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

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

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

ĐKXĐ: \(x\ge1;y\ge1\)

Ta có: \(\frac{x^2-4}{x}+\frac{y^2-4}{y}+8=4\left(\sqrt{x-1}+\sqrt{y-1}\right)\)

\(\Leftrightarrow\frac{x^2-4}{x}+\frac{y^2-4}{y}=4\left[\left(\sqrt{x-1}-1\right)+\left(\sqrt{y-1}+1\right)\right]\)

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

\(\Leftrightarrow\left(x-2\right)\left(\frac{x+2}{x}-\frac{4}{\sqrt{x-1}+1}\right)+\left(y-2\right)\left(\frac{y+2}{y}-\frac{4}{\sqrt{y-1}+1}\right)=0\)

\(\Leftrightarrow\left(x-2\right)\frac{x\sqrt{x-1}+2\sqrt{x-1}+2+x-4x}{x\left(\sqrt{x-1}+1\right)}+\left(y-2\right)\frac{y\sqrt{y-1}+2\sqrt{y-1}+y-4y}{y\left(\sqrt{y-1}+1\right)}=0\)

\(\Leftrightarrow\left(x-2\right)\frac{\left( x-1\right)\sqrt{x-1}+3\sqrt{x-1}-3\left(x-1\right)-1}{x\left(\sqrt{x-1}+1\right)}\)

      \(+\left(y-2\right)\frac{\left(y-1\right)\sqrt{y-1}+3\sqrt{y-1}-3\left(y-1\right)-1}{y\left(\sqrt{y-1}+1\right)}=0\)

\(\Leftrightarrow\left(x-2\right)\frac{\left(\sqrt{x-1}-1\right)^3}{x\left(\sqrt{x-1}+1\right)}+\left(y-2\right)\frac{\left(\sqrt{y-1}-1\right)^3}{y\left(\sqrt{y-1}+1\right)}=0\)

\(\Leftrightarrow\left(x-2\right)\frac{\left(\sqrt{x-1}-1\right)^3\left(\sqrt{x-1}+1\right)^3}{x\left(\sqrt{x-1}+1\right)^4}+\left(y-2\right)\frac{\left(\sqrt{y-1}-1\right)^3\left(\sqrt{y-1}+1\right)^3}{y\left(\sqrt{y-1}+1\right)^4}=0\)

\(\Leftrightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}+\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}=0\)

Vì \(x\ge1;y\ge1\Rightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}\ge0;\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}\ge0\)\(\Rightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}+\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}\ge0\)

Do đó dấu ''='' xảy ra khi \(\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}=\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}=0\Leftrightarrow x-2=y-2=0\Leftrightarrow x=y=2\)

Vậy \(x=y=2\).

Anh tham khảo tại đây:

Câu hỏi của jihoon oppa! I'm May - Toán lớp 8 - Học toán với OnlineMath

zZz Cool Kid zZz cái đó cô chi nhầm nên bị ngược dấu rồi nhé! Mình đang nghĩ cách:(