cdot124}{-36\cdot17+9\cdot\left(-52\right)}\);
c, \(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}+\dfrac{-0,25\cdot\dfrac{-2}{3}-0,75:\left(\dfrac{-1}{2}+\dfrac{2}{3}\right)}{\left|-1\dfrac{1}{2}\right|\cdot\left(\dfrac{-2}{3}-75\%:\dfrac{3}{-2}\right)}\).
a: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{13}{15-2}\cdot\dfrac{-15}{40}=\dfrac{-3}{8}\)
b: \(=\dfrac{18\left(34-124\right)}{36\left(-17-13\right)}=\dfrac{1}{2}\cdot\dfrac{-90}{-30}=\dfrac{3}{2}\)
c: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{\dfrac{-1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}\)
\(=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
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Tìm $x$, biết :
a) $\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}$
b) $\left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}$
c) $\left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}$
d) $\dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)$
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
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\(\dfrac{-5}{9}+1\dfrac{5}{9}\cdot\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\)
\(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{104}{195}+25\%\right)\cdot\dfrac{24}{47}-3\dfrac{12}{13}:3\)
\(1\dfrac{13}{15}\cdot\left(0,5\right)^2\cdot3+\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)
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\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)
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\(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{15}.3+\left(\dfrac{32}{60}+\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{37}{20}:\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{37}{20}.\dfrac{24}{47}\\ =\dfrac{7}{5}+\dfrac{222}{235}\\ =\dfrac{329}{235}+\dfrac{222}{235}\\ =\dfrac{551}{235}\)
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Tính giá trí của biểu thức sau theo cách hợp lí nhất.
a) $mathrm{A}dfrac{3}{5} cdot dfrac{6}{7}+dfrac{3}{7}: dfrac{5}{3}-dfrac{2}{7}: 1 dfrac{2}{3}$;
b) $mathrm{B}left(-13 cdot dfrac{2}{5}+dfrac{-2}{9}: 2 dfrac{1}{2}+dfrac{2}{5}. dfrac{11}{9}right) cdot 2 dfrac{1}{2}$;
c) $mathrm{C}left(dfrac{-4}{5}+dfrac{5}{7}right): dfrac{2}{3}+left(dfrac{-1}{5}+dfrac{2}{7}right): dfrac{2}{3}$;
d) $mathrm{D}dfrac{4}{9}:left(dfrac{1}{15}-dfrac{2}{3}right)+dfrac{4}{9}:left(dfrac{1}{11}-dfrac{5}{22}right)$.
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Tính giá trí của biểu thức sau theo cách hợp lí nhất.
a) $\mathrm{A}=\dfrac{3}{5} \cdot \dfrac{6}{7}+\dfrac{3}{7}: \dfrac{5}{3}-\dfrac{2}{7}: 1 \dfrac{2}{3}$;
b) $\mathrm{B}=\left(-13 \cdot \dfrac{2}{5}+\dfrac{-2}{9}: 2 \dfrac{1}{2}+\dfrac{2}{5}. \dfrac{11}{9}\right) \cdot 2 \dfrac{1}{2}$;
c) $\mathrm{C}=\left(\dfrac{-4}{5}+\dfrac{5}{7}\right): \dfrac{2}{3}+\left(\dfrac{-1}{5}+\dfrac{2}{7}\right): \dfrac{2}{3}$;
d) $\mathrm{D}=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)$.
A= 3/5.6/7+3/5.3/7-3/5.2/7
A= 3/5. (6/7+3/7-2/7) = 3/5
B= (-13.2/5+-2/9.2/5+2/5.11/9).5/2
B= (-13-2/9+11/9).2/5.5/2
B= -13+(11/9-2/9)= -12
C= (-4/5+5/7).3/2+(-1/5+2/7).3/2
C= (-4/5+5/7+-1/5+2/7).3/2
C= ((-4/5+-1/5)+(5/7+2/7)).3/2
C= 0
D= 4/9.-5/3+4/9.-22/3
D= 4/9.110/9
D= 440/81
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