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

`a, x^2-10x+25=0`

`<=>x^2 -2.x.5+5^2=0`

`<=>(x-5)^2=0`

`<=>x-5=0`

`<=>x=5`

__

`x^2 -8x+16=0`

`<=> x^2 - 2.x.4+4^2=0`

`<=>(x-4)^2=0`

`<=>x-4=0`

`<=>x=4`

__

`x^2-49=0`

`<=>x^2 - 7^2=0`

`<=>(x-7)(x+7)=0`

`<=>x-7=0` hoặc `x+7=0`

`<=> x=7` hoặc `x=-7`

__

`4x^2-25=0`

`<=> (2x)^2 -5^2=0`

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

`<=>2x-5=0` hoặc `2x+5=0`

`<=> 2x=5` hoặc `2x=-5`

`<=>x=5/2` hoặc `x=-5/2`

a: =>(x-5)^2=0

=>x-5=0

=>x=5

b: =>(x-4)^2=0

=>x-4=0

=>x=4

c: =>(x-7)(x+7)=0

=>x-7=0 hoặc x+7=0

=>x=7 hoặc x=-7

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

=>2x-5=0 hoặc 2x+5=0

=>x=5/2 hoặc x=-5/2

\(\left(3-x\right)\left(3+x\right)+\left(x-5\right)^2\\ =9-x^2+x^2-10x+25\\ =34-10x\)

Bài 3:

\(C=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}\)

\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}\)

\(=\dfrac{-3}{x-3}\)

Mình cần gấp ạk

A và B

Ta có: x³ - 0,25.x = 0

=> x.(x² - 0,25) = 0

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-0,25=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=0,25=0,5^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=0,5\end{cases}}\)

a)  x³ - 0,25x = 0

 x.x.x - 0,25x = 0

 x. ( x² - 0,25 ) = 0

TH1 : x = 0

TH2 : x² - 0,25 = 0

x²                   = 0 + 0,25

x²                   = 0,25 

=> x = 0,5

Vậy x = 0 ; 0,5

a) x³ - 0,25x= 0

   =>x=0

b) x² - 10x=-25

   -10x =-25=> x=5

Chỉ cần biết 1 x trong phép toán được, chữ x nào cũng được

a) \(25x^2-2=0\)

\(=>\left(5x\right)^2-\left(\sqrt{2}\right)^2=0\)

\(=>\left(5x-\sqrt{2}\right)\left(5x+\sqrt{2}\right)=0\)

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

\(=>\hept{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=-\frac{\sqrt{2}}{5}\end{cases}}\)

b) \(10x-x^2-25=0\)

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

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

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

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

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