1/ Bình phương 2 vế:
\(a+b+2\sqrt{ab}\ge a+b\)
\(\Leftrightarrow\sqrt{ab}\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)
2/ ĐKXĐ: ....
\(\Leftrightarrow\sqrt{x^2-4x+4}+\sqrt{4x-3}+\left(x-2\right)^2=\sqrt{x^2+1}\)
\(VT\ge\sqrt{x^2-4x+4+4x-3}+\left(x-2\right)^2\)
\(VT\ge\sqrt{x^2+1}+\left(x-2\right)^2\ge\sqrt{x^2+1}\)
\(\Rightarrow VT\ge VP\)
\(x=2\)
\(x=2\)
2/Đk: \(x\ge\frac{3}{4}\)
\(PT\Leftrightarrow\left|x-2\right|+\left(x-2\right)^2-\left(\sqrt{x^2+1}-\sqrt{4x-3}\right)=0\)
\(PT\Leftrightarrow\left|x-2\right|\left(1+\left|x-2\right|-\frac{\left|x-2\right|}{\sqrt{x^2+1}+\sqrt{4x-3}}\right)=0\)
\(\Leftrightarrow\left|x-2\right|\left(1+\frac{\left(\sqrt{x^2+1}+\sqrt{4x-3}-1\right)\left|x-2\right|}{\sqrt{x^2+1}+\sqrt{4x-3}}\right)=0\)
Nhìn vô thấy ngay cái ngoặc to > 0
Vậy x = 2
P/s: Is that true?
