\(\Rightarrow x^2-5x-36=0\Rightarrow x\left(x-9\right)+4\left(x-9\right)=0\Rightarrow\left(x-9\right)\left(x+4\right)=0\Rightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

\(\left[\left(2x-11\right):3+1\right].5=20\\\left(2x-11\right):3+1=20:5\\ \left(2x-11\right):3+1=4\\ \left(2x-11\right):3=4-1\\ \left(2x-11\right):3=3\\ 2x-11=3.3\\ 2x-11=9\\ 2x=11+9\\ 2x=20\\ x=20:2\\ x=10 \)

<=> (2X-11):3+1=20:5
<=> (2X-11):3=4-1
<=> 2X-11=3x3
<=> 2X-11=9
<=> 2X=9+11
<=>X=20:2
<=> X=10

`(s:48)-(s:50)=20`

`<=>s/48-s/50=20`

`<=>(25s)/1200-(24s)/1200=20`

`<=>s/1200=20`

`<=>s=24000`

Vậy `s=24000`

\(\dfrac{S}{48}-\dfrac{S}{50}=20\)

\(\Leftrightarrow\dfrac{50S-48S}{48\cdot50}=20\)

\(\Leftrightarrow2S=48\cdot50\cdot20=48000\)

\(\Leftrightarrow S=24000\)

đặt 

\(A=a+b+c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)

\(=>4A=4a+4b+4c+\dfrac{12}{a}+\dfrac{36}{2b}+\dfrac{16}{c}\)

\(=>4A=a+2b+3c+3a+\dfrac{12}{a}+2b+\dfrac{36}{2b}+c+\dfrac{16}{c}\)

áp dụng BDT AM-GM

\(=>\dfrac{12}{a}+3a\ge2\sqrt{12.3}=12\)

\(=>2b+\dfrac{36}{2b}\ge2\sqrt{36}=12\)

\(=>c+\dfrac{16}{c}\ge2\sqrt{16}=8\)

\(=>4A\ge20+12+12+8=52=>A\ge13\)

dấu"=" xảy ra<=>a=2,b=3,c=4

 Điên nhờ...

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!!!!!

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