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Bài 1 : Ta có :
\(A=\sqrt{3x+\sqrt{6x-1}}+\sqrt{3x-\sqrt{6x-1}}\)
\(A\sqrt{2}=\sqrt{6x+2\sqrt{6x-1}}+\sqrt{6x-2\sqrt{6x-1}}\)
\(=\sqrt{6x-1+2\sqrt{6x-1}+1}+\sqrt{6x-1-2\sqrt{6x-1}+1}\)
\(=\sqrt{\left(\sqrt{6x-1}+1\right)^2}+\sqrt{\left(\sqrt{6x-1}-1\right)^2}\)
\(=\left|\sqrt{6x-1}+1\right|+\left|\sqrt{6x-1}-1\right|\)
\(=\sqrt{6x-1}+1+\sqrt{6x-1}-1\)
\(=2\sqrt{6x-1}\)
\(\Rightarrow A=\sqrt{2}\left(\sqrt{6x-1}\right)\)
Thay \(x=4+\sqrt{10}\) vào A ta được :
\(A=\sqrt{2}.\sqrt{6\left(4+\sqrt{10}\right)-1}=\sqrt{2}.\sqrt{24+6\sqrt{10}-1}\)
\(=\sqrt{2}.\sqrt{23+6\sqrt{10}}=\sqrt{46+12\sqrt{10}}\)
\(=\sqrt{36+12\sqrt{10}+10}=\sqrt{\left(6+\sqrt{10}\right)^2}=6+\sqrt{10}\)
Vậy \(A=6+\sqrt{10}\) tại \(x=4+\sqrt{10}\)
\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2-2xy\ge0\)
\(\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow x^2+y^2+2xy\ge4xy\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\Rightarrow1\ge4xy\Leftrightarrow xy\le\frac{1}{4}\)(1)
\(\left(x-y\right)^2\ge0\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge2\Leftrightarrow x+y\ge\sqrt{2}\)
Từ phần a ta có \(x+y\le\sqrt{2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{2x+1}+\sqrt{2y+1}\right)^2\)
\(\le\left(1+1\right)\left(2\left(x+y\right)+2\right)\)
\(=2\cdot\left(2\left(x+y\right)+2\right)\le2\cdot\left(2\sqrt{2}+2\right)\)
\(=4\sqrt{2}+4=VP^2\)
Suy ra \(VT\ge VP\) (ĐPCM)
1/ Đặt \(\hept{\begin{cases}\sqrt{x-2013}=a\\\sqrt{x-2014}=b\end{cases}}\)
Thì ta có:
\(\frac{\sqrt{x-2013}}{x+2}+\frac{\sqrt{x-2014}}{x}=\frac{a}{a^2+2015}+\frac{b}{b^2+2014}\)
\(\le\frac{a}{2a\sqrt{2015}}+\frac{b}{2b\sqrt{2014}}=\frac{1}{2\sqrt{2015}}+\frac{1}{2\sqrt{2014}}\)
2/ \(\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z}\)
\(\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)\)
\(=\frac{3}{4}\)
Ta có \(B=\sqrt{x+3}+\sqrt{5-x}\Leftrightarrow B^2=x+3+5-x+2\sqrt{\left(x+3\right)\left(5-x\right)}=8+2\sqrt{\left(x+3\right)\left(5-x\right)}\) Ta có \(\sqrt{\left(x+3\right)\left(5-x\right)}\ge0\Leftrightarrow2\sqrt{\left(x+3\right)\left(5-x\right)}\ge0\Leftrightarrow8+2\sqrt{\left(x+3\right)\left(5-x\right)}\ge8\Leftrightarrow B^2\ge8\Leftrightarrow B\ge2\sqrt{2}\)Vậy \(2\sqrt{2}\le B\)(1)
Áp dụng bđt Bunhia copski ta có
\(B^2=\left(\sqrt{x+3}+\sqrt{5-x}\right)^2=\left(\sqrt{x+3}.1+\sqrt{5-x}.1\right)^2\le\left[\left(\sqrt{x+3}\right)^2+\left(\sqrt{5-x}\right)^2\right]\left(1^2+1^2\right)=8.2=16\Leftrightarrow B^2\le16\Leftrightarrow B\le4\)(2)
Từ (1),(2)\(\Rightarrow2\sqrt{2}\le B\le4\)
Answer:
a. ĐK để biểu thức có nghĩa
\(\hept{\begin{cases}2-x\ge0\\x+2\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le2\\x\ge-2\end{cases}}\Leftrightarrow-2\le x\le2\left(or\left|x\right|\le2\right)}\)
b. \(f\left(a\right)=\sqrt{2-a}+\sqrt{a+2};f\left(-a\right)=\sqrt{2-\left(-a\right)}+\sqrt{-a+2}=\sqrt{2-a}+\sqrt{a+2}\)
\(\Rightarrow f\left(a\right)=f\left(-a\right)\)
c. \(y^2=\left(\sqrt{2-x}\right)^2+2\sqrt{2-x}.\sqrt{2+x}+\left(\sqrt{2+x}\right)^2=2-x+2\sqrt{4-x^2}+2+x=4+2\sqrt{4-x^2}\ge4\)
Đẳng thức xảy ra khi \(x=\pm2\)
Giá trị nhỏ nhất của y là 2
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq (6x+3y+2z)(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})$
Mà: $6x+3y+2z=3x+(x+y)+2(x+y+z)\leq 3.1+5+2.14=36$
Do đó: $(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq 36.(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})=36$
$\Rightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}\leq 6$ (đpcm)
Dấu "=" xảy ra khi $x=1; y=2; z=3$
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq (6x+3y+2z)(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})$
Mà: $6x+3y+2z=3x+(x+y)+2(x+y+z)\leq 3.1+5+2.14=36$
Do đó: $(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq 36.(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})=36$
$\Rightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}\leq 6$ (đpcm)
Dấu "=" xảy ra khi $x=1; y=2; z=3$
Áp dụng BĐT Bunhiacopxki , ta có :
\(\left(2014-x+x-2012\right)\left(1^2+1^2\right)\ge\left(\sqrt{2014-x}+\sqrt{x-2012}\right)^2\)
\(\Leftrightarrow\left(\sqrt{2014-x}+\sqrt{x-2012}\right)^2\le4\left(2012\le x\le2014\right)\)
\(\Leftrightarrow\sqrt{2014-x}+\sqrt{x-2012}\le2\)
\("="\Leftrightarrow x=2013\left(TM\right)\)