a, x^2-4=8(x-2)

=> x^2 - 4 = 8.x - 16

=> x^2 = (8.x - 16) - 4 

=> x^2 = 8.x - (16+4)

=> x^2 = 8.x - 20

A, \(x^2-4=8\left(x-2\right)\)=> \(\left(x-2\right).\left(x+2\right)=8\left(x-2\right)=>\left(x-2\right).\left(x+2\right)-8\left(x-2\right)=0\)

=>\(\left(x-2\right).\left(x-6\right)=0\)

=> x = 2 hoặc x =6 

B. \(x^2-4x+4=9\left(x-2\right)\)=> \(\left(x-2\right)^2=9\left(x-2\right)=>\left(x-2\right)^2-9\left(x-2\right)=0\)

=>\(\left(x-2\right).\left(x-11\right)=0\)=> x =2 hoặc x =11

C. \(4x^2-12x+9=\left(5-x\right)^2=>\left(2x-3\right)^2=\left(5-x\right)^2\)

=>\(\left(2x-3\right)^2-\left(5-x\right)^2=>\left(3x-8\right).\left(x+2\right)=0\)

=> x = 3/8 hoặc x = - 2

\(x^2-4=8\left(x-2\right)\)

\(\Leftrightarrow x^2-8x+16=4\)

\(\Leftrightarrow\left(x-2\right)^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)

Vậy...

\(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow x^2-13x+22=0\)

\(\Leftrightarrow\left(x+\frac{13}{2}\right)^2=\frac{81}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{-21}{2}\end{cases}}\)

Vậy...

b/ \(x^2-4=8\left(x-2\right)\)

<=> \(x^2-4=8x-16\)

<=> \(x^2-4-8x+16=0\)

<=> \(x^2-8x+12=0\)

<=> \(x^2-8x+16-4=0\)

<=> \(\left(x-4\right)^2-4=0\)

<=> \(\left(x-4\right)^2=4\)

<=> \(\orbr{\begin{cases}x-4=2\\x-4=-2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=6\\x=2\end{cases}}\)

c/ \(x^2-4x+4=9\left(x-2\right)\)

<=> \(\left(x-2\right)^2-9\left(x-2\right)=0\)

<=> \(\left(x-2\right)\left(x-2-9\right)=0\)

<=> \(\left(x-2\right)\left(x-11\right)=0\)

<=> \(\orbr{\begin{cases}x=2\\x=11\end{cases}}\)

d/ \(4x^2-12x+9=\left(5-x\right)^2\)

<=> \(\left(2x-3\right)^2-\left(5-x\right)^2=0\)

<=> \(\left(2x-3-5+x\right)\left(2x-3+5-x\right)=0\)

<=> \(\left(3x-8\right)\left(x+2\right)=0\)

<=> \(\orbr{\begin{cases}3x-8=0\\x+2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}3x=8\\x=-2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{8}{3}\\x=-2\end{cases}}\)

a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x-1+3x^2+2x+1=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x+3x^2+2x-x^3=1-1+8\)
\(\Leftrightarrow5x=8\)
\(\Leftrightarrow x=\dfrac{8}{5}\)
Vậy \(S=\left\{\dfrac{8}{5}\right\}\)

b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)-8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-8\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-6=0\)
:) \(x-2=0\Leftrightarrow x=2\)
:) \(x-6=0\Leftrightarrow x=6\)
Vậy \(S=\left\{2;6\right\}\)

c) \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)-9\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-11=0\)
:) \(x-2=0\Leftrightarrow x=2\)
:) \(x-11=0\Leftrightarrow x=11\)
Vậy \(S=\left\{2;11\right\}\)
(d ko bít lèm)
#IDOL

d, \(4x^2-12x+9=\left(5-x\right)^2\)

\(\Rightarrow4x^2-12x+9=25-10x+x^2\)

\(\Rightarrow4x^2-x^2-12x+10x+9-25=0\)

\(\Rightarrow3x^2-2x-16=0\)

\(\Rightarrow\left(-2-x\right)\left(8-3x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-2-x=0\\8-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-x=2\\-3x=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{8}{3}\end{matrix}\right.\)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

a,\(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1+3\left(x^2+2x+1\right)=x^3+8\)

\(\Leftrightarrow-3x^2+3x-1+3x^2+6x+3=8\)

\(\Leftrightarrow9x=6\)

\(\Leftrightarrow x=\frac{2}{3}\)

b,\(x^2-4=8\left(x-2\right)\)

\(\Leftrightarrow x^2-4=8x-16\)

\(\Leftrightarrow x^2+12x-8x=0\)

\(\Leftrightarrow x^2-2x-6x+12=0\)

\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-6\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

c,\(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow x^2-4x+4=9x-18\)

\(\Leftrightarrow x^2-4x+4-9x+18=0\)

\(\Leftrightarrow x^2-13x+22=0\)

\(\Leftrightarrow x^2-2x-11x+22=0\)

\(\Leftrightarrow x\left(x-2\right)-11\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)

d,\(4x^2-12x+9=\left(5-x\right)^2\)

\(\Leftrightarrow4x^2-12x+9=25-10x+x^2\)

\(\Leftrightarrow4x^2-12x+9-25+10-x^2=0\)

\(\Leftrightarrow3x^2-2x-16=0\)

\(\Leftrightarrow3x^2+6x-8x-16=0\)

\(\Leftrightarrow3x\left(x+2\right)-8\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\)

c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)

\(=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)

d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)

Tương tự với a ; b 

\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)

\(3x^2-6x+3-3x^2+15x-2=0\)

\(9x+1=0\)

\(x=-\frac{1}{9}\)

\(b.4x^2-12x+9=0\)

\(4x^2-6x-6x+9=0\)

\(2x\left(x-3\right)-3\left(x-3\right)=0\)

\(\left(2x-3\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

\(c.\left(2x-3\right)^2-\left(x+5\right)^2=0\)

\(\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)

\(\left(x-8\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

a) 3(x - 1)² - 3x(x - 5) = 2

=> 3(x² - 2x + 1) - 3x² + 15x = 2

=> 3x² - 6x + 3 - 3x² + 15x = 2

=> 9x = 2 - 3

=> 9x = -1

=> x = -1/9

b) 4x² - 12x = -9

=> 4x² - 12x + 9 = 0

=> (2x - 3)² = 0

=> 2x - 3 = 0

=> 2x = 3

=>  x = 3/2

c) (2x - 3)² = (x  + 5)²

=> (2x - 3)² - (x + 5)² = 0

=> (2x - 3 - x - 5)(2x - 3 + x + 5) = 0

=> (x - 8)(3x + 2) = 0

=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

d) \(\left(x^4-2x^3+4x^2-8x\right):\left(x^2+4\right)-2x=-4\)

=> \(\left[x^3\left(x-2\right)+4x\left(x-2\right)\right]:\left(x^2+4\right)-2x=-4\)

=> \(x\left(x-2\right)\left(x^2+4\right):\left(x^2+4\right)-2x=-4\)

=> \(x^2-2x-2x+4=0\)

=> \(\left(x-2\right)^2=0\)

=> x - 2 = 0

=> x = 2

e) khđ