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5(x-1)=125-25
5(x-1)=100
x-1=100:5
x-1=20
x=20+1
x=21
12(x-1):3=64+8
12(x-1):3=72
12(x-1)=72.3
12(x-1)=216
x-1=216:12
x-1=18
x=18+1
x=19
(x-1)^3=5^3
=>x-1=5
x=5+1
x=6
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`5/2 -3(1/3-x)=1/4-7x`
`=> 5/2 - 1 + 3x=1/4 -7x`
`=>3x+7x= 1/4 - 5/2 +1`
`=> 10x= 1/4 - 10/4 +4/4`
`=>10x= -5/4`
`=>x=-5/4 :10`
`=>x=-5/4 xx1/10`
`=>x= -5/40=-1/8`
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a: =2/3+1/5*10/7
=2/3+2/7
=14/21+6/21=20/21
b: \(=\dfrac{1}{2}\cdot\dfrac{-3+2}{4}=\dfrac{1}{2}\cdot\dfrac{-1}{4}=\dfrac{-1}{8}\)
c: \(=\dfrac{3}{4}+\dfrac{9}{5}:\dfrac{3}{2}-1\)
=-1/4+9/5*2/3
=-1/4+18/15
=-1/4+6/5
=-5/20+24/20=19/20
d: \(=\dfrac{3}{2}\cdot\left(\dfrac{7}{3}-\dfrac{5}{3}\cdot4\right)\)
\(=\dfrac{7}{2}-\dfrac{5}{2}\cdot4=\dfrac{7}{2}-\dfrac{20}{2}=\dfrac{-13}{2}\)
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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
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a, | x - 3/4 | = 1/2
=>\(\orbr{\begin{cases}x-\frac{3}{4}=\frac{1}{2}\\x-\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
=>\(\orbr{\begin{cases}x=\frac{1}{2}+\frac{3}{4}\\x=-\frac{1}{2}+\frac{3}{4}\end{cases}}\)
=>\(\orbr{\begin{cases}x=\frac{2}{4}+\frac{3}{4}\\x=-\frac{2}{4}+\frac{3}{4}\end{cases}}\)
=>\(\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)
Vậy....
a) \(|x-\frac{3}{4}|=\frac{1}{2}\)
\(< =>\orbr{\begin{cases}x-\frac{3}{4}=\frac{1}{2}\\x-\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{1}{2}+\frac{3}{4}\\x=-\frac{1}{2}+\frac{3}{4}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)
Vay : x = 5/4 hoặc x = 1/4
b)\(saide\)
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\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)