Question

giải các phương trình ẩn x sau:

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

b) \(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)

c)\(\dfrac{1}{2x}+\dfrac{3}{4x}=\dfrac{5}{2x^2}\)

d) \(\dfrac{2a}{x+a}=1\)

Answer 1

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}

Answer 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!