Trả lời:

Bài 1:

1, 4x² - 25 + ( 2x + 7 )( 5 - 2x ) 

= ( 2x - 5 )( 2x + 5 ) - ( 2x + 7 )( 2x - 5 )

= ( 2x - 5 )( 2x + 5 - 2x - 7 )

= - 2 ( 2x - 5 )

2, 3 ( x + 4 ) - x² - 4x 

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

= ( x + 4 )( 3 - x )

3, 5x² - 5y² - 10x + 10y 

= 5 ( x² - y² - 2x + 2y )

= 5 [ ( x² - y² ) - ( 2x - 2y ) ]

= 5 [ ( x - y )( x + y ) - 2 ( x - y ) ]

= 5 ( x - y )( x + y - 2 )

4, x² - xy + x - y 

= ( x² - xy ) + ( x - y )

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

= ( x - y )( x + 1 )

5, ax - bx - a² + 2ab - b² 

= ( ax - bx ) - ( a² - 2ab + b² )

= x ( a - b ) - ( a - b )² 

= ( a - b )( x - a + b )

6, x² + 4x - y² + 4 

= ( x² + 4x + 4 ) - y² 

= ( x + 2 )² - y² 

= ( x + 2 - y )( x + 2 + y )

\(x^2+2xy-8y^2+2xz+14yz-3z^2\)

\(=\left(x^2+y^2+z^2+2xy+2yz\right)+\left(-9x^2-12yz-4x^2\right)\)

\(=\left(x+y+z\right)^2-[\left(3x\right)2-2.3x2y+\left(2x\right)^2\)

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

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

\(3x^2-22xy-4x+8y+7y^2+1\)

\(=3x^2-21xy-xy-3x-x+7y+y+7y^2+1\)

\(=\left(3x^2-21xy-3x\right)-\left(xy-7y^2-y\right)-\left(x-7y-1\right)\)

\(=3x\left(x-7y-1\right)-y\left(x-7y-1\right)-\left(x-7y-1\right)\)

\(=\left(x-7y-1\right)\left(3x-y-1\right)\)

\(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)

\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)\)

\(=\left(2x-5\right)[2x+5-\left(2x+7\right)]\)

\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)

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

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

\(=-x^2+3x+12-4x\)

\(=-\left(x^2+4x-3x-12\right)\)

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

Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Rightarrow\left(x+\frac{1}{6}y+3\right)^2=x^2+\left(\frac{y}{6}\right)^2+3^2+2.x.\frac{y}{6}+2.\frac{y}{6}.3+2.x.3\)

\(=x^2+\frac{y^2}{36}+9+\frac{xy}{3}+y+6x\)

Giúp em với

d) \(\Leftrightarrow\left(x-2\right)^2=25\)

\(\Leftrightarrow x=7\)hoặc \(x=-3\)

e) \(\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)

\(\Leftrightarrow x=\frac{1}{2}\)hoặc \(x=\frac{9}{2}\)

d) \(x^2-4x+4=25\)

\(\Leftrightarrow\left(x-2\right)^2-5^2=0\)

\(\Leftrightarrow\left(x-2-5\right)\left(x-2+5\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)

Vậy \(S=\left\{7;-3\right\}\)

b) \(\left(5-2x\right)^2-16=0\)

\(\Leftrightarrow\left(5-2x\right)^2-4^2=0\)

\(\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)

\(\Leftrightarrow\left(1-2x\right).\left(9-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}1-2x=0\\9-2x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-2x=-1\\-2x=-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}\)

Vậy \(S=\left\{\frac{1}{2};\frac{9}{2}\right\}\)

a) \(\left(3-xy^2\right)^2-\left(2+xy^2\right)^2\)

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

\(=\left(1-2xy^2\right).5=5-10xy^2\)

b) \(9x^2-\left(3x-4\right)^2\)

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

\(=4.\left(6x-4\right)=24x-16\)

c) \(\left(a-b^2\right)\left(a+b^2\right)\)

\(=a^2-b^{^4}\)

d) \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)

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

\(=a^4+4a^3+4a^2-9\)

b) x³ + 9x² + 27x + 19=0

x³ + 3. x² .1 + 3x . 3² + 3³ - 8 =0

(x + 3)³ - 2³ =0

(x+3)³ = 8

(x+3)³ = 2³

x+3 =2

x= -1

Vậy x  =- 1

b) x(x+5)(x-5) - (x+2)(x² - 2x +4) =3

x(x² - 25) - (x³ + 2³) =3

x³ - 25x - x³ -8 =3

-25x - 8 =3

-25x = 11

x= -11/25

Vậy x = -11/25

9x² - 6x -3 =0

3 (3x² - 2x - 1 ) =0

3x² - 2x -1 =0

3x² - 3x + x -1 =0

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

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

=> x- 1 =0 hoặc 3x + 1=0

=> x= 1 hoặc x = -1/3

Vậy x =1 hoặc x = -1/3

9x² - 6x - 3 = 0 

<=> 9x² - 6x + 1 - 4 = 0 

<=> (3x - 1)² - 4 = 0 

<=> (3x - 3)(3x + 1) = 0 

<=> \(\orbr{\begin{cases}3x-3=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)

Vậy \(x\in\left\{1;-\frac{1}{3}\right\}\)