\(\sqrt{x^2-x+1}+\sqrt{x^2-9x+9}=2x\)

=>\(\sqrt{x^2-x+1}-x+\sqrt{x^2-9x+9}-x=0\)

=>\(\dfrac{x^2-x+1-x^2}{\sqrt{x^2-x+1}+x}+\dfrac{x^2-9x+9-x^2}{\sqrt{x^2-9x+9}+x}=0\)

=>\(\left(-x+1\right)\left(\dfrac{1}{\sqrt{x^2-x+1}+x}+\dfrac{9}{\sqrt{x^2-9x+9}+x}\right)=0\)

=>-x+1=0

=>x=1

a) dat x-1=a

x=a+1

\(a+1+\sqrt{5+\sqrt{a}}=6\)

\(5-a=\sqrt{5+\sqrt{a}}\)

\(25-10a+a^2=5+\sqrt{a}\)

\(20-10a+a^2-\sqrt{a}=0\)

(a - \sqrt{5} - 5) (a + \sqrt{a} - 4) = 0

đúng nhưng b,c,d đâu

ý c)  dk tu viet

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

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

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

\(2x+2=4\)

2x=2

x=1

Lời giải:

ĐK: \(x\geq \frac{-1}{3}\)

Đặt \((\sqrt{x^2-x+1}, \sqrt{3x+1})=(a,b)\)

\(\Rightarrow a^2+b^2=x^2+2x+2\)

PT đã cho trở thành:

\(2xa+4b=a^2+b^2+x^2+4\)

\(\Leftrightarrow a^2+b^2+x^2+4-2xa-4b=0\)

\(\Leftrightarrow (a-x)^2+(b-2)^2=0\)

Điều này xảy ra khi \(\left\{\begin{matrix} (a-x)^2=0\\ (b-2)^2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \sqrt{x^2-x+1}=x\\ \sqrt{3x+1}=2\end{matrix}\right.\)

\(\Rightarrow x=1\)

Thử lại thấy thỏa mãn.

ĐKXĐ: \(x\ge-\frac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}x+1=a>0\\\sqrt{2x+1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a+b=\sqrt{3a^2+b^2}\)

\(\Leftrightarrow a^2+2ab+b^2=3a^2+b^2\)

\(\Leftrightarrow a^2-ab=0\Rightarrow\left[{}\begin{matrix}a=0\left(l\right)\\a=b\end{matrix}\right.\)

\(\Leftrightarrow x+1=\sqrt{2x+1}\)

\(\Leftrightarrow x^2+2x+1=2x+1\)

\(\Leftrightarrow x=0\)