Tìm x,biết:
a) \(\dfrac{2x+1}{x^2-2x+1}-\dfrac{2x+3}{x^2-1}=0\) ; b) \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}=0\)
Trần Thiên KimNguyễn Huy TúPhương An
GIÚP EM VỚI Ạ
a) \(\dfrac{2x+1}{x^2-2x+1}-\dfrac{2x+3}{x^2-1}=0\Leftrightarrow\dfrac{2x+1}{x^2-2x+1}=\dfrac{2x+3}{x^2-1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2-1\right)=\left(2x+3\right)\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1\right)=\left(2x+3\right)\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=\left(2x+3\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x+x+1=2x^2-2x+3x-3\)
\(\Leftrightarrow2x^2+2x+x+1-2x^2+2x-3x+3=0\)
\(\Leftrightarrow2x+4=0\Leftrightarrow2x=-4\Leftrightarrow x=\dfrac{-4}{2}=-2\) vậy \(x=-2\)
\(a.\)
\(\dfrac{2x+1}{x^2-2x+1}-\dfrac{2x+3}{x^2-1}=0\)
\(\Leftrightarrow\dfrac{2x+1}{x^2-2x+1}=\dfrac{2x+3}{x^2-1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2-1\right)=\left(2x+3\right)\left(x^2-2x+1\right)\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)\left(x-1\right)=\left(2x+3\right)\left(x-1\right)^2\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)=\left(2x+3\right)\left(x-1\right)\)
\(\Rightarrow2x^2+2x+x+1=2x^2-2x+3x-3\)
\(\Rightarrow2x^2+2x+x+1-2x^2+2x-3x+3=0\)
\(\Rightarrow2x+4=0\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
a. ĐKXĐ: \(x\ne\pm1\)
\(\dfrac{2x+1}{x^2-2x+1}-\dfrac{2x+3}{x^2-1}=0\Leftrightarrow\dfrac{\left(2x+1\right)\left(x+1\right)-\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}=0\Leftrightarrow\dfrac{2x^2+3x+1-2x^2-x+3}{\left(x-1\right)^2\left(x+1\right)}=0\Leftrightarrow\dfrac{2x+4}{\left(x-1\right)^2\left(x+1\right)}=0\Rightarrow2x+4=0\Leftrightarrow x=-2\left(tmđk\right)\)
Vậy nghiệm của pt là x = -2
b. ĐKSĐ: x \(\ne\pm3\)
\(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}=0\Leftrightarrow\dfrac{3\left(x+3\right)+6x+x\left(x-3\right)}{x^2-9}=0\Leftrightarrow\dfrac{x^2+6x+9}{x^2-9}=0\Leftrightarrow\dfrac{\left(x+3\right)^2}{x^2-9}=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\) (ktmđk)
Vậy pt vô nghiệm
b) điều kiện \(x\ne\pm3\)
\(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}=0\Leftrightarrow\dfrac{3}{x-3}+\dfrac{6}{x^2-9}+\dfrac{x}{x+3}=0\)
\(\Leftrightarrow\dfrac{3}{x-3}+\dfrac{6}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{x+3}=0\Leftrightarrow\dfrac{3\left(x+3\right)+6+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{3x+9+6+x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\Leftrightarrow\dfrac{x^2+15}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow x^2+15=0\)
ta có \(x^2\ge0\) với mọi \(x\) \(\Leftrightarrow x^2+15\ge15>0\) với mọi \(x\)
vậy phương trình vô nghiệm
