Question

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

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

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

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

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

\(\Rightarrow x=-4\)

Answer 3

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

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

Đặt \(x+\frac{1}{x}=y\) ta sẽ có

\(y=y^2-2\)

\(\Leftrightarrow y^2-y-2=0\)

\(\Leftrightarrow\left(y+1\right)\left(y-2\right)=0\Rightarrow\left[{}\begin{matrix}y=-1\\y=2\end{matrix}\right.\)

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

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

Answer 4

b/ĐKXĐ: x khác 0,+-2

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

\(\Leftrightarrow\frac{2x-x-2+x^2-6x+8}{x\left(x^2-4\right)}=0\)

\(\Rightarrow x^2-5x+6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=3\end{matrix}\right.\)

Vậy x=3

Answer 5

c/\(\Leftrightarrow\frac{4x}{\left(x+1\right)\left(x+3\right)}-1-\frac{6}{x+3}+\frac{6}{2\left(x+1\right)}=0\)

\(\Leftrightarrow\frac{-\left(x^2+3\right)}{\left(x+1\right)\left(x+3\right)}-\frac{6\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{3}{\left(x+1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{-x^2-3-6x-6+3}{\left(x+1\right)\left(x+3\right)}=0\)

\(\Rightarrow-x^2-6x-6=0\Leftrightarrow x^2+6x+6=0\)

\(\Leftrightarrow\left(x+3\right)^2-3=0\Leftrightarrow\left(x+3-\sqrt{3}\right)\left(x+3+\sqrt{3}\right)\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}-3\\x=-3-\sqrt{3}\end{matrix}\right.\)

Answer 6

câu c sai dấu tương đương thứ 2, sửa :

\(\frac{-x^2-3}{\left(x+1\right)\left(x+3\right)}-\frac{6\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{3\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{-x^2-3-6x-6+3x+9}{\left(x+1\right)\left(x+3\right)}=0\)

\(\Rightarrow-x^2-3x=0\Rightarrow-x\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=0\end{matrix}\right.\)

Vậy x=0.ĐKXĐ: x khác 3

Answer 7

ĐKXĐ khác +-5,-6

\(\Leftrightarrow\frac{3}{4\left(x-5\right)}-\frac{15}{\left(x-5\right)\left(x+5\right)}-\frac{7}{6\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{9\left(x+5\right)-15.12-14\left(x-5\right)}{12\left(x^2-25\right)}=0\)

\(\Rightarrow-5x-65=0\) \(\Leftrightarrow x+13=0\) vậy x=-13(ghi thừa)