Bài này áp dụng hằng đẳng thức \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\) nha bạn 

Ta có : 

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

\(\Leftrightarrow\)\(x^3-3.x^2.2+3.x.2^2-2^3=0\)

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

\(\Leftrightarrow\)\(x-2=0\)

\(\Leftrightarrow\)\(x=2\)

Vậy \(x=2\)

Chúc bạn học tốt ~ 

\(x^3-6x^2+12x-8=0\) \(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3=0\) \(\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\) \(\Leftrightarrow x=2\)

Ta có \(x^3-6x^2+12-8=0\)

\(\Rightarrow x^3-6x^2+4=0\)

\(\Rightarrow x^3-6x^2=-4\)

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

đề bài sai sai thì phải :v

x³ - 6x² + 12 - 8

<=> (x + 2)³ = 0

<=> x + 2 = 0

<=> x         = 0 - 2

<=> x         = -2

=> x = -2

x³ - 6x² + 12 - 8

(x + 2)³= 0

x + 2 = 0

x         = 0 - 2

x         = -2

x = -2

b . 49x^2 - 81 = 0 

=> 49x^2 = 81 

rồi giải ra nhé 

a ) 2x ( x - 5 ) - x ( 3 + 2x ) = 26

2x² - 10x - 3x - 2x² = 26

- 13x = 26

x = 26 : ( -13 )

x = -2

b) 49x² - 81 = 0

( 7x - 9 )( 7x + 9 ) = 0

Th1 :

7x - 9 = 0

7x = 9

x = \(\frac{9}{7}\)

Th2

7x + 9 = 0

7x = -9

x = \(-\frac{9}{7}\)

Vay x = \(\frac{9}{7}\) hoac x = \(-\frac{9}{7}\)

bn này k bao gio tra cong ng khac, ngu j ma lam

a/ 2x(x-5)-x(3+2x)=26

 =>2x²-10x-2x²-3x=26

=>-13x=26

=>x=-2

b/ 49x²-81=0

=>49x²=81

=>x²=\(\frac{49}{81}\)

\(\Rightarrow x^2=\left(-\frac{9}{7}\right)^2\)hoặc \(\left(\frac{9}{7}\right)^2\)

\(\Rightarrow x=\pm\frac{9}{7}\)

2x(x - 5) - x(3 + 2x) = 26

2x² - 10x - 3x - 2x² = 26

- 13x = 26

x = 26 : (- 13)

x = - 2

49x² - 81 = 0

(7x - 9)(7x + 9) = 0

Th1:

7x - 9 = 0

7x = 9

x = 9/7

Th2:

7x + 9 = 0

7x = - 9 

x = - 9/7

Vậy x = 9/7 hoặc x = - 9/7

\(-y^2-13y+14=0\)

\(\Leftrightarrow y^2+\frac{2.y.13}{2}+\frac{169}{4}=\frac{225}{4}\)

\(\Leftrightarrow\left(y+\frac{13}{2}\right)^2=\left(\frac{15}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}y+\frac{13}{2}=\frac{15}{2}\\y+\frac{13}{2}=-\frac{15}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=-14\end{cases}}\)

xin lỗi , đề là tìm y

\(x^2+4x+y^2-2xy+x^2+4=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)=0\)

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

vì \(\left(x-y\right)^2\ge0;\left(x+2\right)^2\ge0\)nên

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

\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x-2=0\)

\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+6x+12+3x-2=0\)

\(1+1+6x+3x+12-2=0\)

\(9x+12=0\)

\(9x=-12\)

\(x=\frac{-4}{3}\)

\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x-2=0\)

\(\Leftrightarrow\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x=0+2\)

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

\(\Leftrightarrow9x+14=2\)

\(\Leftrightarrow9x=2-14\)

\(\Leftrightarrow9x=-12\)

\(\Leftrightarrow x=\frac{-12}{9}=\frac{-4}{3}\)

\(\Rightarrow x=\frac{-4}{2}\)

Câu 1: 

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

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

Trường hợp 1: x<1

(1) trở thành 1-x+2-x=3

=>3-2x=3

=>x=0(nhận)

Trường hợp 2: 1<=x<2

(1) trở thành x-1+2-x=3

=>1=3(loại)

Trường hợp 3: x>=2

(1) trở thành x-1+x-2=3

=>2x-3=3

=>2x=6

hay x=3(nhận)

Ta có : 

\(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(x-3\right)\left(x+3\right)\)

\(=\)\(\left(3x+2\right)\left[\left(3x\right)^2-3x.2+2^2\right]-\left(x^2-3^2\right)\)

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

\(=\)\(27x^3-x^2+8-9\)

\(=\)\(27x^3-x^2-1\)

Chúc bạn học tốt ~