Câu hỏi của Quoc Tran Anh Le

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Nguyễn Trọng Chiến · Accepted Answer

139:Đặt $x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\left(a,b,c>0\right)$GT $\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{3}{abc}\Rightarrow a+b+c=3$$\Rightarrow\dfrac{y^2}{xy^2+2x^2}=\dfrac{1}{b^2}:\left(\dfrac{1}{ab^2}+\dfrac{2}{a^2}\right)=\dfrac{1}{b^2}:\left(\dfrac{a+2b^2}{a^2b^2}\right)=\dfrac{a^2}{a+2b^2}=a-\dfrac{2ab^2}{a+2b^2}\ge a-\dfrac{2ab^2}{3b\sqrt[3]{ab}}=a-\dfrac{2}{3}\sqrt[3]{a^2b^2}\ge a-\dfrac{2}{9}\left(a+b+ab\right)$ Tương tự ta được: $\dfrac{x^2}{zx^2+2z^2}=\dfrac{c^2}{c+2a^2}=c-\dfrac{2ca^2}{c+2a^2}\ge c-\dfrac{2}{9}\left(c+a+ac\right)$$\dfrac{z^2}{yz^2+2y^2}=\dfrac{b^2}{b+2c^2}=b-\dfrac{2bc^2}{b+2c^2}\ge b-\dfrac{2}{9}\left(b+c+bc\right)$$\Rightarrow\dfrac{y^2}{xy^2+2x^2}+\dfrac{x^2}{zx^2+2z^2}+\dfrac{z^2}{yz^2+2z^2}\ge\left(a+b+c\right)-\dfrac{2}{9}\left(2a+2b+2c+ab+bc+ca\right)$ $\ge3-\dfrac{2}{9}\left[6+\dfrac{\left(a+b+c\right)^2}{3}\right]=3-\dfrac{2}{9}\left(6+\dfrac{9}{3}\right)=3-\dfrac{2}{9}\cdot9=1$Dấu bằng xảy ra $\Leftrightarrow a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=3$Câu trả lời: 139:Đặt $x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\left(a,b,c>0\right)$GT $\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{3}{abc}\Rightarrow a+b+c=3$$\Rightarrow\dfrac{y^2}{xy^2+2x^2}=\dfrac{1}{b^2}:\left(\dfrac{1}{ab^2}+\dfrac{2}{a^2}\right)=\dfrac{1}{b^2}:\left(\dfrac{a+2b^2}{a^2b^2}\right)=\dfrac{a^2}{a+2b^2}=a-\dfrac{2ab^2}{a+2b^2}\ge a-\dfrac{2ab^2}{3b\sqrt[3]{ab}}=a-\dfrac{2}{3}\sqrt[3]{a^2b^2}\ge a-\dfrac{2}{9}\left(a+b+ab\right)$ Tương tự ta được: $\dfrac{x^2}{zx^2+2z^2}=\dfrac{c^2}{c+2a^2}=c-\dfrac{2ca^2}{c+2a^2}\ge c-\dfrac{2}{9}\left(c+a+ac\right)$$\dfrac{z^2}{yz^2+2y^2}=\dfrac{b^2}{b+2c^2}=b-\dfrac{2bc^2}{b+2c^2}\ge b-\dfrac{2}{9}\left(b+c+bc\right)$$\Rightarrow\dfrac{y^2}{xy^2+2x^2}+\dfrac{x^2}{zx^2+2z^2}+\dfrac{z^2}{yz^2+2z^2}\ge\left(a+b+c\right)-\dfrac{2}{9}\left(2a+2b+2c+ab+bc+ca\right)$ $\ge3-\dfrac{2}{9}\left[6+\dfrac{\left(a+b+c\right)^2}{3}\right]=3-\dfrac{2}{9}\left(6+\dfrac{9}{3}\right)=3-\dfrac{2}{9}\cdot9=1$Dấu bằng xảy ra $\Leftrightarrow a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=3$

Khôi Nguyênx · Answer

câu trả lời :Đặt ⇒1ab+1bc+1ca=3abc⇒a+b+c=3⇒1ab+1bc+1ca=3abc⇒a+b+c=3Câu trả lời: câu trả lời :Đặt ⇒1ab+1bc+1ca=3abc⇒a+b+c=3⇒1ab+1bc+1ca=3abc⇒a+b+c=3

Pham Tien Dat · Answer

141. x > -5Đặt $\left\{{}\begin{matrix}2x^2+5x+12=a\2x^2+3x+2=b\end{matrix}\right.\left(a,b>0\right)$pt $\Leftrightarrow a+b=\dfrac{a^2-b^2}{2}$$\Leftrightarrow\left(a+b\right)\left(1-\dfrac{a-b}{2}\right)=0$$\Leftrightarrow a-b=2$ (vì $\sqrt{2x^2+5x+12}+\sqrt{2x^2+3x+2}>0$)$\Rightarrow\sqrt{2x^2+5x+12}=\sqrt{2x^2+3x+2}+2$$\Rightarrow x+3=2\sqrt{2x^2+3x+2}$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\x=-1\end{matrix}\right.$ (thỏa mãn)Vậy ..... Câu trả lời: 141. x > -5Đặt $\left\{{}\begin{matrix}2x^2+5x+12=a\2x^2+3x+2=b\end{matrix}\right.\left(a,b>0\right)$pt $\Leftrightarrow a+b=\dfrac{a^2-b^2}{2}$$\Leftrightarrow\left(a+b\right)\left(1-\dfrac{a-b}{2}\right)=0$$\Leftrightarrow a-b=2$ (vì $\sqrt{2x^2+5x+12}+\sqrt{2x^2+3x+2}>0$)$\Rightarrow\sqrt{2x^2+5x+12}=\sqrt{2x^2+3x+2}+2$$\Rightarrow x+3=2\sqrt{2x^2+3x+2}$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\x=-1\end{matrix}\right.$ (thỏa mãn)Vậy .....

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