Câu hỏi của Quoc Tran Anh Le

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𝓓𝓾𝔂 𝓐𝓷𝓱 · Accepted Answer

Câu 1: PT $\Leftrightarrow\left(x-3\right)\left(x-2\right)=0$ $\Leftrightarrow\left[{}\begin{matrix}x=3\x=2\end{matrix}\right.$ Vậy $S=\left\{2;3\right\}$Câu 2:a) HPT $\Leftrightarrow\left\{{}\begin{matrix}2x+4y=10\3x+4y=5\end{matrix}\right.$ $\Leftrightarrow\left\{{}\begin{matrix}x=-5\y=\dfrac{5-x}{2}\end{matrix}\right.$ $\Leftrightarrow\left\{{}\begin{matrix}x=-5\y=5\end{matrix}\right.$ Vậy $\left(x;y\right)=\left(-5;5\right)$b) HPT $\Leftrightarrow\left\{{}\begin{matrix}5x=10\y=2x-7\end{matrix}\right.$ $\Leftrightarrow\left\{{}\begin{matrix}x=2\y=-3\end{matrix}\right.$ Vậy $\left(x;y\right)=\left(2;-3\right)$Câu trả lời: Câu 1: PT $\Leftrightarrow\left(x-3\right)\left(x-2\right)=0$ $\Leftrightarrow\left[{}\begin{matrix}x=3\x=2\end{matrix}\right.$ Vậy $S=\left\{2;3\right\}$Câu 2:a) HPT $\Leftrightarrow\left\{{}\begin{matrix}2x+4y=10\3x+4y=5\end{matrix}\right.$ $\Leftrightarrow\left\{{}\begin{matrix}x=-5\y=\dfrac{5-x}{2}\end{matrix}\right.$ $\Leftrightarrow\left\{{}\begin{matrix}x=-5\y=5\end{matrix}\right.$ Vậy $\left(x;y\right)=\left(-5;5\right)$b) HPT $\Leftrightarrow\left\{{}\begin{matrix}5x=10\y=2x-7\end{matrix}\right.$ $\Leftrightarrow\left\{{}\begin{matrix}x=2\y=-3\end{matrix}\right.$ Vậy $\left(x;y\right)=\left(2;-3\right)$

HT2k02 · Answer

Câu 5:Đặt $P=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{1}{2xy}$Áp dụng bất đẳng thức Bunhiacopxki dạng phân thức ta có:$\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}\ge4$Áp dụng bất đẳng thức Cosi ta có:$2xy\le\dfrac{\left(x+y\right)^2}{2}\le\dfrac{1}{2}\Rightarrow\dfrac{1}{2xy}\ge2$$\Rightarrow P\ge6$Dấu "=" xảy ra khi $x=y=\dfrac{1}{2}$Câu trả lời: Câu 5:Đặt $P=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{1}{2xy}$Áp dụng bất đẳng thức Bunhiacopxki dạng phân thức ta có:$\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}\ge4$Áp dụng bất đẳng thức Cosi ta có:$2xy\le\dfrac{\left(x+y\right)^2}{2}\le\dfrac{1}{2}\Rightarrow\dfrac{1}{2xy}\ge2$$\Rightarrow P\ge6$Dấu "=" xảy ra khi $x=y=\dfrac{1}{2}$

