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Bài 1:
\(P=x\sqrt{3-x^2}=\sqrt{x^2}\cdot\sqrt{3-x^2}\)
\(=\sqrt{x^2\left(3-x^2\right)}\)\(\le\frac{x^2+3-x^2}{2}=\frac{3}{2}\)
Dấu = khi \(x=\sqrt{\frac{3}{2}}\)
Vậy MaxP=\(\frac{3}{2}\Leftrightarrow x=\sqrt{\frac{3}{2}}\)
a/\(x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{\left(x-3\right)^2}=x+3+\left|x-3\right|=x+3+3-x=6\)
b/ \(\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{\left(x+2\right)^2}-\left|x\right|=\left|x+2\right|-\left|x\right|=-x-2-\left(-x\right)=-x-2+x=-2\)
c/ \(\dfrac{\sqrt{x^2-2x+1}}{x-1}\cdot\left(x-1\right)=\sqrt{x^2-2x+1}=\sqrt{\left(x-1\right)^2}=\left|x-1\right|\)
d/ \(\left|x-2\right|+\dfrac{\sqrt{x^2-4x+4}}{x-2}=2-x+\dfrac{\sqrt{\left(x-2\right)^2}}{x-2}=2-x+\dfrac{\left|x-2\right|}{x-2}=2-x+\dfrac{-\left(x-2\right)}{x-2}=2-x-1=1-x\)
Tacó \(\Delta\)=(-7)2-4x1x2=41>0 =>\(\sqrt{_{ }x1}\)=\(\dfrac{7+\sqrt{41}}{2}\)=>\(_{x1}\)=\(\dfrac{\left(7+\sqrt{41}\right)^2}{4}\)=\(\dfrac{45+7\sqrt{41}}{2}\) =>\(\sqrt{_{ }x2}\)=\(\dfrac{7-\sqrt{41}}{2}\)=>\(_{x_2}\)=\(\dfrac{\left(7-\sqrt{41^{ }}\right)^2}{4}\)=\(\dfrac{45-7\sqrt{41}}{2}\) so sánh với điều kiện X>_0
a) Xét \(x^2-4=\left(\sqrt{\frac{a}{b}}\right)^2+\left(\sqrt{\frac{b}{a}}\right)^2+2-4\)
\(=\left(\sqrt{\frac{a}{b}}\right)^2+\left(\sqrt{\frac{b}{a}}\right)^2-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\ge0\)
b) \(\sqrt{x^2-4}=\sqrt{\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2}=\left|\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right|\)
- Nếu a < b < 0 thì \(\sqrt{\frac{a}{b}}< \sqrt{\frac{b}{a}}\Rightarrow\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}< 0\Rightarrow\sqrt{x^2-4}=\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}\)
- Nếu b < a < 0 thì \(\sqrt{\frac{b}{a}}< \sqrt{\frac{a}{b}}\Rightarrow\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}>0\Rightarrow\sqrt{x^2-4}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\)
a) Vì a<0 , b<0 => \(\frac{a}{b}>0;\frac{b}{a}>0\Rightarrow\sqrt{\frac{a}{b}}>0;\sqrt{\frac{b}{a}}>0\)
Áp dụng bất đẳng thức cô si ta có:
\(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\ge2\sqrt{\sqrt{\frac{a}{b}}\cdot\sqrt{\frac{b}{a}}}=2\)
=> \(\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)^2\ge4\)
Hay \(x^2\ge4\)
a) \(x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{\left(x-3\right)^2}=x+3+x-3=2x\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}=x+2-x=2\)
c) \(\sqrt{\frac{x^2-2x+1}{x-1}}=\sqrt{\frac{\left(x-1\right)^2}{x-1}}=\sqrt{x-1}\)
(Nhớ k cho mình với nhá!)
a)Áp dụng BĐT C-S ta có:
\(A^2=\left(\sqrt{x-2}+\sqrt{4-x}\right)^2\)
\(\le\left(1+1\right)\left(x-2+4-x\right)=4\)
\(\Rightarrow A^2\le4\Rightarrow A\le2\)
Đẳng thức xảy ra khi x=3
b)Tiếp tục áp dụng BĐT C-S
\(B^2=\left(\sqrt{x}+\sqrt{2-x}\right)^2\)
\(\le\left(1+1\right)\left(x+2-x\right)=4\)
\(\Rightarrow B^2\le4\Rightarrow B\le2\)
Xảy ra khi x=1