Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn. 

x^2+2x-3/3+2x/4=x^2/3

1: Ta có: \(x^2-2x+5-\left(x-7\right)\left(x+2\right)\)

\(=x^2-2x+5-x^2-2x+7x-14\)

\(=3x-9\)

2: Ta có: \(-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\)

\(=-5x^2+25x+x^3-7x-3x^2+21\)

\(=x^3-8x^2+18x+21\)

3: Ta có: \(x\left(x^2-x-2\right)-\left(x+5\right)\left(x-1\right)\)

\(=x^3-x^2-2x-x^2-4x+5\)

\(=x^3-2x^2-6x+5\)

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

\(=2x^2-10x-x^2+4x-4-x^2+9\)

\(=\left(2x^2-x^2-x^2\right)+\left(-10x+4x\right)+\left(-4+9\right)\)

\(=-6x+5\)

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

\(=x^2+2x+1-3\left(x^2-25\right)-\left(4x^2-4x+1\right)\)

\(=x^2+2x+1-3x^2+75-4x^2+4x-1\)

\(=-6x^2+6x+75\)

3,\(\left(x-1\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\)

\(=\left(x-1\right)^3-\left(x^3-27\right)\)

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

\(=-3x^2+3x+26\)

4,\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+2\right)^3\)

\(=\left(x^3+125\right)-\left(x^3+6x^2+12x+8\right)\)

\(=x^3+125-x^3-6x^2-12x-8\)

\(=-6x^2-12x+117\)

5,\(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)^2+\left(x+1\right)^2\)

\(=2x^2-14x-\left(x+3\right)\left(x^2-4x+4\right)+x^2+2x+1\)

=\(2x^2-14x-x^3+4x^2-4x-3x^2+12x-12+x^2+2x+1\)

\(=-x^3+4x^2-4x+1\)

6,\(\left(2x+5\right)\left(x-3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)

\(=2x^2-6x+5x-15-x^2+x-5x+5-x^2+8x-16\)

\(=3x-26\)

7,\(\left(x+5\right)\left(x-5\right)\left(x+2\right)-\left(x+2\right)^3\)

=\(\left(x^2-25\right)\left(x+2\right)-x^3-6x^2-12x-8\)

\(=x^3+2x^2-25x-50-x^3-6x^2-12x-8\)

\(=-4x^2-27x-58\)

Nếu đúng thì tick cho mk nha ^_^

1.

$(x-2)(x-5)=(x-3)(x-4)$

$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)

Vậy pt vô nghiệm.

2.

$(x-7)(x+7)+x^2-2=2(x^2+5)$

$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$

$\Leftrightarrow -51=10$ (vô lý)

Vậy pt vô nghiệm.

3.

$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$

$\Leftrightarrow 4x+10=-8$

$\Leftrightarrow 4x=-18$

$\Leftrightarrow x=-4,5$

4.

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

$\Leftrightarrow x^2+2x+1=x^2+x-6$

$\Leftrightarrow x=-7$ 

2:

a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8

=>x^2-x-12-x^2+4x+5=8

=>3x-7=8

=>3x=15

=>x=5

b: =>3x^2+3x-2x-2-3x^2-21x=13

=>-20x=15

=>x=-3/4

c: =>x^2-25-x^2-2x=9

=>-2x=25+9=34

=>x=-17

d: =>x^3-1-x^3+3x=1

=>3x-1=1

=>3x=2

=>x=2/3

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

\(=x^2+4x+4+x^2-6x+9\)

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

=-10x-50

4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)

\(=x^2-6x+9-x^2+16\)

=-6x+25

5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)

\(=y^2-6y+9-y^2+6y-9\)

=0

6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)

\(=4x^2+12x+9-4x^2+9\)

=12x+18

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

đúng vậy