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a, \(\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{5+2\sqrt{6}}=\sqrt{15+2\cdot3\cdot\sqrt{6}}-\sqrt{10+2\cdot2\cdot\sqrt{6}}=\sqrt{9+2\cdot3\cdot\sqrt{6}+6}-\sqrt{6+2\cdot\sqrt{6}\cdot2+4}=\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}+2\right)^2}=3+\sqrt{6}-\sqrt{6}-2=3-2=1\left(đpcm\right)\)
b, đề không rõ ràng
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
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
\(A=\left(\dfrac{4\sqrt{xy}+x-2\sqrt{xy}+y}{2\left(x-y\right)}\right)\cdot\dfrac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\left(x-y\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=1\)
Ta có : \(xy\left(x+y\right)^2\le\frac{1}{64}\)\(\Rightarrow\)\(\sqrt{xy\left(x+y\right)^2}\le\sqrt{\frac{1}{64}}\)
\(\Rightarrow\)\(\sqrt{xy}\left(x+y\right)\le\frac{1}{8}\)
ta cần c/m \(\sqrt{xy}\left(x+y\right)\le\frac{1}{8}\)
Thật vậy, ta có
Áp dụng BĐT : \(ab\le\frac{\left(a+b\right)^2}{4}\). Dấu "=" xảy ra \(\Leftrightarrow\)a = b
\(\sqrt{xy}\left(x+y\right)=\frac{1}{2}.2\sqrt{xy}\left(x+y\right)\le\frac{1}{2}.\frac{\left(x+2\sqrt{xy}+y\right)^2}{4}=\frac{\left(\sqrt{x}^2+2\sqrt{xy}+\sqrt{y}^2\right)^2}{4}.\frac{1}{2}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{8}=\frac{1}{8}\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(x=y=\frac{1}{4}\)
BĐT cần chứng minh tương đương
\(VT\ge4\left(x+y+z\right)\)
\(\Leftrightarrow\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge4\left(x+y+z\right)\)
Theo BĐT Cauchy-Schwarz và AM-GM, ta có:
\(\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge\dfrac{\left(y+z\right)\left(x+\sqrt{yz}\right)}{x}=y+z+\dfrac{\left(y+z\right)\sqrt{yz}}{x}\ge y+z+\dfrac{2yz}{x}\)
Suy ra: \(\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge2\left(x+y+z\right)-2\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\)
Mặt khác, theo AM-GM:
\(\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)^2\ge3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\ge x+y+z\)
\(\Rightarrow\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge4\left(x+y+z\right)\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\dfrac{\sqrt{2}}{3}\)
@Phương An
\( a)\sqrt {4{x^2} - 4x + 1} = 3\\ \Leftrightarrow \sqrt {{{\left( {2x - 1} \right)}^2}} = 3\\ \Leftrightarrow \left| {2x - 1} \right| = 3\\ T{H_1}:2x - 1 \ge 0 \Rightarrow x \ge \dfrac{1}{2}\\ 2x - 1 = 3\\ \Leftrightarrow 2x = 3 + 1\\ \Leftrightarrow 2x = 4\\ \Leftrightarrow x = \dfrac{4}{2} = 2\left( {TM} \right)\\ T{H_2}:2x - 1 < 0 \Rightarrow x < \dfrac{1}{2}\\ - \left( {2x - 1} \right) = 3\\ \Leftrightarrow - 2x + 1 = 3\\ \Leftrightarrow - 2x = 3 - 1\\ \Leftrightarrow - 2x = 2\\ \Leftrightarrow x = - \dfrac{2}{2} = - 1\left( {TM} \right) \)
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 :
\(\left(\sqrt{x}+\sqrt{y}\right)\left(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\right)=\frac{\left(\sqrt{x}+\sqrt{y}\right)\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)
Chỉnh lại đề \sqrt(x) + \sqrt(y)=1
bình phương x+y+2\sqrt(xy) = 1 đặt ( \sqrt(xy) ; x+y)= (a;b)
BĐT cần c/m <=> (ab)^2 <= (2a .b ) ^2 /4 <= ((2a+b)^2/4)^2/4 = 1/64 ĐPCM :))