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a) ĐKXĐ: \(x\ne2y,x\ne-y;x\ne-1\)
b) \(B=\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
\(B=\left[\dfrac{y-x}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{4x^4+4x^2y+y^2-4}{x\left(x+y\right)+\left(x+y\right)}\)
\(B=\left[\dfrac{\left(y-x\right)\left(x+y\right)}{\left(x-2y\right)\left(x+y\right)}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)
\(B=\dfrac{y^2-x^2-x^2-y^2-y+2}{\left(x+y\right)\left(x-2y\right)}:\dfrac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)
\(B=\dfrac{-2x^2-y+2}{\left(x+y\right)\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)\left(x+y\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\)
\(B=\dfrac{-\left(2x^2+y-2\right)}{\left(x+y\right)\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)\left(x+y\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\)
\(B=\dfrac{-\left(x+1\right)}{\left(x-2y\right)\left(2x^2+y+2\right)}\)
Lời giải:
Ta có \(\frac{x^2+y^2}{xy}=\frac{25}{12}\)
\(\Leftrightarrow 12(x^2+y^2)-25xy=0\)
\(\Leftrightarrow (3x-4y)(4x-3y)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-4y=0\\4x-3y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4y}{3}\left(1\right)\\x=\dfrac{3y}{4}\left(2\right)\end{matrix}\right.\)
Với (1):
\(A=\frac{x-y}{x+y}=\frac{\frac{4}{3}y-y}{\frac{4}{3}y+y}=\frac{\frac{1}{3}y}{\frac{7}{3}y}=\frac{1}{7}\)
Với (2)
\(A=\frac{x-y}{x+y}=\frac{\frac{3}{4}y-y}{\frac{3}{4}y+y}=\frac{\frac{-1}{4}y}{\frac{7}{4}y}=\frac{-1}{7}\)
Vậy
\(A=\pm \frac{1}{7}\)
Câu 1:
Áp dụng BĐT Cô-si:
\(x^4+y^2\geq 2\sqrt{x^4y^2}=2x^2y\Rightarrow \frac{x}{x^4+y^2}\leq \frac{x}{2x^2y}=\frac{1}{2xy}=\frac{1}{2}(1)\)
\(x^2+y^4\geq 2\sqrt{x^2y^4}=2xy^2\Rightarrow \frac{y}{x^2+y^4}\leq \frac{y}{2xy^2}=\frac{1}{2xy}=\frac{1}{2}(2)\)
Lấy \((1)+(2)\Rightarrow A\leq \frac{1}{2}+\frac{1}{2}=1\)
Vậy \(A_{\max}=1\). Dấu bằng xảy ra khi \(x=y=1\)
Câu 2:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)(x^2+y^2+2xy)\geq (1+1)^2\)
\(\Rightarrow \frac{1}{x^2+y^2}+\frac{1}{2xy}\geq \frac{4}{x^2+y^2+2xy}=\frac{4}{(x+y)^2}\geq \frac{4}{1}=4(*)\)
(do \(x+y\leq 1\) )
Áp dụng BĐT Cô-si:
\(\frac{1}{4xy}+4xy\geq 2\sqrt{\frac{4xy}{4xy}}=2(**)\)
\(x+y\geq 2\sqrt{xy}\Leftrightarrow 1\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}\)
\(\Rightarrow \frac{5}{4xy}\geq \frac{5}{4.\frac{1}{4}}=5(***)\)
Cộng \((*)+(**)+(***)\Rightarrow B\geq 4+2+5=11\)
Vậy \(B_{\min}=11\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
Ta có : x^2+y^2/xy=12/25
=>12(x^2+y^2)=25xy
=>12(x^2+2xy+y^2)=49xy
=>12(x+y)^2=49xy
=>(x+y)^2=49xy/12 (1)
Ta có : x^2+y^2/xy=12/25
=>12(x^2+y^2)=25xy
=>12(x^2-2xy+y^2)=xy
=>12(x-y)^2=xy
=>(x-y)^2=xy/12 (2)
Từ (1) và (2) suy ra :
(x-y)^2/(x+y)^2=1/49
Vì x<y<0 nên x-y/x=y=-1/7
Tick cho mik nhé