1. a)Tìm x , biết \(\sqrt{4x^2-4x+1}=3\)
b) Chứng minh \(\left(\sqrt{x}+\sqrt{y}\right)\left(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\right)=x-y\) với x > 0 ; y> 0
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2 tháng 10 2017
1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 4 2019
\(A=\frac{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\sqrt{4x-1-2\sqrt{4x-1}+1}}{-\left(\sqrt{4x-1}-1\right).y^2\left(x^2+xy+y^2\right)}=\frac{\left(x^2-y^2\right)\sqrt{\left(\sqrt{4x-1}-1\right)^2}}{-\left(\sqrt{4x-1}-1\right).y^2}\)
Do \(x>1\Rightarrow4x-1>1\Rightarrow\sqrt{4x-1}>1\Rightarrow\sqrt{4x-1}-1>0\)
\(\Rightarrow A=\frac{\left(x^2-y^2\right)\left(\sqrt{4x-1}-1\right)}{-\left(\sqrt{4x-1}-1\right).y^2}=\frac{x^2-y^2}{-y^2}=1-\left(\frac{x}{y}\right)^2\)
\(A=-8\Rightarrow1-\left(\frac{x}{y}\right)^2=-8\Rightarrow\left(\frac{x}{y}\right)^2=9\)
Do \(\left\{{}\begin{matrix}x>1\\y< 0\end{matrix}\right.\) \(\Rightarrow\frac{x}{y}< 0\Rightarrow\frac{x}{y}=-3\)
13 tháng 7 2019
A
Áp dụng BĐT cosi ta có
\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)
\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)
Khi đó
\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)
MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)
13 tháng 7 2019
B
Áp dụng BĐT cosi ta có :
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)
Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)
=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)
\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z
\( a)\sqrt {4{x^2} - 4x + 1} = 3\\ \Leftrightarrow \sqrt {{{\left( {2x - 1} \right)}^2}} = 3\\ \Leftrightarrow \left| {2x - 1} \right| = 3\\ T{H_1}:2x - 1 \ge 0 \Rightarrow x \ge \dfrac{1}{2}\\ 2x - 1 = 3\\ \Leftrightarrow 2x = 3 + 1\\ \Leftrightarrow 2x = 4\\ \Leftrightarrow x = \dfrac{4}{2} = 2\left( {TM} \right)\\ T{H_2}:2x - 1 < 0 \Rightarrow x < \dfrac{1}{2}\\ - \left( {2x - 1} \right) = 3\\ \Leftrightarrow - 2x + 1 = 3\\ \Leftrightarrow - 2x = 3 - 1\\ \Leftrightarrow - 2x = 2\\ \Leftrightarrow x = - \dfrac{2}{2} = - 1\left( {TM} \right) \)
Vậy...
1 a) \(\sqrt{4x^2-4x+1}=3\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\Leftrightarrow\left|2x-1\right|=3\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) Với x > 0 ; y > 0,ta có :
\(\left(\sqrt{x}+\sqrt{y}\right)\left(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\right)=\frac{\left(\sqrt{x}+\sqrt{y}\right)\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)