\(\Leftrightarrow\dfrac{2\left(x+2\right)+x-2-2x+5}{\left(x-2\right)\left(x+2\right)}=0\left(ĐK:x\ne\pm2\right)\)

\(\Leftrightarrow2x+4+x-2-2x+5=0:\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow x+7=0\)

\(\Leftrightarrow x=-7\left(n\right)\)

Vậy \(S=\left\{-7\right\}\)

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

\(=>\dfrac{2\left(x+2\right)+x-2-2x+5}{\left(x+2\right)\left(x-2\right)}=0\)    ( điều kiện x ≠ ± 2)

\(=>2x+4+x+\left(-2\right)-2x+5=\dfrac{0}{\left(x+2\right)\left(x-2\right)}\)

\(=>x+7=0=>x=-7\left(n\right)\)

\(\dfrac{2x+2}{3}-2-\dfrac{x-2}{2}< 0\)

\(\Leftrightarrow\dfrac{4x+4-12-3x+6}{6}< 0\Rightarrow x-2< 0\Leftrightarrow x< 2\)

 \(đkxđ:x\ne-1;x\ne2\)

\(\Leftrightarrow\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{1\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow2x-4-x-1=3x-11\\ \Leftrightarrow2x-4-x-1-3x+11=0\\ \Leftrightarrow-2x+6=0\\ \Leftrightarrow-2x=-6\\ \Leftrightarrow x=3\)

không có cái nào luôn á 

dạng ax+b=0??

Phương trình A nhé bạn

1: Sửa đề: 2/x+2

\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)

=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

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

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

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38

1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)

Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)

\(\Leftrightarrow-3x-12-3+5x-x+4=0\)

\(\Leftrightarrow x=11\left(nhận\right)\)

2. ĐKXĐ: $x\neq \pm 2$

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

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

\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)

\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)

Vậy pt vô nghiệm

3. ĐKXĐ: $x\neq \pm \frac{3}{2}$

PT \(\Leftrightarrow \frac{(x-5)(2x+3)-x(2x-3)}{(2x-3)(2x+3)}=\frac{1-6x}{(2x-3)(2x+3)}\)

\(\Rightarrow (x-5)(2x+3)-x(2x-3)=1-6x\)

\(\Leftrightarrow 2x^2-7x-15-2x^2+3x+6x-1=0\)

\(\Leftrightarrow 2x-16=0\Leftrightarrow x=8\) (thỏa mãn)

1: Ta có: \(\dfrac{3}{x+2}-\dfrac{x-1}{x^2-4}=\dfrac{2}{x-2}\)

Suy ra: \(3x-6-x+1=2x+4\)

\(\Leftrightarrow2x-5=2x+4\left(vôlý\right)\)

2: Ta có: \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)

Suy ra: \(\left(x-5\right)\left(2x+3\right)-x\left(2x-3\right)=1-6x\)

\(\Leftrightarrow2x^2-7x-15-2x^2+6x+6x-1=0\)

\(\Leftrightarrow5x=16\)

hay \(x=\dfrac{16}{5}\)

1.

ĐK: \(x\ne7;x\ne-1;x\ne3\)

\(\dfrac{2x-5}{x^2-6x-7}\le\dfrac{1}{x-3}\left(1\right)\)

TH1: \(x< -1\)

\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\ge x^2-6x-7\)

\(\Leftrightarrow2x^2-11x+15\ge x^2-6x-7\)

\(\Leftrightarrow x^2-5x+22\ge0\)

\(\Leftrightarrow\) Bất phương trình đúng với mọi \(x< -1\)

TH2: \(-1< x< 3\)

\(\left(1\right)\Leftrightarrow\left(3-x\right)\left(2x-5\right)\ge\left(7-x\right)\left(x+1\right)\)

\(\Leftrightarrow-2x^2+11x-15\ge-x^2+6x+7\)

\(\Leftrightarrow-x^2+5x-22\ge0\)

\(\Rightarrow\) vô nghiệm

TH3: \(3< x< 7\)

Khi đó \(\dfrac{2x-5}{x^2-6x-7}\le0\); \(\dfrac{1}{x-3}>0\)

\(\Rightarrow\) Bất phương trình đúng với mọi \(3< x< 7\)

TH4: \(x>7\)

\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\le x^2-6x-7\)

\(\Leftrightarrow2x^2-11x+15\le x^2-6x-7\)

\(\Leftrightarrow x^2-5x+22\le0\)

\(\Rightarrow\) vô nghiệm

Vậy ...

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