Đặt \(a=x^2;b=y^2\left(a;b\ge0\right)\)

\(A=\frac{\left(a-b\right)\left(1-ab\right)}{\left(1+a\right)^2\left(1+b\right)^2}\)

\(\left|A\right|=\frac{\left|\left(a-b\right)\left(1-ab\right)\right|}{\left(1+a\right)^2\left(1+b^2\right)}\le\frac{\left(a+b\right)\left(1+ab\right)}{\left(1+a\right)^2\left(1+b\right)^2}\)

\(\left(1+a\right)\left(1+b\right)=\left(a+b\right)+\left(1+ab\right)\ge2\sqrt{\left(a+b\right)\left(1+ab\right)}\)

\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\ge4\left(a+b\right)\left(1+ab\right)\)

\(\Rightarrow\left|A\right|\le4\)

\(\Rightarrow-4\le A\le4\)

\(A=-4\Leftrightarrow a=0;b=1\Leftrightarrow x=0;y=+1or-1\)

\(A=4\Leftrightarrow a=1;b=0\Leftrightarrow x=+-1;y=0\)

Vậy \(MinA=-4;MaxA=4\)

pt <=> \(\left(x^2+y^2\right)^2-4\left(x^2+y^2\right)+3=-x^2\le0\) (1)

(1)<=> \(A^2-4A+3\le0\Leftrightarrow1\le A\le3\)

Vậy GTNN của A là  1 tại x = 0 y =+- 1

GTLN của  A là 3 tại x = 0 ; y= +-căn3

thank nha thắng 

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\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)

\(y_{min}=-3\) khi \(x=1\)

\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)

\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)

\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)

\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)

\(x^4+2x^2y^2+y^4-3x^2-4y^2+4=1\)

\(\Leftrightarrow\left(x^2+y^2\right)^2-4\left(x^2+y^2\right)+4=1-x^2\)

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

Do \(1-x^2\le1\) \(\forall x\)

\(\Rightarrow-1\le x^2+y^2-2\le1\)

\(\Rightarrow1\le x^2+y^2\le3\)

\(A_{min}=1\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm1\end{matrix}\right.\)

\(A_{max}=3\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm\sqrt{3}\end{matrix}\right.\)

\(A\)xác định \(\Leftrightarrow x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\ne0\)

\(\Leftrightarrow x^2y^2+1+x^2-x^2y-y+y^2\ne0\)

\(\Leftrightarrow\left(x^2y^2+y^2\right)+\left(x^2+1\right)-\left(x^2y+y\right)\ne0\)

\(\Leftrightarrow y^2\left(x^2+1\right)+\left(x^2+1\right)-y\left(x^2+1\right)\ne0\)

\(\Leftrightarrow\left(x^2+1\right)\left(y^2-y+1\right)\ne0\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\)

Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(y-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall y\end{cases}}\)\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\forall x;y\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\forall x;y\)

\(\Leftrightarrow A\ne0\forall x;y\)

1: \(A=\left(x-1\right)^2-10\ge-10\)

Dấu '=' xảy ra khi x=1

2: \(B=-\left|x-1\right|-2\cdot\left(2y-1\right)^2+100\le100\)

Dấu '=' xảy ra khi x=1 và y=1/2

`(x-1)^2 >=0 => (x-1)^2 - 10 >= -10`

Dấu bằng xảy ra khi `x = 1`.

Vì `-|x-1| <=0, -2(2y-1)^2 <= 0`

`=> -|x-1| - 2(2y-1)^2 + 100 <= 100`

Dấu bằng xảy ra `<=> x = 1, y = 1/2`.