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

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

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

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

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

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/ \(4x^2-12x+9=\left(5-x\right)^2\)

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

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

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

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

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

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

x²-4=8(x-2)​

=> ​x²-4=8x-16​​

​=> x²-8x+16-4=0

​=> (x-4)²-4=0

​=>(x-4-2)(x-4+2)=0

​=> (x-2)(x-6)=0

​=> x-2=0 nên x=2

​x-6 =0 nên x=6

​

​

​

​

​

​

​

a) x^2 - 4 = 8(x - 2)​

​<=> (x - 2)(x + 2) - 8(x - 2) = 0

​<=> (x - 2)(x+2-8)=0

​<=>(x-2)(x-6)=0

​<=>x-2=0 hoặc x-6=0​

​<=>x=2 hoặc x=6

​Vậy S={2;6}

​​b)x^2-4x+4=9(x-2)

​<=>(x-2)^2-9(x-2)=0

​<=>(x-2)(x-2-9)=0

​<=>(x-2)(x-11)=0

​<=>x-2=0 hoặc x-11=0

​<=>x=2 hoặc x=11

​Vậy S={2;11}

​c)4x^2-12x+9=(5-x)^2

​<=>(2x)^2-2.2x.3+3^2=(5-x)^2

​<=>(2x-3)^2-(5-x)^2=0

​<=>(2x-3-5+x)(2x-3+5-x)=0

​<=>(3x-8)(x+2)=0

​<=>3x-8=0 hoặc x+2=0

​<=>3x=8 hoặc x= - 2

​<=>x=8:3(8 phần 3) hoặc x= -2

​Vậy S={8:3 ; -2}

B , x^2 - 4x +4 = 9 ( x-2) 
=> (x-2)^2 -9(x-2)  = 0
=> (x-2-9)(x-2) = 0 
=>( x-11 ) ( x-2)=0
=> x= 11
x=2

C, 4x^2 - 12x+9 =(5-x)^2
=>( 2x-3)^2 = (5-x)^2
=> (2x-3-5+x ) (2x-3+5-x ) = 0 
=> 3x-8) ( x-2 ) =0 
=> x= 8/3
x = -2

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)

\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a.

4x(x-5) - (x-1)(4x-3)-5=0

 4x^2-20x-4x^2+3x+4x+3=0

(4x^2-4x^2)+(-20x+3x+4x)+3=0

 13x+3 = 0

13x=-3

x=-3/13

b,

(3x-4)(x-2)-3x(x-9)+3=0

3x^2-6x-4x+8 - 3x^2+27x+3=0

(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0

17x+11=0

17x=-11

x=-11/17

c, 2(x+3)-x^2-3x=0

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

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

TH1: x+3 = 0; x=-3

TH2: 2-x=0;x=2

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) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x-1+3x^2+2x+1=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x+3x^2+2x-x^3=1-1+8\)
\(\Leftrightarrow5x=8\)
\(\Leftrightarrow x=\dfrac{8}{5}\)
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, \(4x^2-12x+9=\left(5-x\right)^2\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7