Question

a) x(x+1)-(x+2)(x-3)=7

b) \(\frac{x-3}{x+1}=\frac{x^2}{x^2-1}\)

c) \(\frac{1}{x+2}-\frac{4}{x-3}=\frac{10}{\left(x+2\right)\left(x-3\right)}\)

d) \(\frac{1}{x-2}+\frac{x-3}{x+2}=1\)

Answer 1

a)

\(x(x+1)-(x+2)(x-3)=7\)

\(\Leftrightarrow (x^2+x)-(x^2-x-6)=7\)

\(\Leftrightarrow 2x+6=7\Leftrightarrow 2x=1\Leftrightarrow x=\frac{1}{2}\)

b) ĐK: \(x\neq \pm 1\)

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

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

\(\Rightarrow x^2-4x+3=x^2\)

\(\Leftrightarrow -4x+3=0\Leftrightarrow x=\frac{3}{4}\) (t/m)

c) ĐK: \(x\neq -2; x\neq 3\)

\(\frac{1}{x+2}-\frac{4}{x-3}=\frac{10}{(x+2)(x-3)}\)

\(\Leftrightarrow \frac{x-3-4(x+2)}{(x+2)(x-3)}=\frac{10}{(x+2)(x-3)}\)

\(\Rightarrow x-3-4(x+2)=10\)

\(\Leftrightarrow -3x-21=0\Leftrightarrow x=-7\) (t/m)

d) ĐK: \(x\neq \pm 2\)

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

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

\(\Rightarrow x+2+(x-3)(x-2)=(x-2)(x+2)\)

\(\Leftrightarrow x^2-4x+8=x^2-4\)

\(\Leftrightarrow -4x=-12\Leftrightarrow x=3\) (t.m)