1;\(Q=5-3\left(2x-1\right)^2\)

Có \(3\left(2x-1\right)^2\ge0\)

\(\Rightarrow Q\le5-0=5\)

Dấu "=" xảy ra khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\)

Vậy Max Q = 5 <=> x = 1/2

2;\(M=\frac{x^2+8}{x^2+2}=1+\frac{6}{x^2+2}\)

Để M đạt GTLN \(\Rightarrow\frac{6}{x^2+2}\)phải lớn nhất

\(\Rightarrow x^2+2\)phải đạt GTNN

Mà \(x^2+2\ge2\Leftrightarrow x=0\)

Vậy \(M\ge1+\frac{6}{2}=1+3=4\)(x = 0)

\(1,\dfrac{1}{1+x}=1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)

Cmtt: \(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(1+x\right)\left(1+z\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)

Nhân VTV

\(\Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\dfrac{x^2y^2z^2}{\left(1+x\right)^2\left(1+y\right)^2\left(1+z\right)^2}}\\ \Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\dfrac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\\ \Leftrightarrow8xyz\le1\Leftrightarrow xyz\le\dfrac{1}{8}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{2}\)

\(2,\\ a,2x^2+y^2-2xy=1\\ \Leftrightarrow\left(x-y\right)^2+x^2=1\\ \Leftrightarrow\left(x-y\right)^2=1-x^2\ge0\\ \Leftrightarrow x^2\le1\Leftrightarrow\sqrt{x^2}\le1\Leftrightarrow\left|x\right|\le1\)

a)x²+2x+4+1=(x+1)²+1

ma (x+1)² >0

nen (x+1)²+1>1

vay x²+2x+5 min la 1 khi x=-1

M=\(\frac{x^2+10x-7}{x^2+2x+1}=\frac{x^2+10x+25-32}{x^2+2x+1}=\frac{\left(x+5\right)^2-32}{\left(x+1\right)^2}\)

\(\Rightarrow\frac{\left(x+5\right)^2-32}{\left(x+1\right)^2}\le-32\)

Vay Max la -32 

Mk cx k chắc lắm đâu .

\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)

\(=3+\dfrac{2}{x^2-2x+5}\)

Mà \(x^2-2x+5\ge4\)

=> \(\dfrac{2}{x^2-2x+5}\le\dfrac{1}{2}\)

=> A ≤ 7/2

Dấu "=" xảy ra ⇔ x = 1

Ta có : \(A=\dfrac{3x^2-6x+17}{x^2-2x+5}=\dfrac{3x^2-6x+15+2}{x^2-2x+5}=\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)

\(=3+\dfrac{2}{x^2-2x+5}\)

- Thấy : \(x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)

Lại có : \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-1\right)^2+4\ge4\forall x\)

\(\Rightarrow\dfrac{2}{x^2-2x+5}\le\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Rightarrow3+\dfrac{2}{x^2-2x+5}\le\dfrac{7}{2}\)

\(HayA\le\dfrac{7}{2}\)

Vậy Max_A = \(\dfrac{7}{2}\) Dấu " = " xảy ra <=> x - 1 = 0

<=> x = 1 .