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

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.

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

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

$\Leftrightarrow x=-7$ 

5.

$x^2-(2x-1)(x+3)=3-x(5+x)$
$\Leftrightarrow x^2-(2x^2+5x-3)=3-(5x+x^2)$
$\Leftrightarro -x^2-5x+3=3-5x-x^2$ (luôn đúng)

Vậy pt có nghiệm $x\in\mathbb{R}$

6.

$3(5-2x)-4(x+2)=5x-18$

$\Leftrightarrow 15-6x-4x-8=5x-18$

$\Leftrightarrow 7-10x=5x-18$

$\Leftrightarrow 25=15x$

$\Leftrightarrow x=\frac{5}{3}$

a: =>9x^2+12x+4-9x^2+12x-4=5x+38

=>24x=5x+38

=>19x=38

=>x=2

e: =>x^3+1-2x=x^3-x

=>-2x+1=-x

=>-x=-1

=>x=1

f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1

=>12x-9=3x+1

=>9x=10

=>x=10/9

b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)

=>-3x+3=3x-9

=>-6x=-12

=>x=2

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

đúng vậy

4 x 2 x 5 = ?

Cách 1: 4 x 2 x 5 = (4 x 2) x 5 = 8 x 5 = 40

Cách 2: 4 x 2 x 5 = 4 x (2 x 5) = 4 x 10 = 40

7 x 2 x 3 = ?

Cách 1: 7 x 2 x 3 = (7 x 2) x 3 = 14 x 3 = 42

Cách 2: 7 x 2 x 3 = 7 x (2 x 3) = 7 x 6 = 42

6 x 3 x 3 = ?

Cách 1: 6 x 3 x 3 = (6 x 3) x 3 = 18 x 3 = 54

Cách 2: 6 x 3 x 3 = 6 x (3 x 3) = 6 x 9 = 54

6 x 2 x 4 = ?

Cách 1: 6 x 2 x 4 = (6 x 2) x 4 = 12 x 4 = 48

Cách 2: 6 x 2 x 4 = 6 x (2 x 4) = 6 x 8 = 48

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

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)

\(\Leftrightarrow6x=-3\)

hay \(x=-\dfrac{1}{2}\)

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

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)

\(\Leftrightarrow2x^3+6x=2x^3+24x\)

\(\Leftrightarrow x=0\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)

\(\Leftrightarrow12x=-11\)

hay \(x=-\dfrac{11}{12}\)