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Lời giải:
Ta có:
$(2x-y)+(2y-z)+(2z-x)=1+2+3$
$2x-y+2y-z+2z-x=6$
$x+y+z=6$
![](https://rs.olm.vn/images/avt/0.png?1311)
TL
=> 3x + 6 - 6x + 30 + 4x = 10
x = 10 - 6 - 30 = -26
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\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{2}+\frac{3}{4}+\frac{3}{4}\right)-2=?\)
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\(15-2.x=25\)
\(2.x=15-25=-10\)
\(x=\left(-10\right)\div2=-5\)
\(31-\left(17+x\right)=18\)
\(\left(17+x\right)=31-18=13\)
\(x\) \(=13-17=-4\)
15−2.x=2515−2.x=25
2.x=15−25=−102.x=15−25=−10
x=(−10)÷2=−5x=(−10)÷2=−5
31−(17+x)=1831−(17+x)=18
(17+x)=31−18=13(17+x)=31−18=13
xx =13−17=−4
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3.(x+2)-6.(x-5)=2.(5-2x)
3x+6-6x+30=10-4x
3x-6x+4x=10-30-6
x=-26
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\(\left[\left(x-2\right)\left(y-3\right)\right]^2=2^2\)
\(\orbr{\begin{cases}\left(x-2\right)\left(y-3\right)=2\left(1\right)\\\left(x-2\right)\left(y-3\right)=-2\left(2\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow\hept{\begin{cases}x-2=\left(-2,-1,1,2\right)\\y-3=\left(-1,-2,2,1\right)\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=5\end{cases}}}\)
\(\left(2\right)\Leftrightarrow\hept{\begin{cases}x=3\\y=1\end{cases}\left(loai\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b/, 3x + 2 + 3x = 90
=> 3x.32 + 3x = 90
=> 3x.9 + 3x = 90
=> 3x.(9+1) = 90
=> 3x.10 = 90
=> 3x = 90:10
=> 3x = 9 = 32
=> x = 2
\(\frac{1}{2}.\left(x-\frac{2}{3}\right)^2=\frac{2}{18}\)
\(\left(x-\frac{2}{3}\right)^2=\frac{2}{18}:\frac{1}{2}\)
\(\left(x-\frac{2}{3}\right)^2=\frac{2}{9}\)
@_@ sr bn nhak đến đoạn này mik tịt r