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.

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

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

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

$\Leftrightarrow 25=15x$

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

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

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

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

\(\Leftrightarrow x^2+4x+4+2x-8=x^2-6x+8\)

\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)

\(\Leftrightarrow12x-12=0\)

\(\Leftrightarrow12x=12\)

hay x=1

Vậy: S={1}

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

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

\(\Leftrightarrow2x^2-4x+3-2x+2=0\)

\(\Leftrightarrow2x^2-6x+5=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{5}{2}\right)=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}=0\)(Vô lý)

Vậy: \(S=\varnothing\)

c) Ta có: \(\left(x+3\right)^2-\left(x-3\right)^2=6x+18\)

\(\Leftrightarrow x^2+6x+9-\left(x^2-6x+9\right)-6x-18=0\)

\(\Leftrightarrow x^2-9-x^2+6x-9=0\)

\(\Leftrightarrow6x-18=0\)

\(\Leftrightarrow6x=18\)

hay x=3

Vậy: S={3}

d) Ta có: \(\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x\left(x^2+2x+1\right)=5x-5x^2-11x-22\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=-5x^2-6x-22\)

\(\Leftrightarrow-5x^2+2x-1+5x^2+6x+22=0\)

\(\Leftrightarrow8x+21=0\)

\(\Leftrightarrow8x=-21\)

hay \(x=-\dfrac{21}{8}\)

Vậy: \(S=\left\{-\dfrac{21}{8}\right\}\)

1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

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

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

Nhiều quá vại :( giải 1,2 câu thôi nhé

a) 

<=> 2x - 4 - 3 = x + 2

<=> 2x - x      = 2 + 4 + 3

<=>   x          = 9

d) (9x +3)^2 = 16

<=> (9x + 3)^2 = 4^2

<=> 9x + 3       = 4

<=> 9x             = 4 - 3

<=> 9x             = 1

<=>  x             = 1 : 9

<=> x              = 1/9

thank

// :là giá trị tuyệt đối nhé