Question

Giúp mik câu 10 , 12 với nha

Answer 1

\(10,ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow x^2+8+x^2+3-2\sqrt{\left(x^2+8\right)\left(x^2+3\right)}=4x^2-4x+1\\ \Leftrightarrow-2\sqrt{x^4+11x^2+24}=2x^2-4x-10\\ \Leftrightarrow-\sqrt{x^4+11x^2+24}=x^2-2x-5\\ \Leftrightarrow x^4+11x^2+24=\left(x^2-2x-5\right)^2\\ \Leftrightarrow x^4+11x^2+24=x^4+4x^2+25-4x^3+20x-10x^2\\ \Leftrightarrow4x^3+17x^2-20x-1=0\\ \Leftrightarrow4x^3-4x^2+21x^2-21x+x-1=0\\ \Leftrightarrow\left(x-1\right)\left(4x^2+21x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\4x^2+21x+1=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-21+5\sqrt{17}}{8}\\x=\dfrac{-21-5\sqrt{17}}{8}\end{matrix}\right.\)