a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)

c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

d) Giống câu c

⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3

c) ${}^{}-10x=-25\Rightarrow {}^{}-10x+$

=(-x^2+2xy+2x)-(4y^2-10y+3)

=-(x^2-2xy-2x)-(4y^2-10y+3)

=-[x^2-2x(y-1)]-(4y^2-10y+3)

=-[x^2-2x(y-1)+(y-1)^2]-[4y^2-10y-(y-1)^2+3]

=-[x^2-2x(y-1)+(y-1)^2]-(4y^2-10y-y^2+2y-1+3)

=-(x-y+1)^2-(3y^2-8y+2)

=-(x-y+1)^2-3(y^2-4/3*2*y+16/9+2/3-16/9

=-(x-y+1)^2-3[(y-4/3)^2-10/9]

=-(x-y+1)^2-3(y-4/3)^2+10/3

a)

 ⇔ \(x^2-16=9\)

⇔ \(x^2=25\)

⇔ \(x=\pm5\)

b)

 ⇔ \(x^2-4x+4-25x^2+20x-4=0\)

⇔ \(16x-24x^2=0\)

⇔ \(8x\left(2-3x\right)=0\)

⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\)   ⇔   \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)

c)  

⇔ \(3x^2-10x-20=0\)

⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)

⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)

⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)

Vậy... 

d) 

⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)

⇔ 7x = 49

⇔ x=7

Vậy...

`#040911`

`a)`

`196 - a^2 + 2ab - b^2`

`= 196 - (a^2 - 2ab + b^2)`

`= 196 - (a - b)^2`

`= 14^2 - (a - b)^2`

`= (14 - a + b)(14 + a - b)`

`b)`

`a^2 + 6a - 4b^2 + 9`

`= (a^2 + 6a + 9) - 4b^2`

`= [ (a)^2 + 2*a*3 + 3^2] - (2b)^2`

`= (a + 3)^2 - (2b)^2`

`= (a + 3 - 2b)(a + 3 + 2b)`

`c)`

`4x - 4 + 9y^2 - x^2`

`=  9y^2 - (x^2 - 4x + 4)`

`= (3y)^2 - [ (x)^2 - 2*x*2 + 2^2]`

`= (3y)^2 - (x - 2)^2`

`= (3y - x + 2)(3y + x - 2)`

`d)`

`5x^2 - 10x + 5 - 45t^2`

`= 5*(x^2 - 2x + 1 - 9t^2)`

`= 5*[ (x^2 - 2x + 1) - 9t^2]`

`= 5*{ [(x)^2 - 2*x*1 + 1^2] - (3t)^2}`

`= 5*[ (x - 1)^2 - (3t)^2]`

`= 5*(x - 1 - 3t)(x - 1 + 3t)`

`e)`

`x^2 - 36y^2t^2 - 10x +25`

`= (x^2 - 10x + 25) - 36y^2t^2`

`= [ (x)^2 - 2*x*5 + 5^2] - (6yt)^2`

`= (x - 5)^2 - (6yt)^2`

`= (x - 5 - 6yt)(x - 5 + 6yt)`

a: =196-(a^2-2ab+b^2)

=196-(a-b)^2

=(14-a+b)(14+a-b)

b: \(=\left(a^2+6a+9\right)-4b^2\)

\(=\left(a+3\right)^2-4b^2\)

\(=\left(a+3-2b\right)\left(a+3+2b\right)\)

c: \(=9y^2-\left(x^2-4x+4\right)\)

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

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

d: \(=5\left(x^2-2x+1-9t^2\right)\)

\(=5\left[\left(x-1\right)^2-\left(3t\right)^2\right]\)

\(=5\left(x-1-3t\right)\cdot\left(x-1+3t\right)\)

e: \(=x^2-10x+25-36y^2t^2\)

\(=\left(x-5\right)^2-\left(6yt\right)^2\)

\(=\left(x-5-6yt\right)\left(x-5+6yt\right)\)

a, 7x + 10x  = 5x 

    17x = 5x

17x - 5x = 0

      12x = 0

          x =0

2; 

a, 4x + 7x = 22

    11x = 22

        x = 2

b, 12x - 8x = 25

     4x = 25

       x = \(\dfrac{25}{4}\)

c,  \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)x = \(\dfrac{4}{5}\) 

     (\(\dfrac{1}{2}-\dfrac{1}{3}\))x = \(\dfrac{4}{5}\)

    \(\dfrac{1}{6}\)x     = \(\dfrac{4}{5}\) 

      x = \(\dfrac{4}{5}\) : \(\dfrac{1}{6}\)

     x = \(\dfrac{24}{5}\)

\(x^2-y^2+7x-7y=\left(x^2-y^2\right)+\left(7x-7y\right)=\left(x-y\right)\left(x+y\right)+7\left(x-y\right)=\left(x-y\right)\left(x+y+7\right)\)

\(x^2-10x+25-9y^2=\left(x^2-10x+25\right)-\left(3y\right)^2=\left(x-5\right)^2-\left(3y\right)^2=\left(x-3y-5\right)\left(x+3y-5\right)\)

\(x^2-y^2+7x-7y=\left(x-y\right)\left(x+y\right)+7\left(x-y\right)=\left(x-y\right)\left(x+y+7\right)\)

\(x^2-10x+25-9y^2=\left(x-5\right)^2-\left(3y\right)^2=\left(x-5-3y\right)\left(x-5+3y\right)\)

1. x² - 10x + 25 - 9y²

<=> x² - 10x + 5² - (3y)²

<=> (x - 5)² - (3y)²

<=> (x - 5 - 3y)(x - 5 + 3y)

2. x² - y² + 7x - 7y

<=> (x - y)(x + y) + 7(x - y)

<=> (x + y + 7)(x - y)