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}

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) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

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

\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

Đặt \(\left[{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\).

Ta có hệ: \(\left[{}\begin{matrix}a+b=\dfrac{1}{16}\\3a+6b=\dfrac{1}{4}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}a=\dfrac{1}{24}\\b=\dfrac{1}{48}\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=24\\y=48\end{matrix}\right.\)

Vậy `(x;y)=(24;48)`.

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)

ĐKXĐ: \(x\ne\left\{2;4\right\}\)

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

\(\Rightarrow\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right)\left(x-4\right)\)

\(\Leftrightarrow x^2-7x+12+x^2-4x+4=-x^2+6x-8\)

\(\Leftrightarrow3x^2-17x+24=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\) (thỏa mãn)

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

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

=>-7x=6

hay x=-6/7

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

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

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

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

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

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

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

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

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

=>-7x=6

hay x=-6/7

ĐKXĐ: \(x\ne\left\{2;4\right\}\)

Đặt \(\left\{{}\begin{matrix}\dfrac{x+1}{x-2}=a\\\dfrac{x-2}{x-4}=b\end{matrix}\right.\) \(\Rightarrow\dfrac{x+1}{x-4}=ab\)

Phương trình trở thành:

\(a^2-12b^2+ab=0\)

\(\Leftrightarrow a^2+4ab-3ab-12b^2=0\)

\(\Leftrightarrow a\left(a+4b\right)-3b\left(a+4b\right)=0\)

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

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

Bạn tự quy đồng và hoàn thành phần còn lại nhé

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

\(\Leftrightarrow\sqrt{x+1+\dfrac{1}{2}\sqrt{4x+3}}+x=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{\dfrac{1}{4}\left(4x+3\right)+2.\dfrac{1}{2}.\dfrac{1}{2}\sqrt{4x+3}+\dfrac{1}{4}}+x=\dfrac{1}{2}\)

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

\(\Leftrightarrow\dfrac{1}{2}\sqrt{4x+3}+\dfrac{1}{2}+x=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{4x+3}=-2x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\4x+3=4x^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\\left(2x-3\right)\left(2x+1\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x=-\dfrac{1}{2}\)

Vậy...