1.thay x=25 vào biểu thức A ta có:

25^3-15.25^2+75.25=8125

2.

a,x^3-3^3-x(x^2-2^2)-1=0

x^3-27-x^3+4x-1=0

4x-28=0

4(x-7)=0

X=7

b,(x^3+3x^2+3x+1)-(x^3-3x^2+3x-1)-6(x^2-2x+1)+10=0

x^3+3x^2+3x+1-X^3+3x^2-3x+1-6x^2+12x-6+10=0

12x+6=0

6(2x+1)=0

2x+1=0

2x=-1

x=-1/2

**** cho mk nha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

bài 1: 

a) ĐKXĐ: x khác 0; x khác -1

 \(\frac{x-1}{x}+\frac{1-2x}{x^2+x}=\frac{1}{x+1}\)

<=> \(\frac{x-1}{x}+\frac{1-2x}{x\left(x+1\right)}=\frac{1}{x+1}\)

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

<=> x^2 - 2x = x

<=> x^2 - 2x - x = 0

<=> x^2 - 3x = 0

<=> x(x - 3) = 0

<=> x = 0 hoặc x - 3 = 0

<=> x = 0 hoặc x = 0 + 3

<=> x = 0 (ktm) hoặc x = 3 (tm)

=> x = 3

b) ĐKXĐ: x khác +-3; x khác -7/2

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)

<=> \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)

<=> 13(x + 3) + (x - 3)(x + 3) = 6(2x + 7)

<=> 13x + 30 + x^2 = 12x + 42

<=> 13x + 30 + x^2 - 12x - 42 = 0

<=> x - 12 + x^2 = 0

<=> (x - 3)(x + 4) = 0

<=> x - 3 = 0 hoặc x + 4 = 0

<=> x = 0 + 3 hoặc x = 0 - 4

<=> x = 3 (ktm) hoặc x = -4 (tm)

=> x = -4

c) ĐKXĐ: x khác +-1

\(\frac{x}{x-1}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)

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

<=> x^2 + x - 2x = 0

<=> x^2 - x = 0

<=> x(x - 1) = 0

<=> x = 0 hoặc x - 1 = 0

<=> x = 0 hoặc x = 0 + 1

<=> x = 0 (tm) hoặc x = 1 (ktm)

=> x = 0

d) \(\frac{x^2+2x}{x^2+1}-2x=0\)

<=> \(\frac{x\left(x+2\right)}{x^2+1}-2x=0\)

<=> x(x + 2) - 2x(x^2 + 1) = 0

<=> x^2 - 2x^3 = 0

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

<=> x^2 = 0 hoặc 1 - 2x = 0

<=> x = 0 hoặc -2x = 0 - 1

<=> x = 0 hoặc -2x = -1

<=> x = 0 hoặc x = 1/2

bài 2: 

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

<=> x^3 + 3x^2 - 2x - x^2 - 3x + 2 - x^2 + 1 = 0

<=> 2x^2 - 2x - 3x + 3 = 0

<=> 2x(x - 1) - 3(x - 1) = 0

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

<=> 2x - 3 = 0 hoặc x - 1 = 0

<=> 2x = 0 + 3 hoặc x = 0 + 1

<=> 2x = 3 hoặc x = 1

<=> x = 3/2 hoặc x = 1

bài 3:

(x^3 + x^2) + (x^2 + x) = 0

<=> x^3 + x^2 + x^2 + x = 0

<=> x^3 + 2x^2 + x = 0

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

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

<=> x = 0 hoặc x + 1 = 0

<=> x = 0 hoặc x = 0 - 1

<=> x = 0 hoặc x = -1

a: \(\Leftrightarrow x\left(16-x^2\right)+x^3-125=3\)

=>16x-125=3

=>16x=128

hay x=8

b: \(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)

=>12x-4=-10

=>12x=-6

hay x=-1/2

c: \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=1\)

\(\Leftrightarrow4x-27=1\)

hay x=7

Chia nhỏ ra bạn ơi!

\(a) x² +3y²+2z²-2x+12y+4z+15=0 \)

\(⇔x²-2x+1+3y²+12y+12+2z²+4z+2=0 \)

\(⇔(x²-2x+1) + 3(y²+4y+4) +2(z²+2z+1)=0 \)

\(⇔(x-1)² +3(y+2)²+2(z+1)²=0 \)

\(⇔ x-1=0 \) và \(y+2=0\) và \(z+1=0\)

Vậy: \(x=1;y=-2;z=-1\)

a) (x-3)(x+3)-(x-1)^2=0

=> (x^2-9)-(x^2-2x+1)=0

=>x^2-9-x^2+2x-1=0

=>(x^2-x^2)-9-1+2x=0

=>-10+2x=0

=>-2.(-5-x)=0

=>-5-x=0

=>-x=0+5

=>x=-5

vậy x=-5

b) x^3-3x^2+3x-1=0

=>(x-1)^3=0

=>x-1=0 

=>x=0+1

=>x=1

vậy x=1

c) 4x^2-28x=0

=>4x.(x-7)=0

=> 2 TH

* 4x=0=>x=0

*x-7=0=>x=0+7=>x=7

vậy x=0 hoặc x=7