@Akai Haruma, Nguyen, Nguyễn Thị Ngọc Thơsvtkvtm

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Câu hỏi của Vũ Sơn Tùng - Toán lớp 9 | Học trực tuyến

a: \(\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+y+2\sqrt{xy}}\)

\(=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\frac{x+y}{xy}\cdot\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\)

\(=\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\cdot\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

\(=\frac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}=\frac{2\left(x+y\right)}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\)

Ta có: \(C=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\)

\(=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\left(x+y\right)}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}=\frac{2\left(x+y\right)+\sqrt{x}-\sqrt{y}}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\)

b: \(\left(\sqrt{x}+\sqrt{y}\right)^2=\left(\sqrt{2-\sqrt3}+\sqrt{2+\sqrt3}\right)^2\)

\(=2-\sqrt3+2+\sqrt3+2\cdot\sqrt{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}=4+2=6\)

\(\sqrt{xy}=\sqrt{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}=\sqrt{4-3}=1\)

\(x+y=2+\sqrt3+2-\sqrt3=4\)

\(\sqrt{x}-\sqrt{y}=\sqrt{2-\sqrt3}-\sqrt{2+\sqrt3}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{4-2\sqrt3}-\sqrt{4+2\sqrt3}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt3-1\right)^2}-\sqrt{\left(\sqrt3+1\right)^2}\right)=\frac{1}{\sqrt2}\left(\sqrt3-1-\sqrt3-1\right)=-\frac{2}{\sqrt2}=-\sqrt2\)

Ta có: \(C=\frac{2\left(x+y\right)+\sqrt{x}-\sqrt{y}}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(=\frac{2\cdot4-\sqrt2}{1\cdot4\cdot6}=\frac{8-\sqrt2}{24}\)

2

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

A= \(\sqrt{9x^2-6x+1}+\sqrt{9x^2-12x+4}\)

A= \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-2\right)^2}=\left|3x-1\right|+\left|3x-2\right|\)

ta có |3x-1|+|3x-2|=|3x-1|+|2-3x| ≥ |3x-1+2-3x|=1

=> A ≥ 1

=> Min A =1 khi 1/3 ≤ x ≤ 2/3

1. ĐKXĐ : \(xy>0\)

Ta có : \(P=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{-\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-2\sqrt{xy}+y+\sqrt{xy}}\right)\)

\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-2\sqrt{xy}+y+\sqrt{xy}}\right)\)

\(=\dfrac{\left(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}\)

\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}\)

\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-\left(x+\sqrt{xy}+y\right)}{x-\sqrt{xy}+y}=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{x-\sqrt{xy}+y}\)

\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

2. Ta thấy : \(x-\sqrt{xy}+y=x-\dfrac{2.\sqrt{x}.\sqrt{y}}{2}+\dfrac{y}{4}+\dfrac{3y}{4}\)

\(=\left(\sqrt{x}-\dfrac{\sqrt{y}}{2}\right)^2+\dfrac{3y}{4}\)

Mà \(\left\{{}\begin{matrix}\left(\sqrt{x}-\dfrac{\sqrt{y}}{2}\right)^2\ge0\\\dfrac{3y}{4}\ge0\end{matrix}\right.\)

\(\Rightarrow x-\sqrt{xy}+y\ge0\)

Lại có : \(\sqrt{xy}\ge0\)

\(\Rightarrow P\ge0\) ( ĐPCM )

Bạn cần làm gì với biểu thức này?
 

b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)

\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)

\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)

\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)

\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)

caau a) binh phuong len ra no x=y tuong tu