Question

[Nguyễn Châu Mỹ Linh]

Giải các phương trình sau:

a) \(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)

b) \(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)

c) \(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+2\right)\)

d) \(\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)^2\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a) ĐKXĐ: \(x\notin\pm\frac{1}{3}\)

Ta có: \(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)

\(\Leftrightarrow\frac{12x+1}{2\left(3x-1\right)}-\frac{9x-5}{3x+1}=\frac{9\left(12x-4x^2-1\right)}{4\left(9x^2-1\right)}\)

\(\Leftrightarrow\frac{2\left(12x+1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\frac{4\left(9x-5\right)\left(3x-1\right)}{4\left(3x+1\right)\left(3x-1\right)}=\frac{9\left(12x-4x^2-1\right)}{4\left(3x+1\right)\left(3x-1\right)}\)

\(\Leftrightarrow72x^2+30x+2-\left(108x^2-96x+20\right)=108x-36x^2-9\)

\(\Leftrightarrow72x^2+30x+2-108x^2+96x-20-108x+36x^2+9=0\)

\(\Leftrightarrow18x-9=0\)

\(\Leftrightarrow9\left(2x-1\right)=0\)

mà 9≠0

nên 2x-1=0

⇔2x=1

hay \(x=\frac{1}{2}\)(tm)

Vậy: \(x=\frac{1}{2}\)

b)ĐKXĐ: x≠0

Ta có: \(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)

\(\Leftrightarrow x+\frac{1}{x}-x^2-\frac{1}{x^2}=0\)

\(\Leftrightarrow\frac{x^3}{x^2}+\frac{x}{x^2}-\frac{x^4}{x^2}-\frac{1}{x^2}=0\)

\(\Leftrightarrow x^3+x-x^4-1=0\)

\(\Leftrightarrow x^3\left(1-x\right)+\left(x-1\right)=0\)

\(\Leftrightarrow x^3\left(1-x\right)-\left(1-x\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(x^3-1\right)=0\)

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

\(\Leftrightarrow-\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0\)(1)

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

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)(2)

Từ (1) và (2) suy ra x-1=0

hay x=1(tm)

Vậy: x=1

c) ĐKXĐ: x≠0

Ta có: \(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+2\right)\)

\(\Leftrightarrow\frac{1}{x}+2-\left(\frac{1}{x}+2\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}+2\right)\left(2-x^2-2\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}+2\right)\cdot\left(-x^2\right)=0\)(3)

Ta có: 1≠0

x≠0

Do đó: \(\frac{1}{x}\ne0\)

\(\Leftrightarrow\frac{1}{x}+2\ne0\)(4)

Từ (3) và (4) suy ra x=0(ktm)

Vậy: x∈∅

d) ĐKXĐ: x≠0

Ta có: \(\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)^2\)

\(\Leftrightarrow\left(x+1+\frac{1}{x}\right)^2-\left(x-1-\frac{1}{x}\right)^2=0\)

\(\Leftrightarrow\left(x+1+\frac{1}{x}+x-1-\frac{1}{x}\right)\left(x+1+\frac{1}{x}-x+1+\frac{1}{x}\right)=0\)

\(\Leftrightarrow2x\cdot\left(2+\frac{2}{x}\right)=0\)

\(\Leftrightarrow4x\left(1+\frac{1}{x}\right)=0\)

mà 4≠0

và x≠0

nên \(1+\frac{1}{x}=0\)

\(\Leftrightarrow\frac{1}{x}=-1\)

hay x=-1(tm)

Vậy: x=-1