\(x+y=4xy\Rightarrow\frac{x+y}{xy}=\frac{1}{x}+\frac{1}{y}=4\)

\(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\Rightarrow4>=\frac{4}{x+y}\Rightarrow x+y>=1\)(bđt svacxo)

\(x^2+y^2>=\frac{\left(x+y\right)^2}{2};xy< =\frac{\left(x+y\right)^2}{4}\)

\(\Rightarrow P=x^2+y^2-xy>=\frac{\left(x+y\right)^2}{2}-\frac{\left(x+y\right)^2}{4}=\frac{\left(x+y\right)^2}{4}>=\frac{1^2}{4}=\frac{1}{4}\)

dấu = xảy ra khi \(x+y=1;x=y\Rightarrow x=y=\frac{1}{2}\left(tm\right)\)

vậy min P là \(\frac{1}{4}\)khi x=y=\(\frac{1}{2}\)

Answer:

3.

\(x^2+2y^2+2xy+7x+7y+10=0\)

\(\Rightarrow\left(x^2+2xy+y^2\right)+7x+7y+y^2+10=0\)

\(\Rightarrow\left(x+y\right)^2+7.\left(x+y\right)+y^2+10=0\)

\(\Rightarrow4S^2+28S+4y^2+40=0\)

\(\Rightarrow4S^2+28S+49+4y^2-9=0\)

\(\Rightarrow\left(2S+7\right)^2=9-4y^2\le9\left(1\right)\)

\(\Rightarrow-3\le2S+7\le3\)

\(\Rightarrow-10\le2S\le-4\)

\(\Rightarrow-5\le S\le-2\left(2\right)\)

Dấu " = " xảy ra khi: \(\left(1\right)\Rightarrow y=0\)

Vậy giá trị nhỏ nhất của \(S=x+y=-5\Rightarrow\hept{\begin{cases}y=0\\x=-5\end{cases}}\)

Vậy giá trị lớn nhất của \(S=x+y=-2\Rightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}\)

Em có cách này không biết có đúng không ạ,em mới lớp 7 thôi.

\(S=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2x^2-2xy+2y^2\)

Đặt \(S=2x^2-2xy+2y^2=a\left(x+y\right)^2+b\left(x-y\right)^2\) (ta đi tìm a, b)

Phân tích ra ta được: \(a\left(x+y\right)^2+b\left(x-y\right)^2\)

\(=ax^2+2xy.a+ay^2+bx^2-2xy.b+by^2\)

\(=\left(a+b\right)x^2+2xy\left(a-b\right)+\left(a+b\right)y^2\)

Đồng nhất hệ số ta được: \(\hept{\begin{cases}a+b=2\\a-b=-1\end{cases}}\).Giải ra ta tìm được: a = ¹/₂ và b = ³/₂

Do đó: \(S=2x^2-2xy+2y^2=\frac{1}{2}\left(x+y\right)^2+\frac{3}{2}\left(x-y\right)^2\ge\frac{1}{2}\left(x+y\right)^2=\frac{1}{2}.4=2\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-y\right)^2=0\\x+y=2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\x+y=2\end{cases}}\Leftrightarrow x=y=1\)

Vậy \(S_{min}=2\Leftrightarrow x=y=1\)

\(S=x^3+y^3\) 

\(=\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=2\left(x^2-xy+y^2\right)\ge2\left[\frac{\left(x+y\right)^2}{2}-\frac{\left(x+y\right)^2}{4}\right]=2\left(2-1\right)=2\)

Dấu '=' xảy ra khi \(\hept{\begin{cases}x=y\\x+y=2\end{cases}\Leftrightarrow x=y=1}\)

Vậy \(S_{min}=2\)khi \(x=y=1\)

:))

Ta co A = 2(x+y)+\(\frac{2}{x+y}\)\(\ge2\sqrt{2\left(x+y\right).\frac{2}{x+y}}\)^{=4          khi x=y =\(\frac{1}{2}\)}

Ta có: \(Q=\dfrac{2}{x^2+y^2}+\dfrac{3}{xy}=\dfrac{2}{x^2+y^2}+\dfrac{6}{2xy}=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{4}{2xy}\)

Áp dụng BĐT phụ: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Rightarrow2\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)\ge2\left(\dfrac{4}{x^2+2xy+y^2}\right)=2\left[\dfrac{4}{\left(x+y\right)^2}\right]=2.\dfrac{4}{4}=2\)

Dấu "=" xảy ra khi x=y=1

Áp dụng BĐT phụ: \(ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Rightarrow xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{2^2}{4}=1\)

Dấu"=" xảy ra khi x=y=1

\(\Rightarrow2xy\le2.1=2\)

\(\Rightarrow\dfrac{4}{2xy}\ge\dfrac{4}{2}=2\)

\(\Rightarrow Q=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{4}{2xy}=\dfrac{2}{x^2+y^2}+\dfrac{3}{xy}\ge2+2=4\)

Dấu"=" xảy ra khi x=y=1

Ta có: 3x + y = 1 => y = 1 - 3x

a, Thay y = 1 - 3x vào M, ta có:

\(\Rightarrow M=3x^2+\left(1-3x\right)^2=3x^2+1-6x+9x^2=12x^2-6x+1=3\left(4x^2-2x+\frac{1}{3}\right)\)

\(=3\left(4x^2-2x+\frac{1}{4}+\frac{1}{12}\right)=3\left(2x-\frac{1}{2}\right)^2+\frac{3}{12}=3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì \(\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-\frac{1}{2}=0\\3x+y=1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=1-3x=1-3.\frac{1}{4}=\frac{1}{4}\end{cases}}\)\(\Leftrightarrow x=y=\frac{1}{4}\)

Vậy GTNN M = 1/4 khi x = y = 1/4

b, Thay y = 1 - 3x vào N

\(\Rightarrow N=x\left(1-3x\right)=x-3x^2=-3\left(x^2-\frac{x}{3}+\frac{1}{36}-\frac{1}{36}\right)\)

\(=-3\left(x-\frac{1}{6}\right)^2-3.\left(-\frac{1}{36}\right)=-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\)

Vì \(\left(x-\frac{1}{6}\right)^2\ge0\forall x\)

\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2\le0\forall x\)

\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\le\frac{1}{12}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{6}=0\\3x+y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{6}\\y=1-3x=1-3.\frac{1}{6}=\frac{1}{2}\end{cases}}\)

Vậy GTLN N = 1/12 khi x = 1/6 và y = 1/2