`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`

`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)

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

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

Bài dài quá, lần sau chia nhỏ câu hỏi nhé!!!!!

đúng vậy

\(\left(2x+1\right)2-4\left(x+2\right)2=9\)

\(4x+2-8x-16=9\)

\(4x-8x=9+16-2\)

\(-4x=23\)

\(x=-\frac{23}{4}\)

a, \(\left(2x+1\right)2-4\left(x+2\right)2=9\)

\(\Leftrightarrow4x+2-8x-16=0\Leftrightarrow-4x-14=0\Leftrightarrow x=-\frac{7}{2}\)

b, \(\left(x+1\right)3-2x\left(x+3\right)=2\)

\(\Leftrightarrow3x+3-2x^2-6x=2\Leftrightarrow-3x+1-2x^2=0\)

bruh

b: =>4x^2+8x-8x^2+5x-10=0

=>-4x^2+13x-10=0

=>x=2 hoặc x=5/4

c: =>2x^2-5x+6x-15=2x^2+8x

=>x-15=8x

=>-7x=15

=>x=-15/7

d: =>3x^2+15x-2x-10-3x^2-12x=5

=>x-10=5

=>x=15

e: =>x^2-3x+2x^2+2x=3x^2-12

=>-x=-12

=>x=12

Bài 10:

a) (x+2)² -x(x+3) + 5x = -20

=> x² + 4x + 4 - x² - 3x + 5x = -20

=> 6x = -20 + (-4)

=> 6x = -24

=> x = -4

b) 5x³-10x²+5x=0   

=>5x(x²-2x+1)=0

=>5x(x-1)² =0

=> 5x=0 hoặc (x-1)²=0

=>x=0 hoặc x=1

c) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

=> (x² - 1)[(x² - 1)² -  (x⁴ + x² + 1)] = 0

<=> (x² - 1)(x⁴ - 2x² + 1 - x⁴ - x² - 1) = 0

<=>  (x² - 1)(-3x²) = 0

<=> (x² - 1)=0 hoặc (-3x²) =0

<=> x²=1 hoặc x²=0

<=> x=−1;1 hoặc x=0

d)

⇔x=-23/12

Học tốt nhé:333

1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)

\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)