a) Ta có: \(P=\left[\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\cdot\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)

\(=\left(\dfrac{2}{\sqrt{xy}}+\dfrac{1}{x}+\dfrac{1}{y}\right):\dfrac{x\sqrt{x}+y\sqrt{x}+x\sqrt{y}+y\sqrt{y}}{x\sqrt{xy}+y\sqrt{xy}}\)

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

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

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

a) Đk:\(x>0;y>0\)

\(P=\left[\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}.\sqrt{y}}.\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{x\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}+\sqrt{y}\right)}{x\sqrt{xy}+y\sqrt{xy}}\)

\(=\left[\dfrac{2}{\sqrt{xy}}+\dfrac{x+y}{xy}\right]:\dfrac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(x+y\right)}\)

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

b) \(xy=16\Leftrightarrow x=\dfrac{16}{y}\)

\(P=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}=\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}=\dfrac{1}{\sqrt{\dfrac{16}{y}}}+\dfrac{1}{\sqrt{y}}=\dfrac{\sqrt{y}}{4}+\dfrac{1}{\sqrt{y}}\)

Áp dụng AM-GM có:

\(\dfrac{\sqrt{y}}{4}+\dfrac{1}{\sqrt{y}}\ge2\sqrt{\dfrac{\sqrt{y}}{4}.\dfrac{1}{\sqrt{y}}}=1\)

\(\Rightarrow P\ge1\)

Dấu "=" xảy ra khi \(y=4\Rightarrow x=4\)

Vậy x=y=4 thì P đạt GTNN là 1

a/ \(\dfrac{x^3}{x^2+1975}\cdot\dfrac{2x+1954}{x+1}+\dfrac{x^3}{x^2+1975}\cdot\dfrac{21-x}{x+1}=\dfrac{x^3\left(2x+1954\right)+x^3\left(21-x\right)}{\left(x^2+1975\right)\left(x+1\right)}=\dfrac{2x^4+1954x^3+21x^3-x^4}{\left(x^2+1975\right)\left(x+1\right)}=\dfrac{x^4+1975x^3}{\left(x^2+1975\right)\left(x+1\right)}\)

b/ \(\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9}{x+1945}+\dfrac{19x+8}{x^2+1945}\cdot\dfrac{x-2}{x-7}=\dfrac{\left(19x+8\right)\left(5x-9\right)+\left(19x+8\right)\left(x-2\right)}{\left(x-7\right)\left(x+1945\right)}=\dfrac{\left(19x+8\right)\left(5x-9+x-2\right)}{\left(x-7\right)\left(x+1945\right)}=\dfrac{114x^2-209x+40x-88}{\left(x-7\right)\left(x+1945\right)}=\dfrac{114x^2-169x-88}{x^2+1938x-13615}\)

c/ \(\dfrac{x+1}{x^2-2x-8}\cdot\dfrac{4-x}{x^2+x}=\dfrac{\left(x+1\right)\left(4-x\right)}{x\left[x^2-4x+2x-8\right]\left(x+1\right)}=-\dfrac{x-4}{x\left(x-4\right)+2\left(x-4\right)}=-\dfrac{x-4}{\left(x-4\right)\left(x+2\right)}=-\dfrac{1}{x+2}\)

a: \(\Leftrightarrow\left(x-1\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

\(\dfrac{x}{3}=\dfrac{y}{4}\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x+y}{3+4}=\dfrac{14}{7}\)=2
* \(\dfrac{x}{3}=2=>x=6\)
*\(\dfrac{y}{4}=2=>y=8\)
Vậy( x, y) ∈{ 6, 8}
Kiểm tra lại nhaa

áp dụng tính chất dãy tỉ số bằng nhau

Ta có:x/3=y/4=x+y/3+4=14/7=2

Vậy x=2.3=6

       y=2.4=8

\(1,\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\ 2,7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-28\end{matrix}\right.\\ 3,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=-\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{45}{2}\\y=-27\\z=-\dfrac{63}{2}\end{matrix}\right.\\ 4,x:y:z=3:5:7\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)

3. Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)

\(x=\dfrac{-45}{2}\)

\(y=-27\)

\(z=\dfrac{-63}{2}\)

1: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{8}=\dfrac{y}{11}=\dfrac{y-x}{11-8}=\dfrac{-42}{3}=-14\)

Do đó: x=-112;y=-154

TÌM X Y Z

1) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{3}{20}\)

2) \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

3) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\)

\(\Leftrightarrow4x=-\dfrac{20}{7}\)

\(\Leftrightarrow x=-\dfrac{5}{7}\)