a) ĐKXĐ: \(x\ne0\)

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

\(\Leftrightarrow\dfrac{4}{12x}+\dfrac{6}{12x}=\dfrac{3x}{12x}\)

Suy ra: \(3x=10\)

\(\Leftrightarrow x=\dfrac{10}{3}\)(thỏa ĐK)

Vậy: \(S=\left\{\dfrac{10}{3}\right\}\)

b) ĐKXĐ: \(x\ne0\)

Ta có: \(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)

\(\Leftrightarrow\dfrac{3x}{8x^2}-\dfrac{4x}{8x^2}=\dfrac{8}{8x^2}\)

Suy ra: \(3x-4x=8\)

\(\Leftrightarrow-x=8\)

hay x=-8(thỏa ĐK)

Vậy: S={-8}

c)ĐKXĐ: \(x\ne0\)

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

\(\Leftrightarrow\dfrac{2x}{4x^2}+\dfrac{3x}{4x^2}=\dfrac{10}{4x^2}\)

Suy ra: 2x+3x=10

\(\Leftrightarrow5x=10\)

hay x=2(thỏa ĐK)

Vậy: S={2}

d, \(\dfrac{2a}{x+a}=1\) (x \(\ne\) -a)

\(\Leftrightarrow\) \(\dfrac{2a}{x+a}-\dfrac{x+a}{x+a}=0\)

\(\Leftrightarrow\) \(\dfrac{a-x}{x+a}=0\)

\(\Leftrightarrow\) a - x = 0 (x + a \(\ne\) 0)

\(\Leftrightarrow\) x = a (TM)

Vậy S = {a}

Chúc bn học tốt!

b: Đặt \(x^2-6x-2=a\)

Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)

=>(a+2)(a+7)=0

\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)

=>x(x-6)(x-1)(x-5)=0

hay \(x\in\left\{0;1;6;5\right\}\)

c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)

\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)

\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)

\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)

=>26x=-3

hay x=-3/26

\(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)

\(\Leftrightarrow\dfrac{8x^2}{3\left(1-2x\right)\left(1+2x\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{1+8x}{4\left(1+2x\right)}\)

\(\Leftrightarrow\dfrac{-32x^2}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x.4\left(1+2x\right)-\left(1+8x\right).3\left(2x-1\right)}{12\left(2x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow8x\left(1+2x\right)-\left(1+8x\right).3.\left(2x-1\right)=-32x^2\)

\(\Leftrightarrow8x+16x^2-6x+3-48x^2+24x+32x^2=0\)

\(\Leftrightarrow26x+3=0\)

\(\Leftrightarrow x=-\dfrac{3}{26}\)

Vậy:......

\(\left(\dfrac{2x-1}{2x+1}-\dfrac{2x+1}{2x-1}\right):\dfrac{8x}{8x-4}\)

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

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

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

\(=\dfrac{-8x}{4x^2-1}:\dfrac{8x}{4\left(2x-1\right)}\)

\(=\dfrac{-8x.4\left(2x-1\right)}{8x\left(2x-1\right)\left(2x+1\right)}\)

\(=\dfrac{-8x.4}{8x\left(2x+1\right)}\)

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

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

4: \(\Leftrightarrow3^{x+4}\cdot\dfrac{1}{3}-4\cdot3^x=3^{16}\left(1-4\cdot3^3\right)\)
=>\(3^x\cdot27-4\cdot3^x=3^{16}\cdot\left(-107\right)\)

=>3^x*23=3^16*(-107)

=>\(x\in\varnothing\)

2: \(\Leftrightarrow2^x\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)=2^{10}\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)\)

=>2^x=2^10

=>x=10

3: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)

=>8^x=8^9

=>x=9

1: \(\Leftrightarrow3^x\cdot\left(4\cdot\dfrac{1}{9}+2\cdot3\right)=3^4\left(4+2\cdot3^3\right)\)

=>3^x=3^4*3^2

=>x=4+2=6

\(\dfrac{2x-1}{4x^2+2x+1}+\dfrac{2}{2x-1}=\dfrac{-8x-2}{\left(2x-1\right)\left(4x^2+2x+1\right)}\)

\(\Leftrightarrow4x^2-4x+1+8x^2+4x+2=-8x-2\)

\(\Leftrightarrow12x^2+3+8x+2=0\)

\(\Leftrightarrow12x^2+8x+5=0\)

\(\text{Δ}=8^2-4\cdot12\cdot5=64-240< 0\)

DO đo: Phương trình vô nghiệm

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

a: \(=2\sqrt{3}-2+10+5\sqrt{3}+3+\sqrt{3}=8\sqrt{3}+11\)