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

TL:

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

\(=\left(2x+1+x-1\right)\left(2x+1-x+1\right)\) 

\(=3x.\left(x+2\right)\) 

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

~Study well~ :)

a) \(x^2-8y^2+6x+9\)

\(=\left(x^2+6x+9\right)-8y^2\)

\(=\left(x+3\right)^2-\left(\sqrt{8}\cdot y\right)^2\)

\(=\left(x+3+\sqrt{8}y\right)\left(x+3-\sqrt{8}y\right)\)

a) 9  -(x-y)²

= 3² - (x-y)²

= (3-x+y).(3+x-y)

b) (x² +4)² - 16x²

= (x²+4)² - (4x)²

= (x² + 4 -4x).(x² + 4 +4x)

      \(9-\left(x-y\right)^2\)

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

\(=\left(3-x+y\right)\left(3+x-y\right)\)

      \(\left(x^2+4\right)^2-16x^2\)

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

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

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

a) 9 - ( x - y ) ^2

= 3 ^ 2 - ( x - y ) ^2

= ( 3 - x +y )( 3 + x -y )

b) (x ^ 2 + 4 ) ^ 2 - 16x ^2

= (x ^ 2 + 4 ) ^ 2 - ( 4x ) ^2

= ( x ^ 2 + 4 - 4x )( x ^ 2 +4 +4x)

k cho mk na ^.^

x²-2.x.1/2 +(1/2)²-9/4

=(x-1/2)²-9/4

=(x-1/2)²-(3/2)²

=(x-1/2-3/2).(x-1/2+3/2)

=(x-2)(x+1)

Ta có (6x+5)²(3x+2)(x+1)-35

= (36x²+60x+25)(3x²+5x+2)-35 (1)

Đặt a=3x²+5x+2

=> 12a+1= 12(3x²+5x+2)+1 =36x²+60x+25

Thay a=3x²+5x+2 vào (1) ta được

(12a+1).a-35=12a²+a-35

= 12a²-20a+21a-35

= 4a(3a-5)+7(3a-5)

= (3a-5)(4a+7) (2)

Thay 3x²+5x+2=a vào (2) ta được

(9x²+15x+6-5)(12x²+20x+8+7)

= (9x²+15x+1)(12x²+20x+15)

Ta có: \(\left(6x+5\right)^2\left(3x+2\right)\left(x+1\right)-35\)

\(=\left(36x^2+60x+25\right)\left(3x^2+5x+2\right)-35\)(1)

Đặt \(3x^2+5x+2=y\)

\(\left(1\right)=\left(12y+1\right)y-35\)

\(=12y^2+y-35\)

\(=\left(3y-5\right)\left(4y+7\right)\)

\(=\left(9x^2+15x+1\right)\left(12x^2+20x+15\right)\)

a: \(x^2-2xy+y^2+3x-3y-4\)

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

\(=\left(x-y+4\right)\left(x-y-1\right)\)

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

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

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

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

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

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

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

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

\(=3xy\left(x^2y-2\right)\)

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

\(=\left(x-y-3\right)\left(x+y+3\right)\)

5: \(\left(3x-y\right)^2-4y^2\)

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

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

\(=\left(3x-3y\right)\left(3x+y\right)\)

\(=3\left(x-y\right)\left(3x+y\right)\)

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

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

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

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-2\right)\)

Đặt: \(x^2-6x+1=a;x^2+1=b\)

Khi đó đa thức này có dạng:

\(2a^2+5ab+2b^2=2a^2+4ab+ab+2b^2\)

\(=2a\left(a+2b\right)+b\left(a+2b\right)=\left(a+2b\right)\left(2a+b\right)\)

Thay lại a và b thì được:

\(\left(a+2b\right)\left(2a+b\right)=\left(x^2-6x+1+2x^2+2\right)\left(2x^2-12x+2+x^2+1\right)\)

\(=\left(3x^2-6x+3\right)\left(3x^2-12x+3\right)\)

\(=9\left(x-1\right)^2\left(x^2-4x+1\right)\)

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