a)\(4\sqrt{5x^3-6x^2+2}+4\sqrt{-10x^3+8x^2+7x-1}=13-x\)
\(\Leftrightarrow4\sqrt{5x^3-6x^2+2}+4\sqrt{-10x^3+8x^2+7x-1}+x-13=0\)
\(\Leftrightarrow4\sqrt{5x^3-6x^2+2}-4+4\sqrt{-10x^3+8x^2+7x-1}-8+x-13+12=0\)
\(\Leftrightarrow\dfrac{16\left(5x^3-6x^2+2\right)-16}{4\sqrt{5x^3-6x^2+2}+4}+\dfrac{16\left(-10x^3+8x^2+7x-1\right)-64}{4\sqrt{-10x^3+8x^2+7x-1}+8}+x-1=0\)
\(\Leftrightarrow\dfrac{16\left(x-1\right)\left(5x^2-x-1\right)}{4\sqrt{5x^3-6x^2+2}+4}+\dfrac{-16\left(x-1\right)\left(10x^2+2x-5\right)}{4\sqrt{-10x^3+8x^2+7x-1}+8}+x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{16\left(5x^2-x-1\right)}{4\sqrt{5x^3-6x^2+2}+4}+\dfrac{-16\left(10x^2+2x-5\right)}{4\sqrt{-10x^3+8x^2+7x-1}+8}+1\right)=0\)
Dễ thấy: \(\dfrac{16\left(5x^2-x-1\right)}{4\sqrt{5x^3-6x^2+2}+4}+\dfrac{-16\left(10x^2+2x-5\right)}{4\sqrt{-10x^3+8x^2+7x-1}+8}+1>0\) (loại)
Nên \(x-1=0\Rightarrow x=1\)
Câu b bạn nhé. Đến đây xét 2 trường hợp thế y theo x rồi thay vào pt* để giải tiếp ha
