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

Mà \(x+y=1\Leftrightarrow x=1-y\)

\(\Leftrightarrow A=1-\left(1-y\right)y=1-y+y^2=\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ A=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\\ A_{min}=\dfrac{3}{4}\Leftrightarrow x=y=\dfrac{1}{2}\)

Ai giúp mik vs

Huhu ai giúp vs

có: \(\dfrac{1}{x^2+y^2}=\dfrac{1}{\left(x+y\right)^2-2xy}=\dfrac{1}{1-2xy}\)(1)

có \(\dfrac{1}{xy}=\dfrac{2}{2xy}\left(2\right)\)

từ(1)(2)=>A=\(\dfrac{1}{1-2xy}+\dfrac{2}{2xy}\ge\dfrac{\left(1+\sqrt{2}\right)^2}{1}=\left(1+\sqrt{2}\right)^2\)

=>Min A=(1+\(\sqrt{2}\))^2

b, ta có : \(x+y=1=>2x+2y=2\)

\(B=\dfrac{1}{x^2+y^2}+\dfrac{3}{4xy}=\dfrac{4}{4x^2+4y^2}+\dfrac{6}{8xy}\)\(\ge\dfrac{\left(2+\sqrt{6}\right)^2}{\left(2x+2y\right)^2}\)

\(=\dfrac{\left(2+\sqrt{6}\right)^2}{2^2}=\dfrac{5+2\sqrt{6}}{2}\)=>\(B\ge\dfrac{5+2\sqrt{6}}{2}\)

=>\(MinB=\dfrac{5+2\sqrt{6}}{2}\)

A>=1/(1+xy)=1/2

Dấu = xảy ra khi x=y=1

Áp dụng bđt Cô-si \(1=x^2+y^2\ge2xy\)

              \(\Rightarrow xy\le\frac{1}{2}\)

Ta có \(A=\frac{-2xy}{1+xy}\ge\frac{-\frac{2.1}{2}}{1+\frac{1}{2}}=-\frac{2}{3}\)

\("="\Leftrightarrow x=y=\frac{1}{\sqrt{2}}\)

Từ gt ta có x^2+y^^2=xy+1

=>P=(x^2+y^2)^2-2x^2y^2-x^2y^2

=(xy+1)²-2x²y²-x²y²

=x²y²+xy+1-3x²y²=-2x²y²+xy+1

=......

\(1=x^2+y^2-xy\ge2xy-xy=xy\Rightarrow xy\le1\)

\(1=x^2+y^2-xy\ge-2xy-xy=-3xy\Rightarrow xy\ge-\dfrac{1}{3}\)

\(\Rightarrow-\dfrac{1}{3}\le xy\le1\)

\(P=\left(x^2+y^2\right)^2-2\left(xy\right)^2-\left(xy\right)^2=\left(xy+1\right)^2-3\left(xy\right)^2=-2\left(xy\right)^2+2xy+1\)

Đặt \(xy=t\in\left[-\dfrac{1}{3};1\right]\)

\(P=f\left(t\right)=-2t^2+2t+1\)

\(f'\left(t\right)=-4t+2=0\Rightarrow t=\dfrac{1}{2}\)

\(f\left(-\dfrac{1}{3}\right)=\dfrac{1}{9}\) ; \(f\left(\dfrac{1}{2}\right)=\dfrac{3}{2}\) ; \(f\left(1\right)=1\)

\(\Rightarrow P_{max}=\dfrac{3}{2}\) ; \(P_{min}=\dfrac{1}{9}\)

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đáp án a

Bài 1:

Ta thấy: $(x+\frac{1}{2})^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow (x+\frac{1}{2})^2+\frac{5}{4}\geq \frac{5}{4}$

Vậy gtnn của biểu thức là $\frac{5}{4}$

Giá trị này đạt tại $x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}$

Bài 2:

$x+y-3=0\Rightarrow x+y=3$
\(M=x^2(x+y)-(x+y)x^2-y(x+y)+4y+x+2019\)

\(=-3y+4y+x+2019=x+y+2019=3+2019=2022\)