Question

[Julian Edward]

giải các phương trình chứa ẩn ở mẫu sau đây dạng \(\frac{a}{q\left(x\right)+m}-\frac{b}{q\left(x\right)+n}=c\)

a) \(\frac{4}{x^2-x-2}+\frac{2}{x^2-x-1}=5\)

b) \(\frac{4}{\sqrt{x-2}+1}-\frac{1}{\sqrt{x-2}}=1\)

c) \(\frac{4}{\sqrt{x}+1}-\frac{1}{2-3\sqrt{x}}=3\)

Mn giup e vs ah, thenk kiu :3333

Answer 1

[Nguyễn Việt Lâm]

a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\x\ne2\\x\ne\frac{1\pm\sqrt{5}}{2}\end{matrix}\right.\)

Đặt \(x^2-x-1=a\) ta được:

\(\frac{4}{a-1}+\frac{2}{a}=5\Leftrightarrow4a+2\left(a-1\right)=5a\left(a-1\right)\)

\(\Leftrightarrow5a^2-11a+2=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x-1=2\\x^2-x-1=\frac{1}{5}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-3=0\\5x^2-5x-6=0\end{matrix}\right.\) (bấm máy)

b/ ĐKXĐ: \(x>2\)

Đặt \(\sqrt{x-2}=a>0\)

\(\frac{4}{a+1}-\frac{1}{a}=1\Leftrightarrow4a-\left(a+1\right)=a\left(a+1\right)\)

\(\Leftrightarrow a^2-2a+1=0\Rightarrow a=1\)

\(\Rightarrow\sqrt{x-2}=1\Rightarrow x=3\)

c/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\frac{4}{9}\end{matrix}\right.\)

\(\Leftrightarrow4\left(2-3\sqrt{x}\right)-\left(\sqrt{x}+1\right)=3\left(\sqrt{x}+1\right)\left(2-3\sqrt{x}\right)\)

\(\Leftrightarrow9x-10\sqrt{x}+1=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=\frac{1}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{81}\end{matrix}\right.\)

Answer 2

[Julian Edward]

Cảm ơn bn nhiều :>