a.
ĐKXĐ: \(\left[{}\begin{matrix}x=0\\x\ge3\end{matrix}\right.\)
- Với \(x=0\) là 1 nghiệm
- Với \(x\ne0\Rightarrow x\ge3\) phương trình tương đương:
\(\sqrt{x-1}+\sqrt{x-2}=2\sqrt{x-3}\)
Do \(\left\{{}\begin{matrix}x-1>x-3\\x-2>x-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}>\sqrt{x-3}\\\sqrt{x-2}>\sqrt{x-3}\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-1}+\sqrt{x-2}>2\sqrt{x-3}\)
\(\Rightarrow\) Pt vô nghiệm khi \(x\ne0\)
Vậy pt có nghiệm duy nhất \(x=0\)
b.
ĐKXĐ: \(x\ge1\)
\(2\left(5x+3\sqrt{\left(x-1\right)\left(x+2\right)}\right)=27+3\sqrt{x-1}+\sqrt{x+2}\)
\(\Leftrightarrow10x-27+6\sqrt{\left(x-1\right)\left(x+2\right)}=3\sqrt{x-1}+\sqrt{x+2}\)
Đặt \(3\sqrt{x-1}+\sqrt{x+2}=t>0\)
\(\Rightarrow t^2=\left(3\sqrt{x-1}+\sqrt{x+2}\right)^2\)
\(\Rightarrow t^2=9\left(x-1\right)+\left(x+2\right)+6\sqrt{\left(x-1\right)\left(x+2\right)}\)
\(\Rightarrow t^2=10x-7\)
\(\Rightarrow10x-27=t^2-20\)
Pt trở thành:
\(t^2-20=t\Leftrightarrow t^2-t-20=0\)
\(\Rightarrow\left[{}\begin{matrix}t=5\\t=-4< 0\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow3\sqrt{x-1}+\sqrt{x+2}=5\)
\(\Leftrightarrow3\sqrt{x-1}-3+\sqrt{x+2}-2=0\)
\(\Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{x-1}+1}+\dfrac{x-2}{\sqrt{x+2}+2}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{3}{\sqrt{x-1}+1}+\dfrac{1}{\sqrt{x+2}+2}\right)=0\)
\(\Leftrightarrow x=2\) (do \(\dfrac{3}{\sqrt{x-1}+1}+\dfrac{1}{\sqrt{x+2}+2}>0\))
4.
\(Q=\dfrac{3}{a^2+4b^2}+\dfrac{3}{4ab}+\dfrac{1}{4ab}\)
\(Q=3\left(\dfrac{1}{a^2+4b^2}+\dfrac{1}{4ab}\right)+\dfrac{2}{4.a.2b}\)
\(Q\ge3.\left(\dfrac{4}{a^2+4b^2+4ab}\right)+\dfrac{2}{\left(a+2b\right)^2}=\dfrac{12}{\left(a+2b\right)^2}+\dfrac{2}{\left(a+2b\right)^2}=\dfrac{12}{1^2}+\dfrac{2}{1^2}=14\)
\(Q_{min}=14\) khi \(\left(a;b\right)=\left(\dfrac{1}{2};\dfrac{1}{4}\right)\)
