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a) Áp dụng bdt cosi schwars ta có
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{a+b+b+c+c+d+d+a}\)
\(=\frac{a+b+c+d}{2}\)
a) \(A=5+\sqrt{-4x^2-4x}\)
\(A==5+\sqrt{-4x\left(x+1\right)}\)
Có: \(-4x\left(x+1\right)\le0\)
\(\Rightarrow\sqrt{-4x\left(x+1\right)}=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy: \(Max_A=5\) tại \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
b) \(B=\sqrt{x-2}+\sqrt{4-x}\)
ĐKXĐ: \(\hept{\begin{cases}x\ge2\\x\le4\end{cases}}\Rightarrow x\in\left\{2;3;4\right\}\)
Thay \(x=2\Rightarrow\sqrt{2-2}+\sqrt{4-2}=\sqrt{2}\)
Thay \(x=3\Rightarrow\sqrt{3-1}+\sqrt{4-3}=2\)
Thay \(x=4\Rightarrow\sqrt{4-2}+\sqrt{4-4}=\sqrt{2}\)
Vậy: \(Max_B=2\) tại \(x=3\)
Bài 2:
a)\(A=\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)
\(\ge x-1+0+3-x=2\)
Dấu = khi \(\hept{\begin{cases}x-1\ge0\\x-2=0\\x-3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x=2\\x\le3\end{cases}}\Leftrightarrow x=2\)
Vậy MinA=2 khi x=2
a ) \(A=\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)
\(\ge\left|x-1+3-x\right|+\left|x-2\right|=\left|x-2\right|+2\ge2\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x-1\right)\left(3-x\right)\ge0\\\left|x-2\right|=0\end{cases}\Rightarrow x=2}\)(TM)
Vậy \(A_{min}=2\Leftrightarrow x=2\)
b ) \(B=\sqrt{x-2\sqrt{x-1}}-\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1}+1}-\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}-\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|-\left|\sqrt{x-1}+1\right|\)
\(\le\left|\sqrt{x-1}-1-\sqrt{x-1}-1\right|=2\)có GTLN là 2
\(B=\dfrac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}\)
\(B=\dfrac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}\)
\(B=\dfrac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}\)
\(B=x-1\)
\(B=A+1\Leftrightarrow\sqrt{x}-1+1=x-1\)
\(\Leftrightarrow x-\sqrt{x}-1=0\)
\(\Leftrightarrow x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}-1=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{5}{4}=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}-\dfrac{\sqrt{5}}{2}\right)\left(\sqrt{x}-\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\left(\sqrt{5}+1\right)^2}{4}\\x=\dfrac{\left(1-\sqrt{5}\right)^2}{4}\end{matrix}\right.\)
câu A sửa lại đề 1 chút
\(A=\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-2}\)
\(A=\dfrac{x-2\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-2}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
\(A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
\(A=\sqrt{x}-1\)
có \(x=4-2\sqrt{3}\)
\(\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)
\(\Leftrightarrow\sqrt{x}=\sqrt{3}-1\)
khi đó \(A=\sqrt{x}-1\Leftrightarrow A=\sqrt{3}-1-1=\sqrt{3}-2\)