x(x - 1) = 4x - 4

<=> x² - x = 4x - 4

<=> x² - x - 4x + 4 = 0

<=> x² - 5x + 4 = 0

<=> (x - 1)(x - 4) = 0

<=> x - 1 = 0 hoặc x - 4 = 0

<=> x = 0 + 1 hoặc x = 0 + 4

=> x = 1 hoặc x = 4

\(x\left(x-1\right)=4x-4\)\(\Leftrightarrow x\left(x-1\right)=4\left(x-1\right)\)

\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}\)

Vậy \(x=1\)hoặc \(x=4\)

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Học tốt nhé 

cảm ơn nhé

vt sai đề nâk

từ gt=> xy+yz+xz=0

áp dụng bdt bunhia

=> A>=0

dấu= xr khi x=y=z

-> dấu = k xr

..........

hoặc: 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\frac{\Rightarrow1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

\(\Rightarrow\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)

\(\Rightarrow xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)

a, =x-z

b,=-9(2-x)^2/4

c,=x+1

d,=(x-1)/(x+1)

hok tốt

a) =\(\frac{x-z}{2}\)

a) Để P xác định \(\Leftrightarrow\hept{\begin{cases}2a-2\ne0\\2-2a^2\ne0\\a+2\ne0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a^2\ne1\\a\ne-2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a\ne-1vâ\ne1\\a\ne-2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a\ne-1\\a\ne2\end{cases}}\)

b) \(P=\left(\frac{a+1}{2a-2}+\frac{1}{2-2a^2}\right).\frac{2a+2}{a+2}\)

\(=\left[\frac{a+1}{2\left(a-1\right)}+\frac{1}{2\left(1-a\right)\left(1+a\right)}\right].\frac{2\left(a+1\right)}{a+2}\)

\(=\left[\frac{\left(a+1\right)^2}{2\left(a-1\right)\left(a+1\right)}-\frac{1}{2\left(a-1\right)\left(1+a\right)}\right].\frac{2\left(a+1\right)}{a+2}\)

\(=\frac{\left(a+1\right)^2-1}{2\left(a-1\right)\left(a+1\right)}.\frac{2\left(a+1\right)}{a+2}\)

\(=\frac{a\left(a+2\right)}{\left(a-1\right)\left(a+2\right)}\)

\(=\frac{a}{a-1}\)

c) \(\left|a\right|=3\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)

+) Với a=3 thỏa mãn \(\hept{\begin{cases}a\ne1\\a\ne-1\\a\ne2\end{cases}}\)nên thay a=3 vào P ta được:

( làm nốt)

TH kia tương tự

1) (x - 5)² - (x + 3)(x - 3) = 14

=> x² - 10x + 25 - x² + 9 = 14

=> -10x + 34 = 14

=> -10x = 14 - 34

=> -10x = -20

=> x = 2

2) (x + 7)² - 3x - 21 = 0

=> x² + 14x + 49 - 3x - 21 = 0

=> x² + 11x + 28 = 0

=> x² + 4x + 7x + 28 = 0

=> x(x + 4) + 7(x + 4) = 0

=> (x + 7)(x + 4) = 0

=> \(\orbr{\begin{cases}x+7=0\\x+4=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-7\\x=-4\end{cases}}\)

a

\(\left(x-5\right)^2-\left(x+3\right)\left(x-3\right)=14\)

\(\Leftrightarrow x^2-10x+25-x^2+9=14\)

\(\Leftrightarrow-10x=-20\)

\(\Leftrightarrow x=2\)

b

\(\left(x+7\right)^2-3x-21=0\)

\(\Leftrightarrow x^2+14x+49-3x-21=0\)

\(\Leftrightarrow x^2+11x+28=0\)

\(\Leftrightarrow\left(x^2+4x\right)+\left(7x+28\right)=0\)

\(\Leftrightarrow x\left(x+4\right)+7\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=0\)

\(\Leftrightarrow x=-4;x=-7\)