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TV
0
9 tháng 7 2018
\(a.x+3+\sqrt{x^2-6x+9}=x+3+\text{ |}x-3\text{ |}=x+3+3-x=6\) \(b.\sqrt{x^2+4x+4}-\sqrt{x^2}=\text{ |}x+2\text{ |}-\text{ |}x\text{ |}=x+2-\left(-x\right)=x+2+x=2x+2\) \(c.\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{x-1}{x-1}=1\)
\(d.\text{ |}x-2\text{ |}+\dfrac{\sqrt{x^2-4x+4}}{x-2}=\text{ |}x-2\text{ |}+\dfrac{\text{ |}x-2\text{ |}}{x-2}=2-x+\dfrac{-\left(x-2\right)}{x-2}=2-x-1=1-x\)
HT
1
28 tháng 3 2019
\(f\left(x;y\right)=x+y+x\sqrt{1-y^2}+y\sqrt{1-x^2}\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)=\frac{\sqrt{3}}{2}\left(x+y\right)+\frac{1}{2}\left(x\sqrt{3-3y^2}+y\sqrt{3-3x^2}\right)\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)\le\frac{\frac{3}{4}+x^2+\frac{3}{4}+y^2}{2}+\frac{1}{2}\left(\frac{-3x^2+y^2+3-3y^2+x^2+3}{2}\right)\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)\le\frac{\frac{3}{2}+x^2+y^2-x^2-y^2+3}{2}=\frac{9}{4}\)
\(\Rightarrow f\left(x;y\right)\le\frac{3\sqrt{3}}{2}\)
Dấu "=" khi x = y = \(\frac{\sqrt{3}}{2}\).
#Kaito#
AT
17 tháng 8 2018
\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)
\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)
\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)
\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)
\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)
\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)