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\(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)=\frac{1}{2}\cdot\left(\frac{6}{27}-\frac{5}{27}+\frac{3}{7}\right)=\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{1}{27}=\frac{3}{14}+\frac{1}{54}=\frac{44}{189}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)=\frac{1}{2}+\frac{3}{4}-\frac{3}{4}+\frac{4}{5}=\frac{1}{2}+\frac{4}{5}=\frac{13}{10}\)
\(\frac{1}{3}\cdot\frac{5}{7}-\frac{7}{27}\cdot\frac{36}{14}=\frac{1}{3}\cdot\frac{5}{7}-\frac{7}{27}\cdot\frac{18}{7}=\frac{1}{3}\cdot\frac{5}{7}-\frac{6}{7}=\frac{5}{21}-\frac{14}{21}=\frac{-3}{7}\)
Câu 1:
\(A=\frac{2^5.7+2^5}{2^5.3-2^5}\)= \(\frac{2^5.8}{2^5.2}\)= 4
Vậy A = 4
Câu 2:
\(B=2^3.5^3-3.\left\{400-\left[673-2^3.\left(7^8:7^6+7^0\right)\right]\right\}\)
\(B=8.125-3.\left\{400-\left[673-8.\left(7^2+1\right)\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-8.\left(49+1\right)\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-8.50\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-400\right]\right\}\)
\(B=1000-3.\left\{400-273\right\}\)
\(B=1000-3.127\)
\(B=1000-381\)
\(B=619\)
Vậy B = 619
a,\(\frac{7}{10}\cdot\frac{4}{9}+\frac{3}{10}\cdot\frac{4}{9}-1\frac{7}{9}\)
\(=\frac{14}{45}+\frac{2}{15}-\frac{16}{9}\)
\(=\frac{14}{45}+\frac{6}{45}-\frac{80}{45}\)
\(=\frac{-60}{45}=\frac{-4}{3}\)
b,\(\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{5}{4}-\frac{2}{3}\right)\cdot\left(-3\right)^2+\frac{5}{9}\cdot30\%\)
\(=\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{7}{12}\right)\cdot9+\frac{5}{9}\cdot\frac{3}{10}\)
\(=\frac{-5}{6}+\frac{7}{3}+\frac{1}{6}\)
\(=\frac{-5}{6}+\frac{14}{6}+\frac{1}{6}\)
=\(=\frac{10}{6}=\frac{5}{3}\)
a, 13/6+5/8 : -3/4 - 7/12.4
= 13/6 + -5/6-7/3
=8/6-7/3
= -6/6
= -1
b, ( 73/5 - 21/3) + ( 4/3-43/5 )
= 73/5-21/3+4/3-43/5
=( 73/5-43/5)-(21/3-4/3)
= 6-17/3
=1/3
c, 7/5.4/9 +7/5: 9/16- 14/10.2/9
= 7/5.4/9 +7/5.16/9 - 14/45
=7/5.(4/9+16/9)-14/45
=7/5.20/9-14/45
= 140/45 - 14/45
= 126/45
Xong rùi nè! Nhưng bạn kiểm tra lại giùm nhé vì làm vào ban đêm nên hơi bất tiện
\(\dfrac{-2}{3}\cdot\dfrac{5}{7}+\dfrac{5}{7}:\dfrac{3}{4}-1=\dfrac{-2}{3}\cdot\dfrac{5}{7}+\dfrac{5}{7}\cdot\dfrac{4}{3}-1=\left(\dfrac{-2}{3}+\dfrac{4}{3}\right)\cdot\dfrac{5}{7}-1=\dfrac{2}{3}\cdot\dfrac{5}{7}-1=\dfrac{10}{21}-1=-\dfrac{11}{21}\)
−2 / 3⋅5 / 7+5 / 7:3 / 4−1=−2 / 3⋅5 / 7+5 / 7⋅4 / 3−1=(−2 / 3+4 / 3)⋅5 /. 7−1=2 / 3⋅5 / 7−1=10 / 21−1=−11 / 21
nha