\(\frac{10}{21}\).\(\frac{-14}{3}\)+\(\frac{5}{9}\)
Tính giá trị biểu thức
\(1.A=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{3}\right)-\frac{1}{7}+\left[\frac{-14}{30}\right]\)
\(2.B=\left(\frac{5}{8}-\frac{4}{12}+\frac{3}{2}\right)-\left(\frac{5}{8}+\frac{9}{13}\right)-\left[\frac{-3}{2}\right]+\frac{7}{-15}\)
\(3.C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)
\(4.D=\frac{1}{9}-\left[\frac{-5}{23}\right]-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{14}-\frac{7}{30}\)
\(5.E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)+\left(\frac{12}{17}+\frac{5}{18}+\frac{7}{5}\right)\)
\(6.F=\frac{15}{14}-\left(\frac{17}{23}-\frac{80}{87}+\frac{5}{4}\right)+\left(\frac{12}{17}-\frac{15}{14}+\frac{1}{4}\right)\)
\(7.G=\frac{1}{25}-\frac{4}{27}+\left(\frac{-23}{27}+\frac{-1}{25}-\frac{5}{43}\right)+\frac{5}{43}-\frac{4}{7}\)
\(8.H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{29}{27}\right)-\frac{2}{27}\)
\(9.K=\frac{1}{16}-\frac{5}{21}+\left(\frac{-1}{16}+\frac{-3}{5}-\frac{-5}{21}\right)+\frac{-2}{5}+\frac{3}{4}\)
\(10.L=\frac{7}{12}+\frac{15}{14}-\left(\frac{14}{22}+\frac{-1}{14}+\frac{5}{21}\right)-\frac{-5}{21}+\frac{3}{5}\)
yutyugubhujyikiu
\(\frac{2}{3}x\frac{5}{6}x\frac{9}{10}x\frac{14}{15}x\frac{20}{21}x\frac{27}{28}\)
\(\frac{2}{3}+\frac{6}{5}+\frac{9}{10}+\frac{14}{15}+\frac{20}{21}+\frac{27}{28}+\frac{35}{36}+\frac{44}{45}+\frac{44}{55}\)
\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)
\(\frac{3}{5}:\left(\frac{-1}{5}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{-1}{3}-1\frac{1}{15}\right)\)
\(\left(\frac{1}{2}-\frac{13}{14}\right):\frac{5}{7}-\left(-\frac{2}{21}+\frac{1}{7}\right):\frac{5}{7}\)
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
a)\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)
\(=1.6:\frac{9^8.3-9^8.9}{9^8.\left(5+7\right)}\)
\(=6:\frac{9^8.\left(3-9\right)}{9^8.12}\)
\(=6:\frac{9^8.\left(-6\right)}{9^8.12}\)
\(=6:\left(-\frac{6}{12}\right)\)
\(=6:\left(-\frac{1}{2}\right)\)
\(=-12\)
b) 3/5 : ( -1/5-1/6)+3/5:(-1/3-16/15) ( mình chuyển về ps luôn )
=3/5: (-11/30) + 3/5 : (-7/5)
=3/5:[-11/30+(-7/5)]
=3/5:53/30
=18/53
c) (1/2-13/14):5/7-(-2/21+1/7):5/7
= -3/7:5/7-1/21:5/7
=(-3/7-1/21):5/7
=-10/21:5/7
=-2/3
câu b vá c mình làm tắt nha. chúc bạn học tốt
Tìm y,biết:
a,\(y-\frac{1}{3}=\frac{10}{21}\div\frac{15}{28}\)
b,\(\frac{2}{7}\div y=\frac{10}{21}\times\frac{9}{14}\)
\(\frac{2}{7}:y=\frac{10}{21}.\frac{9}{14}\)
\(\frac{2}{7}:y=\frac{15}{49}\)
\(y=\frac{2}{7}:\frac{15}{49}\)
\(y=\frac{2}{7}.\frac{49}{15}\)
\(y=\frac{14}{15}\)
\(y-\frac{1}{3}=\frac{10}{21}:\frac{15}{28}\)
\(y-\frac{1}{3}=\frac{10}{21}.\frac{28}{15}\)
\(y-\frac{1}{3}=\frac{8}{9}\)
\(y=\frac{8}{9}+\frac{1}{3}\)
\(y=\frac{8}{9}+\frac{3}{9}\)
\(y=\frac{11}{9}\)
a) Ta có : \(y-\frac{1}{3}=\frac{10}{21}\div\frac{15}{28}\)
\(\Rightarrow\) \(y-\frac{1}{3}=\frac{8}{9}\)
\(\Rightarrow\) \(y\) \(=\frac{8}{9}+\frac{1}{3}\)
\(\Rightarrow\) \(y\) \(=\frac{11}{9}\)
Vậy \(y=\frac{11}{9}\)
b) Ta có : \(\frac{2}{7}\div y=\frac{10}{21}\times\frac{9}{14}\)
\(\Rightarrow\) \(\frac{2}{7}\div y=\frac{15}{49}\)
\(\Rightarrow\) \(y=\frac{2}{7}\div\frac{15}{49}\)
\(\Rightarrow\) \(y=\frac{14}{15}\)
Vậy \(y=\frac{14}{15}\)
Cbht !!!
