5

ko có?

\(x^2=-25\)

Vì \(x^2\ge0\forall x\)

Mà \(x^2=-25\) (vô lí)

Vậy: \(x\in\varnothing\)

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

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

a: \(x^3-4x^2-x+4=0\)

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

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

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

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

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

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

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

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

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

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

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a) 4x(x+1)=8(x+1)

<=>4x(x+1)-8(x+1)=0

<=>(4x-8)(x+1)=0

<=>\(\left[\begin{array}{} 4x-8=0\\ x+1=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=2\\ x=-1 \end{array} \right.\)

Vậy...

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

<=>(x+2)(x-1)=0

<=>\(\left[\begin{array}{} x+2=0\\ x-1=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=-2\\ x=1 \end{array} \right.\)

Vậy...

c)5x(x-2)-(2-x)=0

<=>(5x+1)(x-2)=0

<=>\(\left[\begin{array}{} 5x+1=0\\ x-2 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=-1/5\\ x=2 \end{array} \right.\)

d)5x(x-200)-x+200=0

<=>(5x-1)(x-200)=0

<=>\(\left[\begin{array}{} 5x-1=0\\ x-200=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=1/5\\ x=200 \end{array} \right.\)

e)\(x^3+4x=0 \)

\(\Leftrightarrow x(x^2+4)=0 \)

\(\Leftrightarrow \left[\begin{array}{} x=0\\ x^2+4=0 (loại vì x^2+4>=0 với mọi x) \end{array} \right.\)

Vậy x=0

f)\((x+1)=(x+1)^2\)

\(\Leftrightarrow (x+1)-(x+1)^2=0\)

\(\Leftrightarrow (x+1)(1-x-1)=0\)

\(\Leftrightarrow (x+1)(-x)=0\)

\(\Leftrightarrow \left[\begin{array}{} x=-1\\ x=0 \end{array} \right.\)

Vậy....

a) \(3x\left(x-2\right)-4x+8=0\)

\(\Leftrightarrow3x\left(x-2\right)-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{4}{3}\end{cases}}\)

b) \(3\left(2x-1\right)^2+2-4x=0\)

\(\Leftrightarrow3\left(2x-1\right)^2-2\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(6x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\6x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{6}\end{cases}}\)

(x-2)^2-4x+8=0

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

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

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

=>x=2 hoặc x=6

`(x-2)^2 -4x+8=0`

`<=> (x-2)^2 -(4x-8)=0`

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

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

`<=>(x-2)(x-6)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Bài 1 : 

a) Ta có : x² - 9x + 8 = x² - x - 8x + 8 = x(x - 1) - 8(x - 1) = (x - 8)(x - 1)

b) Ta có : x² + 6x + 8 = x² + 6x + 9 - 1 = (x + 3)² - 1 = (x + 3 - 1)(x + 3 + 1) = (x + 2)(x + 4) 

Bài 2 : 

b) 4x² - 25 = 0

=> 4x² = 25

=>  (2x)² = 5²

=> 2x = -5;5

=> x = -5/2 ; 5/2

b) = x^2 + 2.x.3 + 3^2 - 1

=(x + 3)^2 - 1

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

=(x + 4)(x + 2)

Phần a mk nghĩ bn nên tự lm.

a) Ta có : x² - 9x + 8 = x² - x - 8x + 8 = x(x - 1) - 8(x - 1) = (x - 8)(x - 1)

b) Ta có : x² + 6x + 8 = x² + 6x + 9 - 1 = (x + 3)² - 1 = (x + 3 - 1)(x + 3 + 1) = (x + 2)(x + 4) 

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b) 4x² - 25 = 0

=> 4x² = 25

=>  (2x)² = 5²

=> 2x = -5;5

=> x = -5/2 ; 5/2

Bài 2 : 

a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)

b, \(5x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow5x\left(x-2020\right)-\left(x-2020\right)=0\Leftrightarrow\left(5x-1\right)\left(x-2020\right)=0\)

\(\Leftrightarrow x=\frac{1}{5};2020\)

c, \(\left(4x+5\right)^2-\left(2x-1\right)^2=0\)

\(\Leftrightarrow16x^2+40x+25-\left(4x^2-4x+1\right)=0\)

\(\Leftrightarrow12x^2+44x+24=0\Leftrightarrow4\left(x+3\right)\left(3x+2\right)=0\)

\(\Leftrightarrow x=-3;-\frac{2}{3}\)

a,x²-4x=0

= x.(x-4)=0

=> x=0 hoặc x-4=0

=>x=0 hoặc x=4

a. x² - 4x = 0

<=> x ( x - 4 ) = 0

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

b. 5x ( x - 2020 ) - x + 2020 = 0

<=> 5x ( x - 2020 ) - ( x - 2020 ) = 0

<=> ( 5x - 1 ) ( x - 2020 ) = 0

<=>\(\orbr{\begin{cases}5x-1=0\\x-2020=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{1}{5}\\x=2020\end{cases}}\)

c. ( 4x + 5 )² - ( 2x - 1 )² = 0

<=> 16x² + 40x + 25 - 4x² + 4x - 1 = 0

<=> 12x² + 44x + 24 = 0

<=> 4 ( 3x² + 11x + 6 ) = 0

<=> ( 3x² + 9x ) + ( 2x + 6 ) = 0

<=> 3x ( x + 3 ) + 2 ( x + 3 ) = 0

<=> ( 3x + 2 ) ( x + 3 ) = 0

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

d. x² + 6x - 8 = 0

<=> x² + 6x + 9 = 17

<=> ( x + 3 )² = 17 

<=>\(\orbr{\begin{cases}x+3=\sqrt{17}\\x+3=-\sqrt{17}\end{cases}}\)<=>\(\orbr{\begin{cases}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{cases}}\)

e. 4x² + 2x - 6 = 0

<=> 2 ( 2x² + x - 3 ) = 0

<=> ( 2x² + 3x ) - ( 2x + 3 ) = 0

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

<=> ( x - 1 ) ( 2x + 3 ) = 0

<=>\(\orbr{\begin{cases}x-1=0\\2x+3=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=1\\x=-\frac{3}{2}\end{cases}}\)