Giari các phương trình sau.
a. \(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
b. \(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\)
c. \(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
d. \(\frac{3}{x+2}-\frac{2}{x-3}=\frac{8}{\left(x-3\right)\left(x+2\right)}\)
e. \(\frac{x}{2x+6}-\frac{x}{2x+2}=\frac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
f. \(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\)
g. \(\frac{5}{x+7}+\frac{8}{2x+14}=\frac{3}{2}\)
h. \(\frac{x-1}{x}-\frac{1}{x+1}=\frac{2x-1}{x^2+x}\)
a)
ĐKXĐ: \(x\neq 0; x\neq -10\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
\(\Leftrightarrow \frac{x+10+x}{x(x+10)}=\frac{1}{12}\)
\(\Leftrightarrow \frac{2x+10}{x(x+10)}=\frac{1}{12}\)
\(\Rightarrow 12(2x+10)=x(x+10)\)
\(\Leftrightarrow x^2-14x-120=0\)
\(\Leftrightarrow (x+6)(x-20)=0\Rightarrow \left[\begin{matrix} x=-6\\ x=20\end{matrix}\right.\) (đều thỏa mãn)
b)
ĐKXĐ: \(x\neq 0; x\neq 3\)
PT\(\Leftrightarrow \frac{(x+3).x-(x-3)}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Leftrightarrow \frac{x^2+2x+3}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Rightarrow x^2+2x+3=3\)
\(\Leftrightarrow x^2+2x=0\Leftrightarrow x(x+2)=0\Rightarrow \left[\begin{matrix} x=0\\ x=-2\end{matrix}\right.\) . Kết hợp với đkxđ suy ra $x=-2$
c)
ĐKXĐ: \(x\neq \pm 2\)
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{3(x-2)-2(x+2)}{(x+2)(x-2)}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-10}{x^2-4}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-2}{x^2-4}=0\Leftrightarrow \frac{1}{x+2}=0\) (vô lý)
Vậy pt vô nghiệm.
d)
ĐKXĐ: \(x\neq -2; x\neq 3\)
PT \(\Leftrightarrow \frac{3(x-3)-2(x+2)}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Leftrightarrow \frac{x-13}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Rightarrow x-13=8\Rightarrow x=21\) (thỏa mãn)
Vậy..........
e)
ĐKXĐ: \(x\neq -1; x\neq -3\)
PT \(\Leftrightarrow \frac{x(2x+2)-x(2x+6)}{(2x+6)(2x+2)}=\frac{3x+2}{(x+1)(x+3)}\)
\(\Leftrightarrow \frac{-4x}{2(x+3).2(x+1)}=\frac{3x+2}{(x+1)(x+3)}\)
\(\Leftrightarrow \frac{-x}{(x+3)(x+1)}=\frac{3x+2}{(x+1)(x+3)}\)
\(\Rightarrow -x=3x+2\Rightarrow x=-\frac{1}{2}\) (thỏa mãn)
Vậy..............
f)
ĐKXĐ: \(x\neq \pm 1\)
PT \(\Leftrightarrow \frac{x}{x+1}+\frac{2x-3}{x-1}=\frac{3x^2+5}{x^2-1}\)
\(\Leftrightarrow \frac{x(x-1)+(2x-3)(x+1)}{(x-1)(x+1)}=\frac{3x^2+5}{x^2-1}\)
\(\Leftrightarrow \frac{3x^2-2x-3}{x^2-1}=\frac{3x^2+5}{x^2-1}\)
\(\Rightarrow 3x^2-2x-3=3x^2+5\)
\(\Leftrightarrow x=-4\) (thỏa mãn)
Vậy.........
g)
ĐKXĐ: \(x\neq -7\)
PT \(\Leftrightarrow \frac{5}{x+7}+\frac{8}{2(x+7)}=\frac{3}{2}\)
\(\Leftrightarrow \frac{5}{x+7}+\frac{4}{x+7}=\frac{3}{2}\)
\(\Leftrightarrow \frac{9}{x+7}=\frac{3}{2}\)
\(\Rightarrow 3(x+7)=18\Leftrightarrow x+7=6\Leftrightarrow x=-1\) (thỏa mãn)
Vậy..........
h)
\(x\neq 0; x\neq -1\)
PT \(\Leftrightarrow \frac{(x-1)(x+1)-x}{x(x+1)}=\frac{2x-1}{x^2+x}\)
\(\Leftrightarrow \frac{x^2-1-x}{x^2+x}=\frac{2x-1}{x^2+x}\)
\(\Rightarrow x^2-1-x=2x-1\)
\(\Leftrightarrow x^2-3x=0\Rightarrow \left[\begin{matrix} x=0\\ x=3\end{matrix}\right.\). Kết hợp với đkxđ suy ra $x=3$
