Question

[Kaijo]

1, giải pt sau

a,\(\frac{9}{x}+2=-6\)

b,\(\frac{7}{x+1}+\frac{-18x}{x\left(x^2+4x+3\right)}=\frac{-4}{x+3}\)

c,\(\frac{3x-1}{x-1}-1=\frac{2x+5}{x+3}+\frac{4}{x^2+2x-3}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a) ĐKXĐ: x≠0

Ta có: \(\frac{9}{x}+2=-6\)

⇔\(\frac{9}{x}+2+6=0\)

⇔\(\frac{9}{x}+8=0\)

⇔\(\frac{9}{x}+\frac{8x}{x}=0\)

⇔9+8x=0

⇔8x=-9

hay \(x=-\frac{9}{8}\)

Vậy: \(x=-\frac{9}{8}\)

b) ĐKXĐ: x≠0;x≠-1;x≠-3

Ta có: \(\frac{7}{x+1}+\frac{-18x}{x\left(x^2+4x+3\right)}=\frac{-4}{x+3}\)

⇔\(\frac{7}{x+1}+\frac{-18x}{x\left(x+1\right)\left(x+3\right)}-\frac{-4}{x+3}=0\)

⇔\(\frac{7x\left(x+3\right)}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}+\frac{-18x}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}-\frac{-4x\left(x+1\right)}{\left(x+3\right)\cdot x\cdot\left(x+1\right)}=0\)

⇔\(7x^2+21x-18x+4x\left(x+1\right)=0\)

\(\Leftrightarrow7x^2+21x-18x+4x^2+4x=0\)

⇔\(11x^2+7x=0\)

\(\Leftrightarrow x\left(11x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\11x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\11x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\frac{-7}{11}\end{matrix}\right.\)

Vậy: \(x=\frac{-7}{11}\)

c) ĐKXĐ: x≠1; x≠-3

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

⇔\(\frac{3x-1}{x-1}-1-\frac{2x+5}{x+3}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)

⇔\(\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)

⇔\(\left(3x-1\right)\left(x+3\right)-\left(x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-4=0\)

\(\Leftrightarrow3x^2+9x-x-3-\left(x^2+3x-x-3\right)-\left(2x^2-2x+5x-5\right)-4=0\)

\(\Leftrightarrow3x^2+8x-3-\left(x^2+2x-3\right)-\left(2x^2+3x-5\right)-4=0\)

\(\Leftrightarrow3x^2+8x-3-x^2-2x+3-2x^2-3x+5-4=0\)

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow3x=-1\)

hay \(x=\frac{-1}{3}\)

Vậy: \(x=\frac{-1}{3}\)