ntkhai0708 · Answer

Có $\dfrac{a^2}{5a^2+(b+c)^2}=\dfrac{1}{9}.\dfrac{9a^2}{a^2+b^2+c^2+2a(2a+bc)}$Áp dụng Cauchy-Schwarz có:$\dfrac{a^2}{a^2+b^2+c^2}+\dfrac{4a^2}{2a(2a+bc)} \geq \dfrac{9a^2}{a^2+b^2+c^2+2a(2a+bc)}$Nên $\dfrac{a^2}{5a^2+(b+c)^2} \leq \dfrac{1}{9}.(\dfrac{a^2}{a^2+b^2+c^2}+\dfrac{2a}{2a+bc})$Tương tự $\dfrac{b^2}{5b^2+(a+c)^2} \leq \dfrac{1}{9}.(\dfrac{b^2}{a^2+b^2+c^2}+\dfrac{2b}{2b+ac})$ $\dfrac{c^2}{5c^2+(a+b)^2} \leq \dfrac{1}{9}.(\dfrac{c^2}{a^2+b^2+c^2}+\dfrac{2c}{2c+ab})$Nên $\dfrac{a^2}{5a^2+(b+c)^2}+\dfrac{b^2}{5b^2+(a+c)^2} +\dfrac{c^2}{5c^2+(a+b)^2} \leq \dfrac{1}{9}.(1+3-(\dfrac{bc}{2a+bc}+\dfrac{ca}{2b+ac}+\dfrac{ab}{2c+ab}))$Áp dụng Cauchy Schwarz có:$\dfrac{bc}{2a+bc}+\dfrac{ca}{2b+ac}+\dfrac{ab}{2c+ab} \geq \dfrac{(ab+bc+ca)^2}{(ab)^2+(bc)^2+(ca)^2+2a^2bc+2ab^2c+2abc^2}=\dfrac{(ab+bc+ca)^2}{(ab+bc+ca)^2}=1$Nên $\dfrac{a^2}{5a^2+(b+c)^2}+\dfrac{b^2}{5b^2+(a+c)^2} +\dfrac{c^2}{5c^2+(a+b)^2} \leq \dfrac{1}{9}.(1+3-1)=\dfrac{1}{3}$Dấu $=$ xảy ra khi $a=b=c$Câu trả lời: Có $\dfrac{a^2}{5a^2+(b+c)^2}=\dfrac{1}{9}.\dfrac{9a^2}{a^2+b^2+c^2+2a(2a+bc)}$Áp dụng Cauchy-Schwarz có:$\dfrac{a^2}{a^2+b^2+c^2}+\dfrac{4a^2}{2a(2a+bc)} \geq \dfrac{9a^2}{a^2+b^2+c^2+2a(2a+bc)}$Nên $\dfrac{a^2}{5a^2+(b+c)^2} \leq \dfrac{1}{9}.(\dfrac{a^2}{a^2+b^2+c^2}+\dfrac{2a}{2a+bc})$Tương tự $\dfrac{b^2}{5b^2+(a+c)^2} \leq \dfrac{1}{9}.(\dfrac{b^2}{a^2+b^2+c^2}+\dfrac{2b}{2b+ac})$ $\dfrac{c^2}{5c^2+(a+b)^2} \leq \dfrac{1}{9}.(\dfrac{c^2}{a^2+b^2+c^2}+\dfrac{2c}{2c+ab})$Nên $\dfrac{a^2}{5a^2+(b+c)^2}+\dfrac{b^2}{5b^2+(a+c)^2} +\dfrac{c^2}{5c^2+(a+b)^2} \leq \dfrac{1}{9}.(1+3-(\dfrac{bc}{2a+bc}+\dfrac{ca}{2b+ac}+\dfrac{ab}{2c+ab}))$Áp dụng Cauchy Schwarz có:$\dfrac{bc}{2a+bc}+\dfrac{ca}{2b+ac}+\dfrac{ab}{2c+ab} \geq \dfrac{(ab+bc+ca)^2}{(ab)^2+(bc)^2+(ca)^2+2a^2bc+2ab^2c+2abc^2}=\dfrac{(ab+bc+ca)^2}{(ab+bc+ca)^2}=1$Nên $\dfrac{a^2}{5a^2+(b+c)^2}+\dfrac{b^2}{5b^2+(a+c)^2} +\dfrac{c^2}{5c^2+(a+b)^2} \leq \dfrac{1}{9}.(1+3-1)=\dfrac{1}{3}$Dấu $=$ xảy ra khi $a=b=c$

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