So sánh :
A = \(\frac{9}{11^4}+\frac{5}{11^5}\) và B = \(\frac{5}{11^41}+\frac{9}{11^5}\)
H24
Những câu hỏi liên quan
Tính nhanh:a)frac{{13}}{{23}}.frac{7}{{11}} + frac{{10}}{{23}}.frac{7}{{11}}; b) frac{5}{9}.frac{{23}}{{11}} - frac{1}{{11}}.frac{5}{9} + frac{5}{9}c)left[ {left( { - frac{4}{9}} right) + frac{3}{5}} right]:frac{{13}}{{17}} + left( {frac{2}{5} - frac{5}{9}} right):frac{{13}}{{17}}; d) frac{3}{{16}}:left( {frac{3}{{22}} - frac{3}{{11}}} right) + frac{3}{{16}}:left( {frac{1}{{10}} - frac{2}{5}} right)
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Tính nhanh:
a)\(\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}};\)
b) \(\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\)
c)\(\left[ {\left( { - \frac{4}{9}} \right) + \frac{3}{5}} \right]:\frac{{13}}{{17}} + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}};\)
d) \(\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\)
a)
\(\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}.\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.\frac{23}{23}\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)
b)
\(\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ =\frac{{17}}{{13}}. (\frac{-9}{9}+\frac{5}{5})\\= \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}\)
d)
\(\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ = - 2\end{array}\)
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So sánh A và B nếu :
\(A=-\frac{1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)va \(B=\frac{1}{2022}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
So sánh A và B
\(A=-\frac{1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)
\(B=-\frac{1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\text{ }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
Cho \(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
So sánh S với 10
Ta có :
\(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
\(S=\frac{2+1}{2}+\frac{3+1}{3}+\frac{4+1}{4}+...+\frac{11+1}{11}\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{11}\right)\)
\(S=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)\)
\(S=10+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)>10\)
\(\Rightarrow\)\(S>10\)
Vậy \(S>10\)
Chúc bạn học tốt ~
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So sánh \(A\) và \(B\),biết:
\(A=\frac{-1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)
\(B=\frac{-1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
So sánh A và B biết:
A = \(-\frac{1}{2013}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)
B = \(-\frac{1}{2013}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
\(A=\left(-\frac{1}{2013}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\right)-\frac{4}{11^4};B=\left(-\frac{1}{2013}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{3}{11^2}\right)-\frac{4}{11^2}\)
Vì 114 > 112 nên \(\frac{4}{11^4}-\frac{4}{11^2}\) => A > B
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Thực hiện phép tính
a, \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)
b, \(\frac{5}{9}.\frac{8}{11}+\frac{5}{9}.\frac{9}{11}-\frac{5}{9}.\frac{6}{11}\)
a. \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)
\(=1.\frac{7}{12}:\frac{7}{12}\)
\(=1\)
b.
\(\frac{5}{9}.\frac{8}{11}+\frac{5}{9}.\frac{9}{11}-\frac{5}{9}.\frac{6}{11}\)
\(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
Tk mk nha!
