Sửa đề: \(E=\left(1+\frac{x}{z}\right)\left(1-\frac{y}{x}\right)\left(1-\frac{z}{y}\right)\)

-x+y-z=0

=>x-y+z=0

=>x+z=y; x-y=-z; -y+z=-x

=>x+z=y; x-y=-z; y-z=x

Ta có: \(E=\left(1+\frac{x}{z}\right)\left(1-\frac{y}{x}\right)\left(1-\frac{z}{y}\right)\)

\(=\frac{x+z}{z}\cdot\frac{x-y}{x}\cdot\frac{y-z}{y}\)

\(=\frac{y}{z}\cdot\frac{-z}{x}\cdot\frac{x}{y}=-1\)

Lời giải:
$45-3(7+x)=6$

$3(7+x)=45-6=39$

$7+x=39:3=13$

$x=13-7=6$

a: -28+x=-34+(-11)

=>x-28=-34-11=-45

=>x=-45+28

=>x=-17

b: -12(x+3)=0

=>x+3=0

=>x=0-3

=>x=-3

c: \(\left(7-x\right)\left(-x+2\right)=0\)

=>(x-7)(x-2)=0

=>\(\left[\begin{array}{l}x-7=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=2\end{array}\right.\)

d: \(16x^2-64=0\)

=>\(16x^2=64\)

=>\(x^2=4\)

=>\(\left[\begin{array}{l}x=2\\ x=-2\end{array}\right.\)

\(\Leftrightarrow-\dfrac{2}{5}\left(4x-3\right)^2=-\dfrac{5}{18}\)

\(\Leftrightarrow\left(4x-3\right)^2=\dfrac{25}{36}\)

\(\Leftrightarrow4x-3\in\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)

hay \(x\in\left\{\dfrac{23}{24};\dfrac{13}{24}\right\}\)

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

\(2x^2+2y^2-5xy+x-2y+3=0\)

\(\Leftrightarrow\left(x-2y\right)\left(2x-y\right)+x-2y+3=0\)

\(\Leftrightarrow\left(x-2y\right)\left(2x-y+1\right)=-3\) 

Vậy \(\left(x;y\right)=\left(1;2\right)\) là bộ nghiệm nguyên dương duy nhất

\(a,\Rightarrow4x\left(x^2-9\right)=0\\ \Rightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\\ \Rightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Rightarrow2\left(x-3\right)4\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

a) \(\Rightarrow4x\left(x^2-9\right)=0\)

\(\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

b) \(\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(4x-4\right)=0\)

\(\Rightarrow8\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)