a: ĐKXĐ: x<>-2/3

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

=>5(3x+2)=2x+1

=>15x+10=2x+1

=>13x=-9

=>\(x=-\frac{9}{13}\) (nhận)

b: ĐKXĐ: x∉{1;3}

\(\frac{2x^2-5x+2}{x-1}=\frac{2x^2+x+15}{x-3}\)

=>\(\left(2x^2-5x+2\right)\left(x-3\right)=\left(2x^2+x+15\right)\left(x-1\right)\)

=>\(2x^3-6x^2-5x^2+15x+2x-6=2x^3-2x^2+x^2-x+15x-15\)

=>\(-11x^2+17x-6=-x^2+14x-15\)

=>\(-10x^2+3x+9=0\)

=>\(10x^2-3x-9=0\)

=>\(x^2-\frac{3}{10}x-\frac{9}{10}=0\)

=>\(x^2-2\cdot x\cdot\frac{3}{20}+\frac{9}{400}-\frac{9}{400}-\frac{9}{10}=0\)

=>\(\left(x-\frac{3}{20}\right)^2=\frac{9}{400}+\frac{9}{10}=\frac{9}{400}+\frac{360}{400}=\frac{369}{400}\)

=>\(x-\frac{3}{20}=\pm\frac{3\sqrt{41}}{20}\)

=>\(\left[\begin{array}{l}x=\frac{3\sqrt{41}+3}{20}\left(nhận\right)\\ x=\frac{-3\sqrt{41}+3}{20}\left(nhận\right)\end{array}\right.\)

c: ĐKXĐ: x∉{3;-3}

\(\frac{2x+3}{x-3}-\frac{4}{x+3}=\frac{24}{x^2-9}+2\)

=>\(\frac{\left(2x+3\right)\left(x+3\right)-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{24+2\left(x^2-9\right)}{\left(x-3\right)\left(x+3\right)}\)

=>(2x+3)(x+3)-4(x-3)=\(24+2x^2-18\)

=>\(2x^2+6x+3x+9-4x+12=2x^2+6\)

=>5x+21=6

=>5x=-15

=>x=-3(loại)

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

a: ĐKXĐ: x∈R

\(\frac{5}{x^2-2x+2}-\frac{8}{x^2-2x+5}=3\)

=>\(\frac{5\left(x^2-2x+5\right)-8\left(x^2-2x+2\right)}{\left(x^2-2x+2\right)\left(x^2-2x+5\right)}=3\)

=>\(3\left(x^2-2x+2\right)\left(x^2-2x+5\right)=5x^2-10x+25-8x^2+16x-16=-3x^2+6x+9\)

=>\(3\left\lbrack\left(x^2-2x\right)^2+7\left(x^2-2x\right)+10\right\rbrack=-3\left(x^2-2x\right)+9\)

=>\(\left(x^2-2x\right)^2+7\left(x^2-2x\right)+10=-\left(x^2-2x\right)+3\)

=>\(\left(x^2-2x\right)^2+8\left(x^2-2x\right)+7=0\)

=>\(\left(x^2-2x+1\right)\left(x^2-2x+7\right)=0\)

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

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1(nhận)

b: ĐKXĐ: x<>0

\(\frac{x^2-4x+3}{2x}+\frac{x^2+12x+3}{x^2+3}=4\)

=>\(\frac{x^2+3}{2x}-2+\frac{12x}{x^2+3}+1=4\)

=>\(\frac{x^2+3}{2x}+\frac{12x}{x^2+3}=4+2-1=6-1=5\)

=>\(\frac{\left(x^2+3\right)^2+24x^2}{2x\left(x^2+3\right)}=5\)

=>\(\left(x^2+3\right)^2+24x^2-10x\left(x^2+3\right)=0\)

=>\(\left(x^2+3\right)^2-4x\left(x^2+3\right)-6x\left(x^2+3\right)+24x^2=0\)

=>\(\left(x^2+3\right)\left(x^2+3-4x\right)-6x\left(x^2+3-4x\right)=0\)

=>\(\left(x^2-6x+3\right)\left(x^2-4x+3\right)=0\)

TH1: \(x^2-6x+3=0\)

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

=>\(\left(x-3\right)^2=6\)

=>\(\left[\begin{array}{l}x-3=\sqrt6\\ x-3=-\sqrt6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt6+3\left(nhận\right)\\ x=-\sqrt6+3\left(nhận\right)\end{array}\right.\)

TH2: \(x^2-4x+3=0\)

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

=>(x-1)(x-3)=0

=>x=1(nhận) hoặc x=3(nhận)

Đặt t=x²-2x+3(t\(\ge\)2)

PTTT: \(\dfrac{1}{t-1}+\dfrac{1}{t}=\dfrac{9}{2\left(t+1\right)}\)

<=>2t²+2t+2t²-2=9t²-9

<=>5t²-2t-7=0

<=>(t+1)(5t-7)=0

Do t\(\ge\)2

=>t+1>0 5t-7>0

Vậy pt vô nghiệm

\(\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2-2x+3}=\dfrac{9}{2\left(x^2-2x+4\right)}\)

Đặt \(t=x^2-2x+2=\left(x-1\right)^2+1\ge1\)

Thì ta có:

\(PT\Leftrightarrow\dfrac{1}{t}+\dfrac{1}{t+1}=\dfrac{9}{2\left(t+2\right)}\)

\(\Leftrightarrow5t^2-t-4=0\)

\(\Leftrightarrow\left(5t^2-5t\right)+\left(4t-4\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(5t+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5t+4=0\\t-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{4}{5}\left(l\right)\\t=1\end{matrix}\right.\)

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

\(\Leftrightarrow x=1\)

Vậy PT có 1 nghiệm là x = 1

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}

=>4x-6(2x+1)=2x-3x

=>4x-12x-6+x=0

=>-7x=6

hay x=-6/7

\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{4x}{12}-\dfrac{6\left(2x+1\right)}{12}=\dfrac{2x}{12}-\dfrac{3x}{12}\)

\(\Leftrightarrow4x-6\left(2x+1\right)=2x-3x\)

\(\Leftrightarrow4x-12x-6=-x\)

\(\Leftrightarrow4x-12x-6+x=0\)

\(\Leftrightarrow-7x-6=0\)

\(\Leftrightarrow x=-\dfrac{6}{7}\)

ĐKXĐ: ...

\(\left(\dfrac{x-1}{x+2}\right)^2-4\left(\dfrac{x+2}{x-3}\right)^2+3\left(\dfrac{x-1}{x-3}\right)=0\)

Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)

\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)-\left(x+2\right)^2=0\\\left(x-\right)\left(x-3\right)+4\left(x+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

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

\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3