1.
đkxđ : x khác 5 ; x khác 2
pttđ :
\(\dfrac{\left(2+x\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(2-x\right)\left(x-5\right)}{\left(2-x\right)\left(x-5\right)}-\dfrac{6\left(x-5\right)}{\left(2-x\right)\left(x-5\right)}=0\)
<=> \(\dfrac{2^2-x^2}{....}+\dfrac{\left(6-3x\right)\left(x-5\right)}{....}-\dfrac{6x-30}{...}=0\)
<=> \(\dfrac{4-x^2}{...}+\dfrac{6x-30-3x^2+15x}{...}-\dfrac{6x-30}{...}=0\)
<=> \(4-x^2+6x-30-3x^2+15x-6x+30=0\)
<=> \(-4x^2+15x+4=0\)
\(\Delta=\left(15\right)^2-4.-4.4=289>0\)
=> pt có 2 nghiệm pb
\(x_1=\dfrac{-15+\sqrt{289}}{-8};x_2=\dfrac{-15-\sqrt{289}}{-8}\)
<=> \(x_1=-\dfrac{1}{4}\left(tm\right);x_2=4\left(tm\right)\)
\(\dfrac{x+2}{x-5}\)+3=\(\dfrac{6}{2-x}\)
<=>(x+2) (2-x)+3(x-5) (2-x)=6(x-5)
<=>-26-4x2+21x=6x-30<=>26+4x2-21x+6x-30=0<=>-4+4x2-15x=0<=>4x2+x-16x-4=0<=>x(4x+1)-4(4x+1)=0<=>(4x+1) (x-4)=0<=>x=-\(\dfrac{1}{4}\),42.
\(\Leftrightarrow x^2\left(1-\dfrac{3x^2}{7}\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\1-\dfrac{3x^2}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\\dfrac{3x^2}{7}=1\Rightarrow3x^2=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x^2=\dfrac{7}{3}\Rightarrow x=\sqrt{\dfrac{7}{3}}\end{matrix}\right.\)
