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 - 3 )² - 4 = 0

<=> ( x - 3 )² = 4

<=> \(\orbr{\begin{cases}\left(x-3\right)^2=2^2\\\left(x-3\right)^2=\left(-2\right)\end{cases}}\)

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

<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

Vậy S = { 5 ; 1 }

b) x² - 9 = 0

<=> x² = 9

<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy S = { 3 ; -3 }

c) x( x - 2x ) - x² - 8 = 0

<=> x² - 2x² - x² - 8 = 0

<=> -2x² - 8 = 0

<=> -2x² = 8

<=> x² = -4 ( vô lí )

<=> x = \(\varnothing\)

Vậy S = { \(\varnothing\)}

d) 2x( x - 1 ) - 2x² + x - 5 = 0

<=> 2x² - 2x - 2x² + x - 5 = 0

<=> -x - 5 = 0

<=> -x = 5

<=> x = -5

Vậy S = { -5 }

e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0 

<=> x² - 3x - ( x² - x - 2 ) = 0

<=> x² - 3x - x² + x + 2 = 0

<=> - 2x + 2 = 0

<=> -2x = -2

<=> x = 1

Vậy S = { 1 }

f) x( 3x - 1 ) - 3x² - 7x = 0

<=> 3x² - x - 3x² - 7x = 0

<=> -8x = 0

<=> x = 0

Vậy S = { 0 } 

d) 2x(x - 1) - 2x² + x - 5 = 0

=> 2x² - 2x - 2x² + x - 5 = 0

=> -x = 5

=> x = -5

e) x(x - 3) - (x + 1)(x - 2) = 0

=> x² - 3x - (x² - x - 2) = 0 

=> x² - 3x - x² + x + 2 = 0

=> -2x = - 2

=> x = 1

f) x(3x - 1) - 3x² - 7x = 0

=> 3x² - x - 3x² - 7x = 0

=> -8x = 0

=> x = 0

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

=>x=3 hoặc x=-1

b: =>x(x-3)=0

=>x=0 hoặc x=3

c: =>(x-5)(x+1)=0

=>x=5 hoặc x=-1

d: =>5x^2+7x-5x-7=0

=>(5x+7)(x-1)=0

=>x=1 hoặc x=-7/5

e: =>x^2-4=0

=>x=2 hoặc x=-4

h: =>x^2-4x+4-3=0

=>(x-2)^2=3

=>\(x=2\pm\sqrt{3}\)

a)    x³-x²-21x+45=0

<=> x³+5x²-6x²-30x+9x+45=0

<=> (x+5)(x²-6x+9)=0

<=> (x+5)(x²-3x-3x+9)=0

<=> (x+5)(x-3)²=0

 Vậy S={-5;3}

b)    X³+3X²+4X+2=0

<=>  X³+X²+2X²+2X+2X+2=0

<=> (X+1)(X²+2X+2)=0

VÌ  X²+2X+2 >=0

NÊN S={-1}

C)    X⁴+7X-8=0

<=> X⁴-X³+X³-X²+X²-X+8X-8=0

<=> (X-1)(X³+X²+X+8)=0

VÌ X³+X²+X+8>=0

NÊN S={1}

D)     6X⁴-X³-7X²+X+1=0

<=>  6X⁴-6X³+5X³-5X²-2X²+2X-X+1=0

<=>  (X-1)(6X³+5X²-2X-1)=0

<=> (X-1)(6X³-3X²+8X²-4X+2X-1)=0

<=> (X-1)(2X-1)(3X^2_4X+1)=0

<=>  (X-1)(2X-1)(3X²-3x-x+1)=0

<=> (X-1)²(2X-1)(3x-1)=0

vậy S={1/3;1/2;1}

`#3107.\text {DN01012007}`

\(\left(x-5\right)\cdot\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\3-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0+5\\x=3-0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)

Vậy, \(x\in\left\{3;5\right\}\)

_______

\(\left(2x-8\right)\cdot\left(5-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-8=0\\5-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=8\\x=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\div2\\x=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

Vậy, \(x\in\left\{4;5\right\}\)

_______

\(7x\left(2x-14\right)=0\\ \Rightarrow\left[{}\begin{matrix}7x=0\\2x-14=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\2x=14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=14\div2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

Vậy, \(x\in\left\{0;7\right\}\)

______

\(\left(2x-4\right)\cdot\left(6-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-4=0\\6-2x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\div2\\x=6\div2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy, \(x\in\left\{2;3\right\}.\)

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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