Tìm x biết :
a)\(\sqrt{x^2-3}\le x^2-3\)
b) \(\sqrt{x-1}< x+3\)
Help :'<
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\(f\left(x;y\right)=x+y+x\sqrt{1-y^2}+y\sqrt{1-x^2}\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)=\frac{\sqrt{3}}{2}\left(x+y\right)+\frac{1}{2}\left(x\sqrt{3-3y^2}+y\sqrt{3-3x^2}\right)\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)\le\frac{\frac{3}{4}+x^2+\frac{3}{4}+y^2}{2}+\frac{1}{2}\left(\frac{-3x^2+y^2+3-3y^2+x^2+3}{2}\right)\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)\le\frac{\frac{3}{2}+x^2+y^2-x^2-y^2+3}{2}=\frac{9}{4}\)
\(\Rightarrow f\left(x;y\right)\le\frac{3\sqrt{3}}{2}\)
Dấu "=" khi x = y = \(\frac{\sqrt{3}}{2}\).
#Kaito#
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a) \(\sqrt[3]{x}< 2\Leftrightarrow\left(\sqrt[3]{x}\right)^3< 2^3\Leftrightarrow x< 8\)
b) \(\sqrt[3]{2x-1}>-3\Leftrightarrow\left(\sqrt[3]{2x-1}\right)^3>\left(-3\right)^3\Leftrightarrow2x-1>-27\Leftrightarrow2x>-26\Leftrightarrow x>-13\)
c) \(\sqrt[3]{2-3x}\le1\Leftrightarrow\left(\sqrt[3]{2-3x}\right)^3\le1\Leftrightarrow2-3x\le1\Leftrightarrow3x\ge1\Leftrightarrow x\ge\frac{1}{3}\)
d) \(\sqrt[3]{3-4x}\ge5\Leftrightarrow\left(\sqrt[3]{3-4x}\right)^3\ge5^3\Leftrightarrow3-4x\ge125\Leftrightarrow4x\le-122\Leftrightarrow x\le-\frac{61}{2}\)
Ta có: \(\left(-x^2+4x+21\right)-\left(-x^2+3x+10\right)=x+11>0\Rightarrow B>0\)
\(B^2=\left(x+3\right)\left(7-x\right)+\left(x+2\right)\left(5-x\right)-2\sqrt{\left(x+3\right)\left(7-x\right)\left(x+2\right)\left(5-x\right)}=\left(\sqrt{\left(x+3\right)\left(5-x\right)}-\sqrt{\left(x+2\right)\left(7-x\right)}\right)^2+2\ge2\)
\(\Rightarrow B\ge\sqrt{2}\)
GTNN của B là \(\sqrt{2}\Leftrightarrow x=\dfrac{1}{3}\)