Sai đề rồi bạn ơi. Đề đúng : \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\ge6.\)

Hoặc \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)

a) A= x^2 +3x+2= x2+2.x.32+(32)2−14=(x+32)2−14≥14

Vậy GTNN của A là 1/4

b) tương tự

~Học tốt~

Mik gửi nhầm câu hỏi

Ko phải câu trả lời cho bài này đâu nha

\(\sqrt{36-12\sqrt{5}}\)

\(=\sqrt{9-2\cdot3\cdot2\sqrt{5}+20+7}\)

\(=\sqrt{3^2-2\cdot3\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2+7}\)

\(=\sqrt{\left(3-2\sqrt{5}\right)^2+7}\)

\(\sqrt{29+12\sqrt{5}}\)

= \(\sqrt{\left(2\sqrt{5}\right)^2+2.3.2\sqrt{5}+9}\)

= \(2\sqrt{5}+3\)

\(\sqrt{29+12\sqrt{5}}\)

\(=\sqrt{9+2\cdot3\cdot2\sqrt{5}+20}\)

\(=\sqrt{3^2+2\cdot3\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2}\)

\(=\sqrt{\left(3+2\sqrt{5}\right)^2}\)

\(=\left|3+2\sqrt{5}\right|=3+2\sqrt{5}\)

\(B=x+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right).\)

Vì \(\sqrt{x}\ge0\)\(\Rightarrow B_{min}\)\(=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)

\(\Rightarrow\hept{\begin{cases}\sqrt{x}=0\\\sqrt{x}+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x\in\varnothing\end{cases}}}\)

Vậy \(B_{min}=0\Leftrightarrow x=0\)

\(B=x+\sqrt{x}\)

\(B=\left(\sqrt{x}\right)^2+2\cdot\frac{1}{2}\sqrt{x}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\)

\(B=\left(\sqrt{x}+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\)

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

Có \(\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(\Rightarrow GTNN\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{1}{4}=-\frac{1}{4}\)

\(\Rightarrow GTNNx+\sqrt{x}=-\frac{1}{4}\)

với \(\left(\sqrt{x}+\frac{1}{2}\right)^2=0\)

Đkxđ:\(x\ge0\)

TA có: \(B=x+\sqrt{x}\Rightarrow B=\sqrt{x}\left(\sqrt{x}+1\right)\ge0\) 

Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\sqrt{x=0}\Leftrightarrow x=0\\\sqrt{x}+1=0\Leftrightarrow\sqrt{x}=-1\left(ktm\right)\end{cases}}\) 

Vậy min B=0 tại x=0