\(\left(x^2-4x+3\right)\left(x^2-6x+8\right)=8\) 

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

\(\left[x\left(x-3\right)-1\left(x-3\right)\right]\left[x\left(x-4\right)-2\left(x-4\right)\right]=8\)

\(\left(x-1\right)\left(x-3\right)\left(x-2\right)\left(x-4\right)=8\) 

\(\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)=8\) 

\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)-8=0\)  

Đặt \(t=x^2-5x+4\) 

\(t\left(t+2\right)-8=0\) 

\(t^2+2t-8=0\) 

\(t^2+4t-2t-8=0\) 

\(t\left(t+4\right)-2\left(t+4\right)=0\) 

\(\left(t+4\right)\left(t-2\right)=0\) 

\(\orbr{\begin{cases}t+4=0\\t-2=0\end{cases}}\) 

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

\(\orbr{\begin{cases}x^2-5x+4=-4\\x^2-5x+4=2\end{cases}}\)  

\(\orbr{\begin{cases}x^2-5x+8=0\left(ptvn\right)\\x^2-5x+2=0\end{cases}}\) 

\(x^2-5x+2=0\) 

\(\orbr{\begin{cases}x=\frac{5+\sqrt{17}}{2}\\x=\frac{5-\sqrt{17}}{2}\end{cases}}\)

Đề bài đây là tính à ?

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

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

= \(3x^3-3x^2-x^2+x+x-1+4x^2-3x^3\)

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

a) 5x² + 10y² - 6xy - 4x - 2y + 3 

= ( x² - 6xy + 9y² ) + ( 4x² - 4x + 1 ) + ( y² - 2y + 1 ) + 1

= ( x - 3y )² + ( 2x - 1 )² + ( y - 1 )² + 1 ≥ 1 > 0 ∀ x, y, z

=> đpcm 

b) x² + 4y² + z² - 2x - 6z + 8y + 15 

= ( x² - 2x + 1 ) + ( 4y² + 8y + 4 ) + ( z² - 6z + 9 ) + 1

= ( x - 1 )² + ( 2y + 2 )² + ( z - 3 )² + 1 ≥ 1 > 0 ∀ x, y, z

=> đpcm

2x^2 + 3x - ( x^2 + 3x + 2 ) = 6 

2x^2 + 3x - x^2 - 3x - 2 = 6 

x^2 - 2 = 6 

x^2 = 8 

x = +- căn 8 

Ta có : x(2x + 3) - (x + 2)(x + 1) = 6

=> 2x² + 3x - (x² + 3x + 2) = 6

=> 2x² + 3x - x² - 3x - 2 = 6

=> x² = 8

=> \(x=\pm\sqrt{8}\)

là tự đi mà tìm hiểu nhé

Bài 1.

a) ( 3x + 4y )² = ( 3x )² + 2.3x.4y + ( 4y )² = 9x² + 24xy + 16y²

b) ( x² + 1 )² = ( x² )² + 2.x².1 + 1² = x⁴ + 2x² + 1

c) ( 3 - 2y )² = 3² - 2.3.2y + ( 2y )² = 9 - 12y + 4y²

d) ( xy² - 2 )² = ( xy² )² - 2.xy².2 + 2² = x²y⁴ - 4xy² + 4

Bài 2.

a) x² - 9 = x² - 3² = ( x - 3 )( x + 3 )

b) 25 - 4y² = 5² - ( 2y )² = ( 5 - 2y )( 5 + 2y )

c) 9x⁴ - 4y² = ( 3x² )² - ( 2y )² = ( 3x² - 2y )( 3x² + 2y )

d) ( x + 1 )² - y² = ( x - y + 1 )( x + y + 1 )

B1:

a) \(\left(3x+4y\right)^2=\left(3x\right)^2+2.3x.4y+\left(4y\right)^2=9x^2+24xy+16y^2\)

b) \(\left(x^2+1\right)^2=\left(x^2\right)^2+2.x^2.1+1^2=x^4+2x^2+1\)

c) \(\left(3-2y\right)^2=3^2-2.3.2y+\left(2y\right)^2=9-12y+4y^2\)

d) \(\left(xy^2-2\right)^2=\left(xy^2\right)^2-2.xy^2.2+2^2=xy^4-4xy^2+4\)

B2:

a) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)

b) \(25-4y^2=5^2-\left(2y\right)^2=\left(5-2y\right)\left(5+2y\right)\)

c) \(9x^4-4y^2=\left(3x^2\right)^2-\left(2y\right)^2=\left(3x^2-2y\right)\left(3x^2+2y\right)\)

d) \(\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)

ko bt tự làm đê!!!!