\(A=\frac{2x+1}{x^2+2}\) 

Ta có: \(\hept{\begin{cases}2x+1\ge1\\x^2+2\ge2\end{cases}}\)

Để 2x+ 1 nhỏ nhất => 2x+ 1=1

     x²+ 2 nhỏ nhất => x²+ 2= 2

\(\Rightarrow A=\frac{0+1}{0+1}=\frac{1}{2}=0,5\)

Vậy GTNN của A= 0,5

Ax^2+2A=2x+1

\(\Leftrightarrow Ax^2-2x+2A-1=1\)(*) A=0 <=>-2x-1=0=> luon co nghiem x

\(A\ne0\)(*) co nghiem can

delta(x)=1-A.(2a-1)>=0

\(\Leftrightarrow1-2a^2+a\ge0\Leftrightarrow2a^2-a-1\le0\Leftrightarrow\left(a-1\right)\left(a+\frac{1}{2}\right)\le0\)

\(-\frac{1}{2}\le A\le1\)

A = \(\frac{2x+1}{x^2+2}\)= \(\frac{x^2+2-x^2+2x-1}{x^2+2}\)= \(\frac{x^2+2}{x^2+2}-\frac{x^2-2x+1}{x^2+2}\)= \(1-\frac{\left(x-1\right)^2}{x^2+2}\)

Vì \(\frac{\left(x-1\right)^2}{x^2+2}\)\(\ge\) 0 với mọi x nên để A có GTLN \(\Rightarrow\) \(\frac{\left(x-1\right)^2}{x^2+2}\)=0 \(\Rightarrow\)\(\left(x-1\right)^2=0\Rightarrow x-1=0\Leftrightarrow x=1\)

Max : với x = 0 thì \(A=\frac{x^2}{x^4+x^2+1}=0\)

với x khác 0 thì x⁴ + 1 \(\ge\)2x² > 0 nên x⁴ + x² + 1 \(\ge\)3x² 

\(\Rightarrow\)\(A=\frac{x^2}{x^4+x^2+1}\le\frac{x^2}{3x^2}=\frac{1}{3}\)

Vậy max A = \(\frac{1}{3}\)\(\Leftrightarrow\)x = 1 hoặc -1

Min : Ta có : x⁴ + x² + 1 = ( x²+ 1 )² - x² = ( x² - x + 1 ) ( x² + x + 1 ) > 0 

\(\Rightarrow\)\(A\ge0\)( vì x² \(\ge\)0 )

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

\(x^6+y^6=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=a^2-ab+b^2\)

\(\ge a^2-\frac{a^2+b^2}{2}+b^2=\frac{a^2+b^2}{2}\ge\frac{\left(a+b\right)^2}{4}=\frac{1}{4}\)

Vậy Min = 1/4 khi \(x=y=\frac{1}{\sqrt{2}}\)
Ta có

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

+) Đặt x+y=S, xy = P, ta được: \(S^2-2P=1\)
+)\(x^6+y^6=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)=x^4-x^2y^2+y^4=\left(x^2+y^2\right)^2-3x^2y^2\)

\(=\left[\left(x+y\right)^2-2xy\right]^2-3x^2y^2=\left(S^2-2P\right)^2-3P^2=S^4-4S^2P+4P^2-3P^2\)

\(=S^4-4S^2P+P^2=\left(2P+1\right)^2-4\left(2P+1\right)P+P^2\)

\(=4P^2+4P+1-8P^2-4P+P^2=-3P^2+1\le1\)

Dấu = xảy ra khi \(\hept{\begin{cases}P=0\\S=1\end{cases}}\), khi đó x=1, y=0 hoặc x=0, y=1

a) Đặt  \(A=-x^2+9x-12\)

\(-A=x^2-9x+12\)

\(-A=\left(x^2-9x+\frac{81}{4}\right)-\frac{33}{4}\)

\(-A=\left(x-\frac{9}{2}\right)^2-\frac{33}{4}\)

Mà  \(\left(x-\frac{9}{2}\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge-\frac{33}{4}\Leftrightarrow A\le\frac{33}{4}\)

Dấu "=" xảy ra khi :  \(x-\frac{9}{2}=0\Leftrightarrow x=\frac{9}{2}\)

Vậy  \(A_{Max}=\frac{33}{4}\Leftrightarrow x=\frac{9}{2}\)

b) Đặt \(B=2x^2+10x-1\)

\(B=2\left(x^2+5x+\frac{25}{4}\right)-\frac{29}{4}\)

\(B=2\left(x+\frac{5}{2}\right)^2-\frac{29}{4}\)

Mà  \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\Rightarrow2\left(x+\frac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow B\ge-\frac{29}{4}\)

Dấu "=" xảy ra khi :  \(x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy  \(B_{Min}=-\frac{29}{4}\Leftrightarrow x=-\frac{5}{2}\)

c) Đặt  \(C=\left(2x+6\right)\left(x-1\right)\)

\(C=2x^2-2x+6x-6\)

\(C=2x^2+4x-6\)

\(C=2\left(x^2+2x+1\right)-8\)

\(C=2\left(x+1\right)^2-8\)

Mà  \(\left(x+1\right)^2\ge0\forall x\Rightarrow2\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow C\ge-8\)

Dấu "=" xảy ra khi :  \(x+1=0\Leftrightarrow x=-1\)

Vậy  \(C_{Min}=-8\Leftrightarrow x=-1\)

d) Đặt  \(D=3x-2x^2\)

\(-2D=4x^2-6x\)

\(-2D=\left(4x^2-6x+\frac{9}{4}\right)-\frac{9}{4}\)

\(-2D=\left(2x-\frac{3}{2}\right)^2-\frac{9}{4}\)

Mà  \(\left(2x-\frac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow-2D\ge-\frac{9}{4}\)

\(\Leftrightarrow D\le\frac{9}{8}\)

Dấu "=" xảy ra khi :  \(2x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{4}\)

Vậy  \(D_{Max}=\frac{9}{8}\Leftrightarrow x=\frac{3}{4}\)

\(P=\dfrac{\sqrt{x-2}}{x}\)ĐK:\(x\ne0\)

\(P^2=\dfrac{x-2}{x^2}\)

\(\Rightarrow P^2x^2-x+2=0\)

Để pt có ng₀ thì:

\(1^2-8P^2\ge0\)

\(\Leftrightarrow P\le\dfrac{1}{2\sqrt{2}}\)

Vậy P max =\(\dfrac{1}{2\sqrt{2}}.\)

(x+1)^2>=0 và (y-1)^2>=0

=>C>=-10

Dấu = xảy ra khi x+1=0,y-1=0

=>x=-1,y=1

Vậy C=-10 khi x=-1,y=1

k cho mk nha

\(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{cases}\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2-10\ge-10}\)

Dấu ''='' xảy ra <=> x = -1 ; y = 1

D lón nhất <=> (2x-1)^2+3 nhỏ nhất

Vì (2x-1)^2>=0

=>(2x-1)^2+3>=3

Dấu = xảy ra khi 2x-1=0

=>2x=1=>x=1/2

k cho mk nha