2(5) 11;3 (3) 10;8(5)41;20(?) 101
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\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)
\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)
= \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{6}{5}\)
= \(\dfrac{1}{4}-\dfrac{6}{5}\)
= \(-\dfrac{19}{20}\)
\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)
\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)
\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)
\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)
\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)
\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)
\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)
\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)
\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)
\(=-\dfrac{5}{11}\)
\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)
b: \(27D=3^{14}+3^{17}+...+3^{2024}\)
\(\Leftrightarrow26D=3^{2024}-3^{11}\)
hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)
c: \(25E=-5^4-5^6-...-5^{1002}\)
\(\Leftrightarrow24E=-5^{1002}+5^2\)
hay \(E=\dfrac{-5^{1002}+5^2}{24}\)
\(\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\times-\dfrac{7}{11}=\dfrac{-5}{11}\)
8/3 . 2/5 . 3/8 . 10 .19/92 trên 5/7 . 5/11 +5/7 .2/11 - 5/7.11/11
=16/15 .15/4 .19/92 trên 5/7(5/11 +2/11 -11/11)
= .4 .19/92 trên 5/7.(-4/11)
=19/2 trên -20/77
\(\frac{\frac{8}{3}.\frac{2}{5}.\frac{3}{8}.10.\frac{19}{92}}{\frac{5}{7}.\frac{5}{11}+\frac{5}{7}.\frac{2}{11}-\frac{5}{7}.\frac{11}{11}}\)
=\(\frac{\frac{2}{5}.10.\frac{19}{92}}{\frac{5}{7}.\left(\frac{5}{11}+\frac{2}{11}-1\right)}\)
=\(\frac{\frac{19}{23}}{\frac{-20}{77}}\)
=\(\frac{19}{23}.\frac{-77}{20}\)
=\(\frac{-1463}{460}\)
5/7×5/11+5/7×2/11-5/7×14/11
= 5/7 x ( 5/11 + 2/11 - 14/11 )
= 5/7 x -7/11
= -35/77
a) Ta có: \(A=\dfrac{2}{3}-\left(-\dfrac{5}{7}+\dfrac{2}{3}\right)\)
\(=\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{2}{3}\)
\(=\dfrac{5}{7}\)
a ,b
cách 1 : cộng 2 số trong ngặc rồi tính bình thường
cách 2 : nhân phân phối ra rồi tính nhân chia trước cộng trừ sau
c,d
cách 1 : tính bình thường, nhân chia trc cộng trừ sau
cách 2 : đặt nhân tử ( thừa số ) chung
Cách 1 : \((\frac{6}{11}+\frac{5}{11})\cdot\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}\)
Cách 2 : \(\frac{6}{11}\cdot\frac{3}{7}+\frac{5}{11}\cdot\frac{3}{7}=\frac{18}{77}+\frac{15}{77}=\frac{33}{77}=\frac{3}{7}\)
Cách 1 dễ rồi nha
Cách 2 : \(\frac{6}{7}:\frac{2}{5}-\frac{4}{7}:\frac{2}{5}=\frac{15}{7}-\frac{10}{7}=\frac{5}{7}\)
Cách 1 tự làm
Cách 2 : \(\frac{3}{5}\cdot(\frac{7}{9}-\frac{2}{9})=\frac{3}{5}\cdot\frac{5}{9}=\frac{1}{3}\)
Cách 1 : ...
Cách 2 : \((\frac{8}{15}+\frac{7}{15}):\frac{2}{11}=1:\frac{2}{11}=\frac{2}{11\cdot1}=\frac{2}{11}\)
5 nha
mk chắc 100% luôn
k mk nha mk nhanh nhất