a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

ĐKXĐ: x+3>=0

=>x>=-3

\(x+\left(x+1\right)\sqrt{x+3}=5\)

=>\(x+\sqrt{\left(x+3\right)\left(x+1\right)^2}=5\)

=>\(x+\sqrt{\left(x+3\right)\left(x^2+2x+1\right)}=5\)

=>\(x+\sqrt{x^3+2x^2+x+3x^2+6x+3}=5\)

=>\(x+\sqrt{x^3+5x^2+7x+3}=5\)

=>\(x-1+\sqrt{x^3+5x^2+7x+3}-4=0\)

=>\(\left(x-1\right)+\dfrac{x^3+5x^2+7x+3-16}{\sqrt{x^3+5x^2+7x+3}+4}=0\)

=>\(\left(x-1\right)+\dfrac{x^3-x^2+6x^2-6x+13x-13}{\sqrt{x^3+5x^2+7x+3}+4}=0\)

=>\(\left(x-1\right)+\dfrac{\left(x-1\right)\left(x^2+6x+13\right)}{\sqrt{x^3+5x^2+7x+3}+4}=0\)

=>\(\left(x-1\right)\left(1+\dfrac{x^2+6x+13}{\sqrt{x^3+5x^2+7x+3}+4}\right)=0\)

=>x-1=0

=>x=1(nhận)

=>x^2+8x+15=x^2+6x+8

=>8x+15=6x+8

=>2x=-7

=>x=-7/2

Đề: 

`=> x^2 + 3x + 5x + 15 = 2x + 8 + x^2 + 4x`

`=> x^2 + 8x + 15 = x^2 + 6x + 8`

`=> x^2 - x^2 + 8x - 6x + 15 - 8 = 0`

`=> 2x + 7 = 0`

`=> 2x = -7`

`=> x = -7/2`

Vậy `x = -7/2`

\(\left(x+3\right)\left(x+5\right)=\left(x+4\right)\left(2+x\right)\)

\(\Leftrightarrow\left(x+3\right)\left(x+5\right)\left(4+x\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\\4+x=0\\x+2=0\end{matrix}\right.\)                 \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\\x=-4\\x=-2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{-3;-5;-4;-2\right\}\)

Với x < -5, ta có: x +5 < 0; x -2 < 0 => |x+5| = - x - 5; |x-2| = 2 - x

=> - x - 5 + 3. (2-x) = 14 + x => x = -2,6 ( ko thỏa mãn các giá trị x đang xét)

Với \(-5\le x< 2\), ta có: x + 5 \(\le\)0; x - 2 < 0 => |x+5| = x+5; |x-2| = 2-x

=> x+5 + 3.(2-x) = 14 + x=> x = -1 (thỏa mãn các giá trị x đang xét)

Với \(x\ge2\), ta có: x+ 5 > 0; x - 2 \(\ge\)0 => |x+5| = x+5; |x-2| = x-2

=> x+5 + 3.(x-2) = 14 + x => x = 5 (thỏa mãn các giá trị x đang xét)

Vậy phương trình đã cho có tập nghiệm là \(S=\left\{-1;5\right\}\)

nha.. Chúc bn hc tốt

mấy cái đoạn với hơi khó hiểu 1 chút

bạn có thể giúp giải rõ ràng hơn ko

Lời giải:

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

$\Leftrightarrow x^2+2x-15=x^2-x-12$

$\Leftrightarrow 3x=3\Rightarrow x=1$

\(\Leftrightarrow x-1-4=5\left(x-5\right)\)

=>x-5=5(x-5)

=>x-5-5x+25=0

=>-4x+20=0

hay x=5(loại)

`1/[x-5]-4/[(x-5)(x-1)]=5/[x-1]`       `ĐK: x \ne 5,x \ne 1`

`<=>[x-1-4]/[(x-5)(x-1)]=[5(x-5)]/[(x-5)(x-1)]`

   `=>x-5=5x-25`

`<=>4x=20`

`<=>x=5` (ko t/m)

Vậy ptr vô nghiệm

lap bang ra roi xet tung truong hop mot

a. ĐKXĐ \(x\ge2\)

\(\sqrt{x+3}-3+\sqrt{x-2}-2=0\)

\(\Leftrightarrow\dfrac{x-6}{\sqrt{x+3}+3}+\dfrac{x-6}{\sqrt{x-2}+2}=0\)

\(\Leftrightarrow\left(x-6\right)\left(\dfrac{1}{\sqrt{x+3}+3}+\dfrac{1}{\sqrt{x-2}+2}\right)=0\)

\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x^2-x-1=\left(1-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^2-x-1=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x=2\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow\) Pt vô nghiệm 

\(a.\sqrt{x+3}=5-\sqrt{x-2}\)

\(\sqrt{x+3}+\sqrt{x-2}=5\)

\(\sqrt{\left(x+3\right)^2}+\sqrt{\left(x-2\right)^2}=5^2\)

\(x+3+x-2=25\)

\(2x+1=25\)

\(x=12\)

\(b.\sqrt{x^2-x-1}=1-x\)

\(\sqrt{\left(x^2-x-1\right)^2}=\left(1-x\right)^2\)

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

\(x^2-x-1-1+2x-x^2=0\)

\(x-2=0\)

\(x=2\)

\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)

=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)

=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)

=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)

=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)

=> x =  - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)

Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)

\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)

\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)

\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)

mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)

nên x+100=0

hay x=-100

Vậy: S={-100}