'=Bài 3:

\(Y=\left(x^{100}+1+1+1+1+1+1+1+1+1\right)-10x^{10}+1\)

Áp dụng BĐT Cauchy cho 10 số không âm ta có:

\(x^{100}+1+1+1+1+1+1+1+1+1\ge10\sqrt{x^{100}}=10x^{10}\)

\(Y\ge10x^{10}-10x^{10}+1=1\)

\(\Rightarrow maxY=1\)

Dấu "=" xảy ra\(\Leftrightarrow x^{100}=1\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Bài 2 : 

Tìm min : Bình phương 

Tìm max : Dùng B.C.S ( bunhiacopxki )

Bài 3 : Dùng B.C.S

KP9

nói thế thì đừng làm cho nhanh bạn ạ

Người ta cũng có chút tôn trọng lẫn nhau nhé đừng có vì dăm ba cái tích 

toàn 1 lũ hãm điểm

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 \(MinA=-4;MaxA=4\)

Ta có : A = x(x + 1)(x² +  x - 4)

= (x² + x)(x² + x - 4)

Đặt x² + x = t

Khi đó A = t(t - 4)

= t² - 4t = t² - 4t + 4 - 4 = (t - 2)² - 4 \(\ge\)-4

 Dấu "=" xảy ra <=> t - 2 = 0

=> t = 2

=> x² + x = 2

=> x² + x - 2 = 0

=> x² + 2x - x - 2 = 0

=> x(x + 2) - (x + 2) = 0

=> (x - 1)(x + 2) = 0

=> \(\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Vậy Min A = -4 <=> x \(\in\left\{1;-2\right\}\)

A = x( x + 1 )( x² + x - 4 )

= ( x² + x )( x² + x - 4 )

Đặt t = x² + x

A <=> t( t - 4 )

      = t² - 4t

      = ( t² - 4t + 4 ) - 4

      = ( t - 2 )² - 4 

      = ( x² + x - 2 )² - 4 ≥ -4 ∀ x

Đẳng thức xảy ra <=> x² + x - 2 = 0

                             <=> x² - x + 2x - 2 = 0

                             <=> x( x - 1 ) + 2( x - 1 ) = 0

                             <=> ( x - 1 )( x + 2 ) = 0

                             <=> \(\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

=> MinA = -4 <=> x = 1 hoặc x = -2

a,\(A=x\left(x+1\right)\left(x^2+x-4\right)\)

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

Đặt \(x^2+x=t\)ta có:

\(A=t\left(t-4\right)\)

\(=t^2-4t\)

\(=\left(t^2-4t+4\right)-4\)

\(=\left(t-2\right)^2-4\ge-4\forall t\)

Dấu "="xảy ra khi \(\left(t-2\right)^2=0\Rightarrow t=2\)

\(\Rightarrow Min_A=-4\Leftrightarrow t=2\)

\(\Leftrightarrow x^2+x=2\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow x=1;x=2\)

b,\(B=-x^2-y^2+xy+2x+2y\)

\(\Leftrightarrow-2B=2x^2+2y^2-2xy-4x-4y\)

\(=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2-4y+4\right)-8\)

\(=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\ge-8\Leftrightarrow B\le4\)

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

Vậy \(Max_B=4\Leftrightarrow x=y=2\)

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

\(y_{min}=-3\) 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 \(x=-2\)