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a. 3/5 . 15/7 - 15/7 . 8/5
= 15/7(3/5-8/5)
=15/7. -\(\frac{1}{1}\)
=22/7
b. 4/5 . 1 3/7 + 4/5 . 4/7
=4/5(13/7+4/7)
=4/5.17/7
= 68/35
\(a,\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\cdot(-2)^2\)
\(=\frac{6}{7}+\frac{5}{8}:\frac{5}{1}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{5}{8}\cdot\frac{1}{5}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3\cdot4}{16}\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3\cdot1}{4}\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}=\frac{48+7-42}{56}=\frac{13}{56}\)
\(b,\frac{2}{3}+\frac{1}{3}\cdot\left[\frac{-2}{3}+\frac{5}{6}\right]:\frac{2}{3}\)
\(=\frac{2}{3}+\frac{1}{3}\cdot\left[\frac{-4+5}{6}\right]:\frac{2}{3}\)
\(=\frac{2}{3}+\frac{1}{3}\cdot\frac{1}{6}:\frac{2}{3}=\frac{2}{3}+\frac{1}{3}\cdot\frac{1}{6}\cdot\frac{3}{2}=\frac{2}{3}+\frac{1}{12}=\frac{8}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}\)
c, Xem lại đề
d, \(\frac{-3}{5}+\left[\frac{-2}{5}-99\right]\)
\(=\frac{-3}{5}+\frac{-497}{5}=\frac{-500}{5}=-100\)
b, Tìm x
\(\left[\frac{2}{11}+\frac{1}{3}\right]\cdot x=\left[\frac{1}{7}-\frac{1}{8}\right]\cdot56\)
\(\Rightarrow\left[\frac{2}{11}+\frac{1}{3}\right]\cdot x=\left[\frac{8}{56}-\frac{7}{56}\right]\cdot56\)
\(\Rightarrow\left[\frac{6}{33}+\frac{11}{33}\right]\cdot x=1\)
\(\Rightarrow\frac{17}{33}\cdot x=1\)
\(\Rightarrow x=1:\frac{17}{33}=1\cdot\frac{33}{17}=\frac{33}{17}\)
c) -12 . ( x - 5 ) + 7 . ( 3 -x ) = 5
<=> -12.x + 60 + 21 -7.x = 5
<=> -19 .x + 81 = 5
<=> -19.x = 5 - 81
<=> -19.x = -76
<=> x = -76 : -19
<=> x = 4
Vậy x = 4
d) 30(x+2)-6(x-5)-24x=100
<=> 30.x + 60 - 6.x + 30 -24.x = 100
<=> 0 + 90 = 100
<=> 90 = 100
<=> x \(\in\varnothing\)
Vậy x \(\in\varnothing\)
c: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
d: \(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{2}{3}-\dfrac{3}{5}=\dfrac{1}{15}\)
hay \(x=\dfrac{4}{9}:\dfrac{1}{15}=\dfrac{4}{9}\cdot15=\dfrac{20}{3}\)
f: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
VD: 1/2 là 1 phần 2 đó nha.
a: x/3-1/6=1/5
=>x/3=11/30
hay x=11/90
b: =>1/2x=2
hay x=4
c: =>2/3:x=-7-1/3=-22/3
=>x=-1/11