kẻ lười biếng nạp card, đi ô tô

Ta có : \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\) \(\Rightarrow\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=x+y+z\)

\(\Rightarrow\frac{x^2}{y+z}+\frac{xy+xz}{y+z}+\frac{y^2}{z+x}+\frac{xy+yz}{z+x}+\frac{z^2}{x+y}+\frac{zx+zy}{x+y}\)\(=x+y+z\)

\(\Rightarrow P+\frac{x\left(y+z\right)}{y+z}+\frac{y\left(x+z\right)}{x+z}+\frac{z\left(x+y\right)}{x+y}=x+y+z\)

\(\Rightarrow P+x+y+z=x+y+z\Rightarrow P=0\)

Vậy P = 0

Đề  sai rồi nếu là vầy thì mình làm dc    x+y+z=1 và x/(y+z)+y/(z+x)+z/(x+y)=1.Tính x^2/(y+z)+y^2/(x+z)+z^2/(x+y)+?

gaaaaaaaaaaaaaaaaaaaaaa

à

a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)

\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)

\(P=\dfrac{4x}{\sqrt{x}-3}\)

b) \(P=\dfrac{4x}{\sqrt{x}-3}\)

\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)

Theo BĐT côsi ta có:

\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)

Vậy: \(P_{min}=36\Leftrightarrow x=36\) 

\(P=\dfrac{1}{2xy}+\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\ge\dfrac{1}{\dfrac{2.\left(x+y\right)^2}{4}}+\dfrac{4}{2xy+x^2+y^2}=\dfrac{6}{\left(x+y\right)^2}=6\)

\(P_{min}=6\) khi \(a=b=\dfrac{1}{2}\)

Cách khác:

Đặt $xy=t$. Bằng $AM-GM$ dễ thấy $t\leq \frac{1}{4}$

\(P=\frac{1}{xy}+\frac{1}{(x+y)^2-2xy}=\frac{1}{xy}+\frac{1}{1-2xy}=\frac{1}{t}+\frac{1}{1-2t}\)

\(=\frac{1}{t}-4+\frac{1}{1-2t}-2+6=\frac{(1-4t)(1-3t)}{t(1-2t)}+6\geq 6\) với mọi $t\leq \frac{1}{4}$

Vậy $P_{\min}=6$ khi $x=y=\frac{1}{2}$

a, \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)

\(=\frac{1}{\sqrt{x}+1}.\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}\)

\(=\frac{1}{\sqrt{x}+1}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(1-x\right)^2}{2}\)

\(=\frac{1}{\sqrt{x}+1}.\frac{-2\sqrt{x}}{x-1}.\frac{\left(1-x\right)^2}{2}=-x+\sqrt{x}\)

b, \(x=7-4\sqrt{3}=\left(\sqrt{3}-2\right)^2\Rightarrow\sqrt{x}=2-\sqrt{3}\)

Khi đó \(P=-x+\sqrt{x}=-\left(7-4\sqrt{3}\right)+\left(2-\sqrt{3}\right)=-5+3\sqrt{3}\)

c, \(P=-x+\sqrt{x}=-\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

\(\Rightarrow MaxP=\frac{1}{4}\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)

Mik lm đc r

\(a\text{) }\dfrac{x\sqrt{x}-8}{x+2\sqrt{x}+4}-3\left(1-\sqrt{x}\right)\\ =\dfrac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{x+2\sqrt{x}+4}-3\left(1-\sqrt{x}\right)\\ =\sqrt{x}-2-3+3\sqrt{x}\\ =4\sqrt{x}-5\)

\(b\text{) }Q=\dfrac{2P}{1-P}=\dfrac{2\left(4\sqrt{x}-5\right)}{1-4\sqrt{x}-5}\\ =\dfrac{2\left(5-4\sqrt{x}\right)}{4+4\sqrt{x}}=\dfrac{5-4\sqrt{x}}{2+2\sqrt{x}}\\ =\dfrac{9-4-4\sqrt{x}}{2+2\sqrt{x}}=\dfrac{9}{2+2\sqrt{x}}-2\)

\(\Rightarrow\)Để Q nhận giá trị nguyên

thì \(\dfrac{9}{2+2\sqrt{x}}\in Z\)

\(\Rightarrow9⋮2\sqrt{x}+2\\ \Rightarrow2\sqrt{x}+2\inƯ_{\left(9\right)}\\ Mà\text{ }2\sqrt{x}+2>2\\ \Rightarrow2\sqrt{x}+2\in\left\{3;9\right\}\)

Lập bảng giá trị:

Vậy không có giá trị nguyên nào của x để Q nhận giá trị nguyên.

a. Để P được xđ thì MT phải khác 0.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9\ne0\\x^2+3x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(x+3\right)\ne0\\x\left(x+3\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne0\end{matrix}\right.\)

b. \(P=\left(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\right).\dfrac{x-3}{x+3}\)

\(P=\left(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\right).\dfrac{x-3}{x+3}\)

\(P=\left(\dfrac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}\right).\dfrac{x-3}{x+3}\)

\(P=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+3}\)

\(P=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+3}\)

\(P=\dfrac{1}{x}\)