a) ( x + 2 )( x + 3 ) - ( x - 2 )( x + 5 ) = 16

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 16

<=> x² + 5x + 6 - x² - 3x + 10 = 16

<=> 2x + 16 = 16

<=> 2x = 0

<=> x = 0

b) 3x( 2x - 4 ) - 2x( 3x + 5 ) = 44

<=> 6x² - 12x - 6x² - 10x = 44

<=> -22x = 44

<=> x = -2

c) 2( 5x - 8 - 3 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 5x - 11 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 20x² - 69x + 55 ) = 12x - 16

<=> 40x² - 138x + 110 = 12x - 16

<=> 40x² - 138x + 110 - 12x + 16 = 0

<=> 40x² - 150 + 126 = 0 ( chưa học nghiệm vô tỉ nên để vô nghiệm nha :) )

=> Vô nghiệm

a. (x+2) (x+3)-(x-2) (x+5)= 16

x²+5x+6-x²-3x+10=16

2x+16=16

2x=0

x=0

b,3x (2x-4)-2x (3x+5)= 44

6x²-12x-6x²-10x=44

-22x=44

x=-2

Ý c bạn tự lm,tương tự nhưa,b

a, \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=16\)

\(\Leftrightarrow x^2+3x+2x+6-\left(x^2+5x-2x-10\right)=16\)

\(\Leftrightarrow8x-4=16\Leftrightarrow x=\frac{20}{8}=\frac{5}{2}\)

b, \(3x\left(2x-4\right)-2x\left(3x+5\right)=44\)

\(\Leftrightarrow6x^2-12x-\left(6x^2-10x\right)=44\Leftrightarrow-2x=44\Leftrightarrow x=-22\)

c, \(2\left(5x-8-3\right)\left(4x-5\right)=4\left(3x-4\right)\)

\(\Leftrightarrow\left(10x-22\right)\left(4x-5\right)=4\left(3x-4\right)\Leftrightarrow40x^2-50x-88x+110=0\)

\(\Leftrightarrow40x^2-138x+110=0\)( vô nghiệm ) 

b) \(\left(x+3\right)^2-5x-15=0\\ \Leftrightarrow\left(x+3\right)^2-5\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+3-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)

c) \(2x^5-4x^3+2x=0\\ \Leftrightarrow2x\left(x^4-2x^2+1\right)=0\\ \Leftrightarrow2x\left(x^2-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\\left(x^2-1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

Vậy tập nghiệm của pt là : \(S=\left\{0;1;-1\right\}\)

a) x=3/2

b)x=-1

c) x=5

d) x= 5/2

/5x-4/=/x+2/

\(\orbr{\begin{cases}5x-4=x+2\\5x-4=-x+2\end{cases}}suyra\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)

vậy x=3/2 hoặc x=1/2

Ta có : \(\left|5x-4\right|=\left|x+2\right|\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)

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

\(\Rightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}2x-3x=2+3\\2x+3x=-2+3\end{cases}\Rightarrow}\orbr{\begin{cases}-x=5\\5x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}}\)

c)/2+3x/=/4x-3/

\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-\left(4x-3\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}\)

d)/7x+1/-/5x+6|=0

\(\Rightarrow\left|7x+1\right|=\left|5x+6\right|\)

\(\Rightarrow\orbr{\begin{cases}7x+1=5x+6\\7x+1=-\left(5x+6\right)\end{cases}\Rightarrow\orbr{\begin{cases}7x-5x=6-1\\7x+1=-5x-6\end{cases}\Rightarrow}\orbr{\begin{cases}2x=5\\7x+5x=-6-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}}\)

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

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

-2x^4 + 3x^3-2x^2=0

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

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

Bài 1 :

1) 4x² - y² = ( 2x + y ) ( 2x - y )
2) 9x² - 4y² = ( 3x - 2y ) ( 3x + 2y )

3) 4x² + y² + 4xy = ( 2x + y )²

Bài 2:

1) 2x² + 8x = 0

=> 2x ( x + 4 ) = 0

=> \(\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\) 

=> \(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

2) 3 ( x - 4 ) + x² - 4x = 0

=> 3 ( x - 4 ) + x ( x - 4 ) = 0

=> ( x - 4 ) ( 3 + x ) = 0

=> \(\orbr{\begin{cases}x-4=0\\3+x=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)

3) 3 ( x - 2 ) = x² - 2x 

=> 3 ( x - 2 ) - x² + 2x = 0

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

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

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

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

4) x ( x - 2 ) - 6 ( 2 - x ) = 0

=> x ( x - 2 ) + 6 ( x - 2 ) = 0

=> ( x - 2 ) ( x + 6 ) = 0

=> \(\orbr{\begin{cases}x-2=0\\x+6=0\end{cases}}\)

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

5) 2x ( x + 5 ) = x² + 5x

=> 2x ( x + 5 ) - x² - 5x = 0

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

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

=> \(\orbr{\begin{cases}x+5=0\\2x-x=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)

6 ) ( x - 2 )² - x ( x + 3 ) = 9

=> x² - 4x + 4 - x² - 3x = 9

=> - 7x + 4 = 9

=> - 7x = 5

=> x = \(-\frac{5}{7}\)

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

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

\(3,4x^2+y^2+4xy=\left(2x\right)^2+2.2x.y+y^2=\left(2x+y\right)^2\)

\(1,2x^2+8x=0\Rightarrow2x\left(x+4\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

\(2,3\left(x-4\right)+x^2-4x=0\)

\(\Rightarrow3\left(x-4\right)+x\left(x-4\right)=0\)

\(\Rightarrow\left(3+x\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3+x=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)

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

\(\Rightarrow3\left(x-2\right)-x^2+2x=0\)

\(\Rightarrow3\left(x-2\right)-x\left(x-2\right)=0\)

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

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

\(4,x\left(x-2\right)-6\left(2-x\right)=0\)

\(\Rightarrow x\left(x-2\right)+6\left(x-2\right)=0\)

\(\Rightarrow\left(x+6\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)

Pt đa thức thành nhân tử :

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

2) \(9x^2-4y^{22}=\left(3x-2y^{11}\right)\left(3x+2y^{11}\right)\)

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

Tìm x :

1) \(2x^2+8x=0\Leftrightarrow x^2+4x=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

2) \(3\left(x-4\right)+x^2-4x=0\Leftrightarrow3\left(x-4\right)+x\left(x-4\right)=0\Leftrightarrow\left(3+x\right)\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)