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.

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

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

$\Leftrightarrow x=-7$ 

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: ĐKXĐ: x∉{3;-1}

\(\frac{2}{x+1}-\frac{1}{x-3}=\frac{3x-11}{x^2-2x-3}\)

=>\(\frac{2}{x+1}-\frac{1}{x-3}=\frac{3x-11}{\left(x-3\right)\left(x+1\right)}\)

=>\(\frac{2\left(x-3\right)-x-1}{\left(x-3\right)\left(x+1\right)}=\frac{3x-11}{\left(x-3\right)\left(x+1\right)}\)

=>3x-11=2(x-3)-x-1

=>3x-11=2x-6-x-1=x-7

=>3x-x=-7+11

=>2x=4

=>x=2(nhận)

b: ĐKXĐ: x<>0; x<>2

\(\frac{3}{x-2}+\frac{1}{x}=\frac{-2}{x\left(x-2\right)}\)

=>\(\frac{3x+x-2}{x\left(x-2\right)}=\frac{-2}{x\left(x-2\right)}\)

=>\(\frac{4x-2}{x\left(x-2\right)}=\frac{-2}{x\left(x-2\right)}\)

=>4x-2=-2

=>4x=0

=>x=0(loại)

c: ĐKXĐ: x<>3; x<>-3

\(\frac{x-3}{x+3}-\frac{2}{x-3}=\frac{3x+1}{9-x^2}\)

=>\(\frac{\left(x-3\right)^2-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-3x-1}{\left(x-3\right)\left(x+3\right)}\)

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

=>\(x^2-6x+9-2x-6+3x+1=0\)

=>\(x^2-5x+4=0\)

=>(x-1)(x-4)=0

=>x=1(nhận) hoặc x=4(nhận)

d: ĐKXĐ: x<>2; x<>-1

\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-5}{x^2-x-2}\)

=>\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-5}{\left(x-2\right)\left(x+1\right)}\)

=>\(\frac{2\left(x-2\right)-x-1}{\left(x-2\right)\left(x+1\right)}=\frac{3x-5}{\left(x-2\right)\left(x+1\right)}\)

=>3x-5=2x-4-x-1=x-5

=>2x=0

=>x=0(nhận)

e: ĐKXĐ: x<>2; x<>-2

\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)

=>\(\frac{\left(x-2\right)^2+3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)

=>\(\left(x-2\right)^2+3\left(x+2\right)=x^2-11\)

=>\(x^2-4x+4+3x+6=x^2-11\)

=>-x+10=-11

=>-x=-21

=>x=21(nhận)

f: ĐKXĐ: x<>-1;x<>0

\(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)

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

=>2x(x+1)=x(x+3)+(x-2)(x+1)

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

=>0=-2(vô lý)

=>Phương trình vô nghiệm

g: ĐKXĐ: x<>5; x<>-5

\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)

=>\(\frac{\left(x+5\right)^2-\left(x-5\right)^2}{\left(x+5\right)\left(x-5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\)

=>\(\left(x+5\right)^2-\left(x-5\right)^2=20\)

=>\(x^2+10x+25-x^2+10x-25=20\)

=>20x=20

=>x=1

h: ĐKXĐ: x<>1; x<>-1

\(\frac{x+4}{x+1}+\frac{x}{x-1}=\frac{2x^2}{x^2-1}\)

=>\(\frac{\left(x+4\right)\left(x-1\right)+x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{2x^2}{\left(x-1\right)\left(x+1\right)}\)

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

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

=>4x-4=0

=>4x=4

=>x=1(loại)

i: ĐKXĐ: x<>1; x<>-1

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

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

=>\(\left(x+1\right)^2-\left(x-1\right)=x^2+2\)

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

=>x+2=2

=>x=0(nhận)

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: \(S=\left\{-\dfrac{21}{8}\right\}\)

Xl nhưng câu b) mik ghi sai đề bại ạ

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

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

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