Question

Giai phuong trinh: x²  - 10x + 27= \(\sqrt{6-x}+\sqrt{x-4}\)

Answer 1

ĐKXĐ : \(4\le x\le6\)

Xét \(VP^2=6-x+x-4+2\sqrt{\left(6-x\right)\left(x-4\right)}=2+2\sqrt{\left(6-x\right)\left(x-4\right)}\)

Áp dụng bđt Cauchy ta có : \(2+2\sqrt{\left(6-x\right)\left(x-4\right)}\le2+6-x+x-4=4\)

\(\Rightarrow VP\le2\forall x\)(1)

Xét \(VT=x^2-10x+27=\left(x^2-10x+25\right)+2=\left(x-5\right)^2+2\ge2\forall x\)(2)

Từ (1);(2) \(\Rightarrow VT\ge2\ge VP\)

Dấu "=" xảy ra \(\hept{\begin{cases}6-x=x-4\\\left(x-5\right)^2=0\end{cases}\Rightarrow x=5\left(TMĐKXĐ\right)}\)

Vậy nghiệm pt là x = 5