Question

a)\(\frac{5}{x+7}+\frac{8}{2x+14}=\frac{3}{2}\)

b)\(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\)

c)\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)

d)\(\frac{90}{x}-\frac{36}{x-36}=2\)

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

Answer 1

a, \(\frac{5}{x+7}+\frac{8}{2x+14}=\frac{3}{2}\) Đkxđ : \(x\ne-7\)

⇔ \(\frac{5}{x+7}+\frac{8}{2\left(x+7\right)}=\frac{3}{2}\)

⇔ \(\frac{10}{2\left(x+7\right)}+\frac{8}{2\left(x+7\right)}=\frac{3\left(x+7\right)}{2\left(x+7\right)}\)

⇒ \(10+8=3\left(x+7\right)\)

⇔ \(10+8=3x+21\)

⇔ \(-3x=21-10-8\)

⇔ \(-3x=3\)

⇔ \(x=-1\) ( tm )

Ptr có tập nhiệm : S \(=\left\{-1\right\}\)

b, \(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\) Đkxđ : \(x\ne3;x\ne0\)

⇔ \(\frac{x\left(x+3\right)}{x\left(x-3\right)}-\frac{1\left(x-3\right)}{x\left(x-3\right)}=\frac{3}{x\left(x-3\right)}\)

⇒ \(x\left(x-3\right)-1\left(x-3\right)=3\)

⇔ \(x^2-3x-x+3=3\)

⇔ \(x^2-4x=0\)

⇔ \(x\left(x-4\right)=0\)

⇔ \(\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=0\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

Ptr có tập nhiệm : S \(=\left\{4\right\}\)