\(VT=\sqrt{5-x}+\sqrt{x+3}\le\sqrt{2\left(5-x+x+3\right)}=4\)
\(VP=3\left(x-1\right)^2+4\ge4\)
\(\Rightarrow VT\le VP\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}5-x=x+3\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow x=1\)
Vậy pt có nghiệm duy nhất \(x=1\)
gpt:\(\sqrt{3x^2+6x+4}+\sqrt{2x^2+4x+11}=\left(1-x\right)\left(x+3\right)\)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-x^2-2x\)
\(\sqrt{x^2-x+2}+\sqrt{x^2-3x+6}=2x\)
GPT: \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)
GPT: x3+7x2+6x+1=4x2+3x
GPT : \(\sqrt[3]{3x^2-x+2001}-\sqrt[3]{3x^2-7x+2002}-\sqrt[3]{6x-2003}=\sqrt[3]{2002}\)
Gpt: \(5\sqrt{x-1}-\sqrt{x+7}=3x-4\) (2 cách)
GPt: \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=-5-x^2+6x\)
GPT: \(x\sqrt{3-2x}=3x^2-6x+4\)
GPT: \(2\sqrt{x^2-3x-1}=3x^2-6x-8\)
GPT
\(x^2+6x+1=\left(2x+1\right)\sqrt{x^2+2x+3}\)3
\(\sqrt{3x-2}-\sqrt{x-1}=2x^2-x-3\)
GPT:
\(\sqrt[6]{6x-5}=\frac{x^7}{8x^2-10x+3}\)
