Question

 giải pt : \(x^2-x-2\sqrt{1+16x}=2\)

Answer 1

<=> \(2\frac{\left(x^2+8x+16-1-16x\right)}{x+4+\sqrt{1+16x}}+\left(x-5\right)\left(x+2\right)=0\)

<=> \(2\frac{\left(x^2-8x+15\right)}{x+4+\sqrt{1+16x}}+\left(x-5\right)\left(x+2\right)=0\)

<=> \(2\frac{\left(x-5\right)\left(x-3\right)}{x+4+\sqrt{1+16x}}+\left(x-5\right)\left(x+2\right)=0\)

<=> \(\left(x-5\right)\left(\frac{2x-6}{x+4+\sqrt{1+16x}}+x+2\right)=0\)

<=> \(x=5\)( vì theo đk nên cái ngoặc thứ 2 khác 0) 

vậy x=5

Answer 2

đkxđ: \(x\ge-\frac{1}{16}\)

pt<=> \(2\left(x+4\right)-2\sqrt{1+16x}+x^2-3x-10\)=0