5x² - 4(x² - 2x + 1) - 5 = 0

=> 5x² - 4x² + 8x - 4 - 5 = 0 

=> x² + 8x - 9 = 0

=> x² + 9x - x - 9 = 0 

=> x(x + 9) - (x + 9) = 0

=> (x + 9)(x - 1) = 0

=> x + 9 = 0 => x = -9

hoặc x - 1 = 0 = > x = 1

                                                                       Vậy x = -9, x = 1

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

\(\left(5x^2-5\right)-4\left(x^2-2.1.x+1^2\right)=0\)

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

\(5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)\left(x-1\right)=0\)

\(\left[5\left(x+1\right)-4\left(x-1\right)\right]\left(x-1\right)=0\)

\(\left(5x+5-4x+4\right)\left(x-1\right)=0\)

\(\left(x+9\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=-9\\x=1\end{cases}}.\)

\(5x\div21=0\)

\(5x=0\times21\)

\(5x=0\)

\(\Rightarrow x=0\)

À thui mình nghĩ ra roài

a/ (do (x-3)^2 + 1 ≠0 vs mọi x

a) (x-3)³-3+x=0

=> (x-3)³+(x-3)=0

=> (x-3)(x²-6x+10)

=> \(\left[{}\begin{matrix}x-3=0\\x^2-6x+10=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)

\(5\left(x-3\right)+\left(x-2\right)\left(5x-1\right)=5x^2\)

\(\Leftrightarrow5x-15-\left(5x^2-11x+2\right)=5x^2\)

\(\Leftrightarrow5x-15-5x^2+11x-2=5x^2\)

\(\Leftrightarrow-10x^2+16x-17=0\)

\(\cdot\Delta=16^2-4.\left(-10\right).\left(-17\right)=-304< 0\)

Vậy pt vô nghiệm

Bài 2:

\(2x^3+6x^2=x^2+3x\)

\(\Rightarrow2x^3+6x^2-x^2-3x=0\)

\(\Rightarrow2x^3+5x^2-3x=0\)

\(\Rightarrow x\left(2x^2+5x-3\right)=0\)

\(\Rightarrow x\left(2x^2-x+6x-3\right)=0\)

\(\Rightarrow x\left[x\left(2x-1\right)+3\left(2x-1\right)\right]=0\)

\(\Rightarrow x\left(x+3\right)\left(2x-1\right)=0\)

=>x=0 hoặc x=-3 hoặc x=1/2

1)

\(\Leftrightarrow x^2-2x-3x+6=0\)

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

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

Tới đây b cho từng cái = 0 rồi giải ra tìm x nha :)

`a)2x^2+3(x-1)(x+1)=5x(x+1)`

`<=>2x^2+3x^2-3=5x^2+5x`

`<=>5x=-3`

`<=>x=-3/5`

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`b)(x-3)^3+3-x=0` nhỉ?

`<=>(x-3)^3-(x-3)=0`

`<=>(x-3)(x^2-1)=0`

`<=>[(x=3),(x^2=1<=>x=+-1):}`

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`c)5x(x-2000)-x+2000=0`

`<=>5x(x-2000)-(x-2000)=0`

`<=>(x-2000)(5x-1)=0`

`<=>[(x=2000),(x=1/5):}`

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

`<=>6x-9+4-2x=-3`

`<=>4x=2`

`<=>x=1/2`

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`e)x+6x^2=0`

`<=>x(1+6x)=0`

`<=>[(x=0),(x=-1/6):}`