số bé là:

   (1995-5)/(200-1)=10

số lớn là:

    10+1995=2005

           Đáp số: 10 và 2005

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)

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.

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

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

$\Leftrightarrow x=-7$ 

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

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