Question

[Nguyen Phuong Anh]

Giair các phương trình sau :

a) \(\frac{x+4}{x^2-3x+2}\)+\(\frac{x+1}{x^{2^{ }}-4x+3}\)=\(\frac{2x+5}{x^2-4x+3}\) e)\(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)

b)\(\frac{2}{x^{2^{ }}-4}\)-\(\frac{1}{x\left(x-2\right)}\)+\(\frac{x-4}{x\left(x+2\right)}\)=0 f)\(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+2\right)\)

c)\(\frac{4x}{x^{2^{ }}+4x+3}\)-1=6(\(\frac{1}{x+3}-\frac{1}{2x+2}\))

d)\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{7}{6x+30}\)

MONG CÁC BẠN CÓ THỂ GIÚP TỚ TRONG NGÀY HÔM NAY VÌ TỚ ĐANG CẦN GẤP !!THANKS YOU^-^

Answer 1

[Trần Quốc Khanh]

f/ ĐKXĐ: x khác 0

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

\(\Leftrightarrow2x^2+x+2+\frac{1}{x}=0\)

\(\Leftrightarrow x\left(2x+1+\frac{2}{x}+\frac{1}{x^2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\2x+1+\frac{2x+1}{x^2}=0\end{matrix}\right.\)

\(\Rightarrow\left(2x+1\right)\left(1+\frac{1}{x^2}\right)=0\Rightarrow x=-\frac{1}{2}\)( vì 1+1/x^2>0)

Answer 2

[Trần Quốc Khanh]

a/\(\Leftrightarrow\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{x+4}{\left(x-1\right)\left(x-2\right)}-\frac{x+4}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\left(x+4\right)\left(\frac{1}{\left(x-1\right)\left(x-2\right)}-\frac{1}{\left(x-1\right)\left(x-3\right)}\right)=0\)

\(\Rightarrow x=-4\)