Question

[Trần Khánh Hoài]

a, cho x+y=2 . tìm GTNN của A=x²+2y²+x - 2y+1

b, cho x+2y=1 . tìm GTLN của B=x²-5y²+3x - y - 2

c, cho x²+y²=4. tìm GTNN của D=x⁶+y⁶+x⁴+y⁴

^{PLZ ! HELP MEE!}

giúp mình với! 2 tuần nữa thi cấp 3 rồi :((

Answer 1

[Nguyễn Việt Lâm]

Câu 1:

\(x+y=2\Rightarrow y=2-x\)

\(\Rightarrow A=x^2+2\left(2-x\right)^2+x-2\left(2-x\right)+1\)

\(A=x^2+2x^2-8x+8+x-4+2x+1\)

\(A=3x^2-5x+5\)

\(A=3\left(x^2-2.\frac{5}{6}x+\frac{25}{36}\right)+\frac{35}{12}\)

\(A=3\left(x-\frac{5}{6}\right)^2+\frac{35}{12}\ge\frac{35}{12}\)

\(\Rightarrow A_{min}=\frac{35}{12}\) khi \(x=\frac{5}{6}\) ; \(y=\frac{7}{6}\)

Câu 2:

\(x+2y=1\Rightarrow x=1-2y\)

\(\Rightarrow B=\left(1-2y\right)^2-5y^2+3\left(1-2y\right)-y-2\)

\(B=4y^2-4y+1-5y^2+3-6y-y-2\)

\(B=-y^2-11y+2\)

\(B=-\left(y^2+11y+\frac{121}{4}\right)+\frac{129}{4}\)

\(B=-\left(y+\frac{11}{2}\right)^2+\frac{129}{4}\le\frac{129}{4}\)

\(\Rightarrow B_{max}=\frac{129}{4}\) khi \(\left\{{}\begin{matrix}y=-\frac{11}{2}\\x=12\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

Câu 3:

Ta có:

\(x^2+y^2\ge2\sqrt{x^2y^2}=2\left|xy\right|\Rightarrow2\left|xy\right|\le4\Rightarrow\left|xy\right|\le2\Rightarrow x^2y^2\le4\)

\(D=\left(x^2\right)^3+\left(y^2\right)^3+x^4+y^4\)

\(D=\left(x^2+y^2\right)\left[\left(x^2+y^2\right)^2-3x^2y^2\right]+\left(x^2+y^2\right)^2-2x^2y^2\)

\(D=4\left(16-3x^2y^2\right)+16-2x^2y^2\)

\(D=80-14x^2y^2\ge80-14.4=24\)

\(\Rightarrow D_{min}=24\) khi \(\left\{{}\begin{matrix}x^2=2\\y^2=2\end{matrix}\right.\)