ĐK: x>= -1/3

Ta có: \(pt\Leftrightarrow2x\sqrt{x^2-x+1}+4\sqrt{3x+1}=2x^2+2x+6\)

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

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

Mà : \(\left(x-\sqrt{x^2-x+1}\right)^2\ge0;\left(\sqrt{3x+1}-2\right)^2\ge0\)

Khi đó: \(\left(x-\sqrt{x^2-x+1}\right)^2+\left(\sqrt{3x+1}-2\right)^2\ge0\)

Dấu "=" xảy ra khi và chỉ khi: 

\(\hept{\begin{cases}\left(x-\sqrt{x^2-x+1}\right)^2=0\\\left(\sqrt{3x+1}-2\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2=x^2-x+1,x\ge0\\3x+1=4\end{cases}}\Leftrightarrow x=1\)tm đk

Vậy x=1

Ta có thể dùng cô si chăng?

ĐK: \(x\ge-\frac{1}{3}\)

\(VT=\sqrt{x^2\left(x^2-x+1\right)}+\sqrt{4\left(3x+1\right)}\)

\(\le\frac{x^2+x^2-x+1}{2}+\frac{4+3x+1}{2}=\frac{2x^2+2x+6}{2}=x^2+x+3=VP\)

Để đẳng thức xảy ra, tức là xảy ra đẳng thức ở phương trình thì:

\(\hept{\begin{cases}x^2=x^2-x+1\\4=3x+1\end{cases}}\Leftrightarrow x=1\)

Vậy...

Is it true??

bình phương 2 vế lên

\(\sqrt{2x-1}+\sqrt{3x-1}=\sqrt{5x-2}\)

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

\(\Leftrightarrow2x-1+2\sqrt{\left(2x-1\right)\left(3x-1\right)}+3x-1=5x-2\)

\(\Leftrightarrow5x-2+2\sqrt{\left(2x-1\right)\left(3x-1\right)}=5x-2\)

\(\Leftrightarrow2\sqrt{\left(2x-1\right)\left(3x-1\right)}=0\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(3x-1\right)}=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\3x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)

a, đặt \(\sqrt{x+5}=t\Rightarrow\)\(t^2-5=x\) ta có pt \(\left(t^2-5\right)^2-4\left(t^2-5\right)-3=t\)

Giải ra t=2 thay vào x=-1

b, đăt \(x=a,\sqrt{x^2+1}=b\)ta có pt

\(b^2+3a=\left(a+3\right)b\)

\(b^2-ab+3a-3b=b\left(b-a\right)+3\left(a-b\right)\)

\(=\left(b-3\right)\left(b-a\right)=0\)

\(TH:b=3,a=b\)\(\sqrt{x^2+1}=3\Rightarrow x^2+1=9\Rightarrow x=\mp2\sqrt{2}\)

\(x=\sqrt{x^2+1}\Rightarrow x^2=x^2+1\left(L\right)\)

3,