Giá trị của \(x\left(2x+y\right)+xy+\frac{1}{2}y^2\) biết \(8x^3+12x^2y+6xy^2+y^3=27\)
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\(A=8x^3+12x^2y+6xy^2+y^3=27\)
\(\Leftrightarrow\left(2x+y\right)^3=27\)
=>2x+y=3
\(B=x\left(2x+y\right)+xy+\dfrac{1}{2}y^2\)
\(=3x+\dfrac{1}{2}y\left(2x+y\right)=3x+\dfrac{1}{2}y\cdot3=3x+\dfrac{3}{2}y\)
\(=\dfrac{3}{2}\left(2x+y\right)=\dfrac{3}{2}\cdot3=\dfrac{9}{2}\)
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\(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\le0\)
Nhận thấy: \(\left|2x+1\right|\ge0\); \(\left|x+y-\frac{1}{2}\right|\ge0\)
=> \(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\ge0\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x+1=0\\x+y-\frac{1}{2}=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=-\frac{1}{2}\\y=1\end{cases}}\)
đến đây bạn thay x,y tìm đc vào A để tính nhé
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Ta có :\(\frac{3\left(x+y\right)^2}{3\left(x-y\right)^2}=\frac{\left(x+y\right)^2}{\left(x-y\right)^2}=\frac{\left(x+y\right)\left(x+y\right)}{\left(x-y\right)\left(x-y\right)}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}\)
Thay x.y 1/2 vào ta được:
\(\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{x^2+1+y^2}{x^2-1+y^2}=\frac{x^2+2-1+y^2}{x^2-1+y^2}=\frac{x^2-1+y^2}{x^2-1+y^2}+\frac{2}{x^2-1+y^2}\)
\(=1+\frac{2}{x^2-1+y^2}\)
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\(\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1}{x^3+1}-\frac{3}{x^3+1}+\frac{3\left(x+1\right)}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1-3+3x+3}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn
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Áp dụng BĐT \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)
\(\Rightarrow P\ge\frac{1}{2}\left(2x+\frac{1}{x}+2y+\frac{1}{y}\right)^2=\frac{1}{2}\left[2\left(x+y\right)+\frac{1}{x}+\frac{1}{y}\right]^2\)
\(\Rightarrow P\ge\frac{1}{2}\left[2\left(x+y\right)+\frac{4}{x+y}\right]^2=18\)
\(\Rightarrow P_{min}=18\) khi \(x=y=\frac{1}{2}\)
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Đề: Biết \(8x^3+12x^2y+6xy^2+y^3=27\) . Tính \(A=x\left(2x+y\right)+xy+\frac{1}{2}y^2\)
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Ta có:
\(8x^3+12x^2y+6xy^2+y^3=27\)
\(\Leftrightarrow\) \(\left(2x+y\right)^3=27\)
\(\Leftrightarrow\) \(2x+y=3\)
Do đó:
\(A=3x+xy+\frac{1}{2}y^2\)
\(=3x+\frac{1}{2}y\left(2x+y\right)\)
\(=3x+\frac{3}{2}y\)
\(=\frac{3}{2}\left(2x+y\right)\)
\(A=\frac{9}{2}\)
\(8x^3+12x^2y+6xy^2+y^3=27\Leftrightarrow\left(2x+y\right)^3=27\Leftrightarrow2x+y=3\)
\(x\left(2x+y\right)+xy+\frac{1}{2}y^2=2x^2+2xy+\frac{1}{2}y^2=\left(\sqrt{2}x+\frac{1}{\sqrt{2}}y\right)^2\)
\(=\frac{1}{2}.2\left(\sqrt{2}x+\frac{1}{\sqrt{2}}y\right)^2=\frac{1}{2}.\left(2x+y\right)^2=\frac{1}{2}.3^2=\frac{9}{2}\)