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b) \(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
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a,\(=\frac{2}{3}+\left(\frac{-4}{27}+\frac{5}{18}\right):\frac{7}{12}\)=\(\frac{2}{3}+\frac{-16}{63}+\frac{10}{21}\)=\(\frac{8}{9}\)
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Xem thêm câu trả lời
Bài 1 : Thưc Hiên Phep tínha, Afrac{5}{3}+frac{5}{7}+frac{-20}{41}+frac{8}{13}+frac{-21}{41}b, Bfrac{5}{7}.frac{2}{11}+frac{5}{7}.frac{12}{11}-frac{5}{7}.frac{7}{11}c, Gfrac{left(-2right)}{3}+frac{left(-5right)}{7}+frac{2}{3}+frac{left(-2right)}{7}e, Hfrac{left(-5right)}{7}.frac{2}{11}+frac{left(-5right)}{7}.frac{9}{11}g, Nfrac{-5}{13}+frac{5}{7}+frac{20}{41}+frac{-8}{13}+frac{21}{41}j, Efrac{5}{7}.frac{12}{11}+frac{5}{7}.frac{12}{11}-frac{5}{7}.frac{17}{11}
Đọc tiếp
Bài 1 : Thưc Hiên Phep tính
a, \(A=\frac{5}{3}+\frac{5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
b, \(B=\frac{5}{7}.\frac{2}{11}+\frac{5}{7}.\frac{12}{11}-\frac{5}{7}.\frac{7}{11}\)
c, \(G=\frac{\left(-2\right)}{3}+\frac{\left(-5\right)}{7}+\frac{2}{3}+\frac{\left(-2\right)}{7}\)
e, \(H=\frac{\left(-5\right)}{7}.\frac{2}{11}+\frac{\left(-5\right)}{7}.\frac{9}{11}\)
g, \(N=\frac{-5}{13}+\frac{5}{7}+\frac{20}{41}+\frac{-8}{13}+\frac{21}{41}\)
j, \(E=\frac{5}{7}.\frac{12}{11}+\frac{5}{7}.\frac{12}{11}-\frac{5}{7}.\frac{17}{11}\)
thực hiện phép tính
a.frac{5}{7}.1^9_{13}-frac{5}{7}.frac{9}{13}
b.frac{11}{24}-frac{5}{41}+frac{13}{24}+0,5-frac{36}{41}
c.frac{-4}{13}.frac{5}{17}+frac{-12}{13}.frac{4}{17}+frac{4}{13}
d.left(frac{4}{3}-frac{3}{2}right)^2-2.left|frac{-1}{9}right|+left(frac{-5}{18}right)
e.left(frac{-3}{4}+frac{2}{3}right):frac{5}{11}+left(frac{-1}{4}+frac{1}{3}right):frac{5}{11}
Đọc tiếp
thực hiện phép tính
a.\(\frac{5}{7}.1^9_{13}-\frac{5}{7}.\frac{9}{13}\)
b.\(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
c.\(\frac{-4}{13}.\frac{5}{17}+\frac{-12}{13}.\frac{4}{17}+\frac{4}{13}\)
d.\(\left(\frac{4}{3}-\frac{3}{2}\right)^2-2.\left|\frac{-1}{9}\right|+\left(\frac{-5}{18}\right)\)
e.\(\left(\frac{-3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(\frac{-1}{4}+\frac{1}{3}\right):\frac{5}{11}\)
A = 5/7.(1+9/13) − 5/7.9/13
A= 5/7.(1+9/13 - 9/13)
A = 5/7.1
A = 5/7
B = 11/24 − 5/41 + 13/24 + 0.5 − 36/41
B = (11/24 + 13/24) - (5/41 + 36/41) + 0.5
B = 1 - 1 + 0.5
B = 0.5
C = −4/13.5/17 + (−12/13).4/17 + 4/13
C = 4/13.(-5/17) + (−12/13).4/17 + 4/13
C = 4/13.(-5/17 + 1) + (−12/13).4/17
C = 4/13.(−12/17) + (−12/13).4/17
C = (4.-12)/(13.17) + (−12/13).4/17
C = 4/17.(−12/13) + (−12/13).4/17
C = 4/17.(−12/13).2
C = 96/221
D = (4/3 − 3/2)2 − 2.∣−1/9∣ + (−5/18)
D = (4/3 − 3/2)2 − 2.1/9+ (−5/18)
D = -1/62 - 2/9+ (−5/18)
D = -1/12 - ( 2/9+ (−5/18) )
D = -1/12 - ( 4/18+ (−5/18) )
D = -1/12 - (-1/18)
D = -1/12 + 1/18
D = -3/36 + 2/36
D = -1/36
E = (−3/4 + 2/3):5/11 + (−1/4 + 1/3):5/11
E = (−3/4 + 2/3 + (−1/4) + 1/3):5/11
E = ((−3/4 + (−1/4)) + (2/3 + + 1/3)):5/11
E = ( - 1 + 1):5/11
E = 0:5/11
E = 0
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