\(A=\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+...+\frac{9901}{9900}\)
\(B=\frac{2}{3}+\frac{5}{6}+\frac{9}{10}+\frac{14}{15}+...+\frac{4949}{4950}\)
\(A=\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+...+\frac{9901}{9900}=\left(1+\frac{1}{2.3}\right)+\left(1+\frac{1}{3.4}\right)+\left(1+\frac{1}{4.5}\right)+...+\left(1+\frac{1}{99.100}\right)\)\(=\left(1+1+1+...+1\right)+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(=98+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)=98+\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=98+\frac{49}{100}=98\frac{49}{100}\)
\(\frac{10}{x-5}=\frac{6}{y-9}=\frac{14}{z-21}\) và\(xyz=6720\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
Sửa đề \(\frac{10}{x-15}=\frac{6}{y-9}=\frac{14}{z-21}\)
=>\(\frac{x-15}{10}=\frac{y-9}{6}=\frac{z-21}{14}\)
=> \(\frac{x}{10}-\frac{15}{10}=\frac{y}{6}-\frac{9}{6}=\frac{z}{14}-\frac{21}{14}\)
=> \(\frac{x}{10}-\frac{3}{2}=\frac{y}{6}-\frac{3}{2}=\frac{z}{14}-\frac{3}{2}\)
=> \(\frac{x}{10}=\frac{y}{6}=\frac{z}{14}=k\Rightarrow\hept{\begin{cases}x=10k\\y=6k\\z=14k\end{cases}}\)
Mà xyz=6720 => 10k.6k.14k = 6720 => 840k3 = 6720 => k3 = 8 => k = 2
=> x = 20; y=12; z = 28
tính bằng cách thuận tiện nhất
a,\(\frac{6}{7}\cdot\frac{16}{15}\cdot\frac{7}{6}\cdot\frac{21}{32}\)b, \(\frac{21}{17}\cdot\frac{13}{14}\cdot56\cdot\frac{3}{42}\) c,\(\frac{7}{4}\cdot\frac{11}{21}+\frac{11}{21}\cdot\frac{5}{4}\) d,\(\frac{23}{14}\cdot\frac{6}{14}-\frac{9}{14}\cdot\frac{6}{13}\)
a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)
Phần b T2
c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)
1 tính
a \(\left(\frac{9}{14}-\frac{9}{21}\right):\left(\frac{4}{5}+\frac{5}{6}\right)\)
b \(\frac{7}{10}-\frac{-3}{4}+\frac{-5}{6}-\frac{1}{5}+\frac{-2}{3}\)
c \(\left(-2\right)^2.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
d \(-\frac{1}{7}.\left(9\frac{1}{2}-0,75\right):\frac{2}{7}+0,625:1\frac{2}{3}\)
a) 45/343
b) -1/4
c) 24/13
d) -4
bạn cho mình đi
cậu làm cả các bước cho tớ thầy tớ ko có làm kết quả ko đâu