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b2 \(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}=\sqrt{x}.\sqrt{1-\frac{1}{x}}+\sqrt{y}.\)\(\sqrt{y}.\sqrt{1-\frac{1}{y}}+\sqrt{z}.\sqrt{1-\frac{1}{z}}\)rồi dung bunhia là xong
A= \(\frac{1}{a^3}\)+ \(\frac{1}{b^3}\)+ \(\frac{1}{c^3}\)+ \(\frac{ab^2}{c^3}\)+ \(\frac{bc^2}{a^3}\)+ \(\frac{ca^2}{b^3}\)
Svacxo:
3 cái đầu >= \(\frac{9}{a^3+b^3+c^3}\)
3 cái sau >= \(\frac{\left(\sqrt{a}b+\sqrt{c}b+\sqrt{a}c\right)^2}{a^3+b^3+c^3}\)
Cô-si: cái tử bỏ bình phương >= 3\(\sqrt{abc}\)
=> cái tử >= 9abc= 9 vì abc=1
Còn lại tự làm
a, ĐKXĐ : \(\left[{}\begin{matrix}x\ge0\\ y>0\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x>0\\y\ge0\end{matrix}\right.\)
Ta có :\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\sqrt{x^2}\sqrt{x}+\sqrt{y^2}\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
= \(\left(x-\sqrt{xy}+y\right)-\left(x-2\sqrt{xy}+y\right)\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
= \(\sqrt{xy}\)
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}\) \(=\sqrt{\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}}\)\(=\sqrt{\frac{a^2-1}{b^2-1}}\) (*)
Thay a=7,25 và b= 3,25 vào (*) ta có:
\(\sqrt{\frac{7,25^2-1}{3,25^2-1}}\) \(=\frac{5\sqrt{33}}{4}:\frac{3\sqrt{17}}{4}=\frac{5\sqrt{33}}{3\sqrt{17}}=\frac{5\sqrt{561}}{51}\)
Ừa, nhầm 1 xíu
\(B=\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\ge1+\frac{2}{0+1}=3\)
Lý giải:
\(\sqrt{x}\ge0;\forall x\Rightarrow\sqrt{x}+1\ge0+1\)
\(\Rightarrow\frac{2}{\sqrt{x}+1}\le\frac{2}{0+1}\Rightarrow1+\frac{2}{\sqrt{x}+1}\ge1+\frac{2}{0+1}\)
Anh Mai
\(B=2-\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+3}{\sqrt{x}+1}\)
\(B=1+\frac{2}{\sqrt{x}+1}\le1+\frac{2}{0+1}=3\)
\(B_{max}=3\) khi \(x=0\)
\(C=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+7\right)}\)
\(\Rightarrow\frac{1}{C}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+7\right)}{\sqrt{x}-1}=\frac{x+8\sqrt{x}+7}{\sqrt{x}-1}=\sqrt{x}+9+\frac{16}{\sqrt{x}-1}\)
\(\frac{1}{C}=\sqrt{x}-1+\frac{16}{\sqrt{x}-1}+10\ge2\sqrt{\frac{16\left(\sqrt{x}-1\right)}{\sqrt{x}-1}}+10=18\)
\(\Rightarrow\frac{1}{C}\ge18\Rightarrow C\le\frac{1}{18}\)
\(\Rightarrow C_{max}=\frac{1}{18}\) khi \(\sqrt{x}-1=4\Leftrightarrow x=25\)