Question

Giải các phương trinhg sau:

a) \(x+1=\sqrt{2\left(x+1\right)+2\sqrt{2\left(x+1\right)+2\sqrt{4\left(x+1\right)}}}\)

b) \(\sqrt{x+2\sqrt{x+1}+2}+\sqrt{x-2\sqrt{x+1}+2}=\frac{x+5}{2}\)

c) \(\sqrt{x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}}=x+\frac{3}{4}\)

Answer 1

a) \(x+1=\sqrt{2\left(x+1\right)+2\sqrt{2\left(x+1\right)+2\sqrt{4\left(x+1\right)}}}\)

<=> \(\left(x+1\right)^2=\left[\sqrt{2\left(x+1\right)+2\sqrt{2\left(x+1\right)+2\sqrt{4\left(x+1\right)}}}\right]^2\)

<=> \(x^2+2x+1=2x+2+2\sqrt{2x+2+4\sqrt{x+1}}\)

<=> \(x^2+1=2x+2+2\sqrt{2x+2+4\sqrt{x+1}}-2x\)

<=> \(x^2+1=2\sqrt{2x+2+4\sqrt{x+1}}+2\)

<=> \(x^2+1-2=2\sqrt{2x+2+4\sqrt{x+1}}\)

<=> \(x^2-1=2\sqrt{2x+2+4\sqrt{x+1}}\)

<=> \(\left(x^2-1\right)^2=\left(2\sqrt{2x+2+4\sqrt{x+1}}\right)^2\)

<=> \(x^4-2x^2+1=8x+8+16\sqrt{x+1}\)

<=> \(x^4-2x^2+1-8x=16\sqrt{x+1}+8\)

<=> \(x^4-2x^2-8x-7=16\sqrt{x+1}\)

<=> \(\left(x^4-2x^2-8x-7\right)^2=\left(16\sqrt{x+1}\right)^2\)

<=> \(x^8-4x^6-16x^5-10x^4+32x^3+92x^2+112x+49=256x+256\)

<=> \(x^8-4x^6-16x^5-10x^4+32x^3+92x^2+112x-144x-207=0\)

<=> \(\left(x+1\right)\left(x-2\right)\left(x^6+2x^5+3x^4-4x^3-9x^2+2x+69\right)=0\)

<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Vì: \(x^6+2x^5+3x^4-4x^3-9x^2+2x+69\ne0\)

=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)