Question

Giải phương trình:

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

Answer 1

đk: x ≥ -1/3

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

\(\Leftrightarrow\frac{x\left(x^2-x+1-1\right)}{\sqrt{x^2-x+1}+1}+\frac{2\left(3x+1-4\right)}{\sqrt{3x+1}+2}-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\frac{x^2\left(x-1\right)}{\sqrt{x^2-x+1}+1}+\frac{6\left(x-1\right)}{\sqrt{3x+1}+2}-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\frac{x^2}{\sqrt{x^2-x+1}+1}+\frac{6}{\sqrt{3x+1}+2}-x-1\right]=0\)

.....

Answer 2

Lời giải:
ĐKXĐ: $x\geq \frac{-1}{3}$

Áp dụng BĐT AM-GM:
$x\sqrt{x^2-x+1}\leq \sqrt{x^2(x^2-x+1)}\leq \frac{x^2+(x^2-x+1)}{2}(1)$

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

Lấy $(1)+(2)$ suy ra:

$x\sqrt{x^2-x+1}+2\sqrt{3x+1}\leq x^2+x+3$

Dấu "=" xảy ra khi \(\left\{\begin{matrix} x=\sqrt{x^2}\\ x^2=x^2-x+1\\ 4=3x+1\end{matrix}\right.\) hay $x=1$ (thỏa mãn)

Vậy..........