a)đặt y=\(\sqrt{x+2}...">
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a) Do: \(y=\sqrt{x+2}\) <=> \(y^2=x+2\) <=> \(x=y^2-2\) Khi đó: \(A=y^2-2-2y\) Vậy \(A=y^2-2y-2\) b) \(A=y^2-2y-2\left(cmt\right)\) \(A=\left(y^2-2y+1\right)-3\) \(A=\left(y-1\right)^2-3\) Do \(\left(y-1\right)^2\ge0\forall y\) => \(\left(y-1\right)^2-3\ge-3\) => \(A\ge-3\) Vậy A MIN = -3 <=> \(\left(y-1\right)^2=0\) <=> \(y=1\) Do: \(y=\sqrt{x+2}\) <=> \(\sqrt{x+2}=1\) <=> \(x+2=1\) <=> \(x=-1\) áp dụng bdt cauchy -schửat dạng engel ta có \(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{x+z}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)\(\ge\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2}=\frac{1}{2}\) (do \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\) bn tự cm nhé) dau = xay ra \(\Leftrightarrow x=y=z=\frac{1}{3}\) 2. \(P=x^2-x\sqrt{3}+1=\left(x^2-x\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=\left(x-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\) Dấu '=' xảy ra khi \(x=\frac{\sqrt{3}}{2}\) Vây \(P_{min}=\frac{1}{4}\)khi \(x=\frac{\sqrt{3}}{2}\) 3. \(Y=\frac{x}{\left(x+2011\right)^2}\le\frac{x}{4x.2011}=\frac{1}{8044}\) Dấu '=' xảy ra khi \(x=2011\) Vây \(Y_{max}=\frac{1}{8044}\)khi \(x=2011\) 4. \(Q=\frac{1}{x-\sqrt{x}+2}=\frac{1}{\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{7}{4}}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}\le\frac{4}{7}\) Dấu '=' xảy ra khi \(x=\frac{1}{4}\) Vậy \(Q_{max}=\frac{4}{7}\)khi \(x=\frac{1}{4}\) a) \(ĐKXĐ:x>0\) \(Y=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-1-\frac{2x+\sqrt{x}}{\sqrt{x}}\) \(\Leftrightarrow Y=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{\left(x-\sqrt{x}+1\right)}-1-2\sqrt{x}-1\) \(\Leftrightarrow Y=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(x-\sqrt{x}+1\right)}-2\sqrt{x}-2\) \(\Leftrightarrow Y=x+\sqrt{x}-2\sqrt{x}-2\) \(\Leftrightarrow Y=x-\sqrt{x}-2\) b) Ta có \(Y=x-\sqrt{x}-2=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\) Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\) \(\Leftrightarrow x=\frac{1}{4}\) Vậy \(Min_Y=-\frac{9}{4}\Leftrightarrow x=\frac{1}{4}\) c) Để \(Y-\left|Y\right|=0\) \(\Leftrightarrow Y=\left|Y\right|\) \(\Leftrightarrow Y\ge0\) \(\Leftrightarrow x-\sqrt{x}-2\ge0\) \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\ge0\) \(\Leftrightarrow\sqrt{x}-2\ge0\) (Vì \(\sqrt{x}+1\ge0\)) \(\Leftrightarrow\sqrt{x}\ge2\) \(\Leftrightarrow x\ge4\) (ĐPCM) \(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\) ĐK : \(\hept{\begin{cases}x,y>0\\x\ne y\end{cases}}\) \(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\) \(=\frac{x+2\sqrt{xy}+y}{x-y}-\frac{x-2\sqrt{xy}+y}{x-y}\) \(=\frac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}=\frac{4\sqrt{xy}}{x-y}\) Với \(\hept{\begin{cases}x=7+2\sqrt{3}\\y=7-2\sqrt{3}\end{cases}}\)( tmđk ) => \(A=\frac{4\sqrt{\left(7+2\sqrt{3}\right)\left(7-2\sqrt{3}\right)}}{7+2\sqrt{3}-\left(7-2\sqrt{3}\right)}\) \(=\frac{4\sqrt{7^2-\left(2\sqrt{3}\right)^2}}{7+2\sqrt{3}-7+2\sqrt{3}}\) \(=\frac{4\sqrt{49-12}}{4\sqrt{3}}\) \(=\frac{4\sqrt{37}}{4\sqrt{3}}=\frac{\sqrt{37}}{\sqrt{3}}=\frac{\sqrt{37}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{\sqrt{111}}{3}\) \(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\) \(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\) \(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\) \(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\) \(=-3\) \(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\) \(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\) \(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\) \(=\frac{\sqrt{x}-1}{\sqrt{x}}\) b, Ta có \(B< A\) \(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\) \(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\) \(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\) \(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\) \(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\) \(\Leftrightarrow\sqrt{x}< \frac{1}{4}\) \(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ) Vậy ...
K
Khách
NT
Ngô Thị Thu Trang
15 tháng 7 2018
Đúng(0)
Các câu hỏi dưới đây có thể giống với câu hỏi trên
10 tháng 8 2020
10 tháng 8 2020
19 tháng 10 2017
16 tháng 5 2019
MN
1 tháng 8 2020
23 tháng 10 2020
27 tháng 4 2019
