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

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

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 .

a/ Ta có ;

\(B=\left(\frac{2}{3}x-\frac{1}{5}\right)^2+3\)

Mà \(\left(\frac{2}{3}x-\frac{1}{5}\right)^2\ge0\)

\(\Leftrightarrow A\ge3\)

Để B đạt GTNN thì \(\left(\frac{2}{3}x-\frac{1}{5}\right)^2\) đạt GTNN

Dấu "=" xảy ra khi :

\(\left(\frac{2}{3}x-\frac{1}{5}\right)^2=0\)

\(\Leftrightarrow\frac{2}{3}x-\frac{1}{5}=0\)

\(\Leftrightarrow x=\frac{3}{10}\)

Vậy B đạt GTNN = 3 khi x = 3/10

b, tương tự

a) \(A=\sqrt{x-2}+\sqrt{6-x}\)

\(\Rightarrow A^2=x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}\)

Ta có \(\sqrt{\left(x-2\right)\left(6-x\right)}\ge0,\forall x\)

Do đó \(A^2=4+2\sqrt{\left(x-2\right)\left(6-x\right)}\ge4\)

Mà A không âm \(\Leftrightarrow A\ge2\)

Dấu "=" \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

Áp dụng BĐT Bunhiacopxky:

\(A^2=\left(\sqrt{x-2}+\sqrt{6-x}\right)^2\le\left(x-2+6-x\right)\left(1+1\right)=4\cdot2=8\)

\(\Leftrightarrow A\le\sqrt{8}\)

Dấu "=" \(\Leftrightarrow x-2=6-x\Leftrightarrow x=4\)

Mấy bài còn lại y chang nha 

Tick hộ nha

\(B=3x^2+3x-1\)

\(=3\left(x^2+x-\dfrac{1}{3}\right)\)

\(=3\left(x^2+x+\dfrac{1}{4}-\dfrac{7}{12}\right)\)

\(=3\left(x+\dfrac{1}{2}\right)^2-\dfrac{7}{4}>=-\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x+1/2=0

=>\(x=-\dfrac{1}{2}\)

\(C=-2x^2+7x+3\)

\(=-2\left(x^2-\dfrac{7}{2}x-\dfrac{3}{2}\right)\)

\(=-2\left(x^2-2\cdot x\cdot\dfrac{7}{4}+\dfrac{49}{16}-\dfrac{73}{16}\right)\)

\(=-2\left(x-\dfrac{7}{4}\right)^2+\dfrac{73}{8}< =\dfrac{73}{8}\forall x\)

Dấu '=' xảy ra khi x-7/4=0

=>x=7/4

a đưa về bình phương r` dùng cauchy, b cx v