hết cứu đi mà làm

Rảnh rỗi thật sự .-.

a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

Cảm ơn nhìu ạ :3

\(o,x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

\(n,3x^3-3x^2-6x=0\)

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

\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)

\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)

\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)

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

`a) x^3 - 9x^2 + 14x = 0`

\(\Rightarrow\) `x^3 - 7x^2 - 2x^2 + 14x = 0`

\(\Rightarrow\) `(x^3 - 2x^2) - (7x^2 - 14x) =0`

\(\Rightarrow\) `x^2.(x - 2) - 7x.(x - 2) =0`

\(\Rightarrow\) `(x^2 - 7x)(x-2)=0`

\(\Rightarrow\) `x.(x-7)(x-2)=0`

\(\Rightarrow\left[\begin{array}{l}x=0\\ x-7=0\\ x-2=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=0+7\\ x=0+2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=7\\ x=2\end{array}\right.\)

Vậy \(x\in\left\lbrace0;7;2\right\rbrace\)

`3.x^3 - 5x^2 + 8x - 4 = 0`

\(\Rightarrow\) `x^3 - x^2 - 4x^2 + 4x + 4x - 4 =0`

\(\Rightarrow\) `x^2 . (x-1) - 4x(x-1) + 4.(x-1) =0`

\(\Rightarrow\) `(x^2 - 4x + 4)(x-1)=0`

\(\Rightarrow\) `(x-2)^2(x-1)=0`

\(\Rightarrow\left[\begin{array}{l}x-2=0\\ x-1=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=2\\ x=1\end{array}\right.\)

Vậy \(x\in\left\lbrace2;1\right\rbrace\)

2: \(x^2-2xy+y^2-2x+2y\)

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

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

3: \(3x^2-2x-5\)

\(=3x^2-5x+3x-5\)

=x(3x-5)+(3x-5)

=(3x-5)(x+1)

4: \(16-x^2+4xy-4y^2\)

\(=16-\left(x^2-4xy+4y^2\right)\)

\(=4^2-\left(x-2y\right)^2\)

=(4-x+2y)(4+x-2y)

5: \(x^2-2x+1-y^2\)

\(=\left(x-1\right)^2-y^2\)

=(x-1-y)(x-1+y)

6: \(x^2+8x+15\)

\(=x^2+3x+5x+15\)

=x(x+3)+5(x+3)

=(x+3)(x+5)

7: \(\left(x^2+6x+8\right)\left(x^2+14x+48\right)-9\)

=(x+2)(x+4)(x+6)(x+8)-9

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)-9\)

\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+384-9\)

\(=\left(x^2+10x\right)^2+15\left(x^2+10x\right)+25\left(x^2+10x\right)+375\)

\(=\left(x^2+10x+25\right)\left(x^2+10x+15\right)=\left(x+5\right)^2\cdot\left(x^2+10x+15\right)\)

8: \(\left(x^2-8x+15\right)\left(x^2-16x+60\right)-24x^2\)

\(=\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)

\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)

\(=\left(x^2+30\right)^2-24x\left(x^2+30\right)+143x^2-24x^2\)

\(=\left(x^2+30\right)^2-24x\left(x^2+30\right)+119x^2\)

\(=\left(x^2-7x+30\right)\left(x^2-17x+30\right)\)

\(=\left(x^2-7x+30\right)\left(x-2\right)\left(x-15\right)\)

a) 5 - 4x = 3x - 9

\(\Leftrightarrow5-4x-3x+9=0\)

\(\Leftrightarrow14-7x=0\)

\(\Leftrightarrow7x=14\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

b) \(\left(x-4\right)\left(3x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{-3;4\right\}\)

c) \(\dfrac{x}{x+4}+\dfrac{12}{x-4}=\dfrac{4x+48}{x\cdot x-16}\)(1)

ĐKXĐ: \(x\ne\pm4\)

\(\left(1\right)\Leftrightarrow\dfrac{x\left(x-4\right)+12\left(x+4\right)-4x-48}{\left(x+4\right)\left(x-4\right)}=0\)

\(\Leftrightarrow x^2-4x+12x+48-4x-48=0\)

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

\(\Leftrightarrow x\left(x+4\right)=0\)

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

Vậy \(S=\left\{0\right\}\)

d) \(4-2x=7-x\)

\(\Leftrightarrow4-2x-7+x=0\)

\(\Leftrightarrow-x-3=0\)

\(\Leftrightarrow-x=3\Leftrightarrow x=-3\)

Vậy \(S=\left\{-3\right\}\)

e) \(\left(x+4\right) \left(8-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\8-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{-4;2\right\}\)

f) \(\dfrac{x}{x+5}+\dfrac{11}{x-5}=\dfrac{x+55}{x\cdot x-25}\left(2\right)\)

ĐKXĐ: \(x\ne\pm5\)

\(\left(2\right)\Leftrightarrow\dfrac{x\left(x-5\right)+11\left(x+5\right)-x-55}{\left(x+5\right)\left(x-5\right)}=0\)

\(\Leftrightarrow x^2-5x+11x+55-x-55=0\)

\(\Leftrightarrow x^2+5x=0\)

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

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

Vậy \(S=\left\{0\right\}\)

g) \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)-3x-1-10-12x}{6}=0\)

\(\Leftrightarrow9x+6-3x-1-10-12x=0\)

\(\Leftrightarrow-6x-5=0\)

\(\Leftrightarrow-6x=5\)

\(\Leftrightarrow x=-\dfrac{5}{6}\)

Vậy \(S=\left\{-\dfrac{5}{6}\right\}\)

h) \(2x-\left(3-5x\right)=4\left(x+3\right)\)

\(\Leftrightarrow2x-3+5x-4x-12=0\)

\(\Leftrightarrow3x-15=0\)

\(\Leftrightarrow x=5\)

Vậy \(S=\left\{5\right\}\)

i) \(3x-6+x=9-x\)

\(\Leftrightarrow3x-6+x-9+x=0\)

\(\Leftrightarrow5x-15=0\)

\(\Leftrightarrow x=3\)

Vậy \(S=\left\{3\right\}\)

k)\(2t-3+5t=4t+12\)

\(\Leftrightarrow2t-3+5t-4t-12=0\)

\(\Leftrightarrow3t-15=0\)

\(\Leftrightarrow t=5\)

Vậy \(S=\left\{5\right\}\)

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