Question

[Lục Thiên Nguyên]

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

Answer 1

[Akai Haruma]

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 $x=-2$

Answer 2

[Akai Haruma]

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..........