Question

1. \( x^4 - 11x^2 - 8 = 13\sqrt{x^2 + 1} \).

2. \( \sqrt{x^2 - 2x + 5} + \sqrt{x - 1} = 2 \).

3. \( \frac{4x^2 - 2x - 20}{2x - 11} = \sqrt{2x + 3} \).

4. \( x^3 + 3x^2 + 4x + 2 = (3x + 2)\sqrt{3x + 1} \).

5. \( x + \sqrt{5 + \sqrt{x - 1}} = 6 \).

Answer 1

2: ĐKXĐ: x>=1

Ta có: \(\sqrt{x^2-2x+5}+\sqrt{x-1}=2\)

=>\(\sqrt{x^2-2x+5}-2+\sqrt{x-1}=0\)

=>\(\frac{x^2-2x+5-4}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0\)

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

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

=>\(\sqrt{x-1}=0\)

=>x-1=0

=>x=1(nhận)

5: ĐKXĐ: x>=1

Ta có: \(x+\sqrt{5+\sqrt{x-1}}=6\)

=>\(x-1+\sqrt{5+\sqrt{x-1}}=5\) (1)

Đặt \(\sqrt{x-1}=a\)

Phương trình (1) sẽ trở thành: \(a^2+\sqrt{5+a}=5\)

=>\(a^2+\sqrt{5+a}=5+a-a\)

=>\(a^2+\sqrt{5+a}-\left(\sqrt{5+a}\right)^2+a=0\)

=>\(\left(a-\sqrt{a+5}\right)\left(a+\sqrt{a+5}\right)+\left(\sqrt{5+a}+a\right)=0\)

=>\(\left(a+\sqrt{a+5}\right)\left(a-\sqrt{a+5}+1\right)=0\)

=>\(a+1-\sqrt{a+5}=0\)

=>\(\sqrt{a+5}=a+1\)

=>\(\begin{cases}a+5=\left(a+1\right)^2\\ a+1\ge0\end{cases}\Rightarrow\begin{cases}a^2+2a+1=a+5\\ a+1\ge0\end{cases}\)

=>\(\begin{cases}a^2+a-4=0\\ a\ge-1\end{cases}\Rightarrow a=\frac{1+\sqrt{17}}{2}\)

=>\(x-1=\left(\frac{1+\sqrt{17}}{2}\right)^2=\frac{18+2\sqrt{17}}{4}=\frac{9+\sqrt{17}}{2}\)

=>\(x=\frac{11+\sqrt{17}}{2}\) (nhận)