a: Sửa đề: \(7x^2-6\sqrt{2}x+2=0\)
\(\text{Δ}=\left(-6\sqrt{2}\right)^2-4\cdot7\cdot2=72-56=16>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{6\sqrt{2}-4}{14}=\dfrac{3\sqrt{2}-2}{7}\\x_2=\dfrac{3\sqrt{2}+2}{7}\end{matrix}\right.\)
b: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{2x}{x-2}-\dfrac{3x+10}{x^2-4}=\dfrac{x}{x+2}\)
=>\(\dfrac{2x}{x-2}-\dfrac{3x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
=>\(2x\left(x+2\right)-3x-10=x\left(x-2\right)\)
=>\(2x^2+4x-3x-10-x^2+2x=0\)
=>\(x^2+3x-10=0\)
=>(x+5)(x-2)=0
=>\(\left[{}\begin{matrix}x=-5\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
