\(B=\frac{x^2-6x+14}{x^2-6x+12}=\frac{x^2-6x+12+2}{x^2-6x+12}=1+\frac{2}{x^2-6x+12}\)

ta có: \(x^2-6x+12=x^2-2.3.x+3^2+4=\left(x-3\right)^2+4\ge4\)

để Bmax => \(\left(\frac{2}{x^2-6x+12}\right)max\Rightarrow x^2-6x+12min\)và lớn hơn 0 vì 2>0

mà \(\left(x-3\right)^2+4\) \(\ge\)4

dấu = xảy ra khi x-3=0

=> x=3

Vậy \(MaxB=\frac{3}{2}\)khi x=3

\(B=\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\Rightarrow B_{max}=\frac{3}{4}\) khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\)

2/ Xem lại đề bài, đề bài này thì ko có max, 12 ở mẫu là dấu + thì may ra làm được

1, B=\(\frac{3}{4x^2-4x+5}\)

=\(\frac{3}{\left(4x^2-2.2x+4\right)+5-4}\)

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

Để B=3 thì : (2x-2)²=0

\(\Leftrightarrow2x-2=0\)

\(\Leftrightarrow x=1\)

Vậy Max B =3 \(\Leftrightarrow x=1\)

a) \(A=\sqrt[]{x^2-2x+5}\)

\(\Leftrightarrow A=\sqrt[]{x^2-2x+1+4}\)

\(\Leftrightarrow A=\sqrt[]{\left(x+1\right)^2+4}\)

mà \(\left(x+1\right)^2\ge0,\forall x\in R\)

\(A=\sqrt[]{\left(x+1\right)^2+4}\ge\sqrt[]{4}=2\)

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

Vậy \(GTNN\left(A\right)=2\left(khi.x=-1\right)\)

b) \(B=5-\sqrt[]{x^2-6x+14}\)

\(\Leftrightarrow B=5-\sqrt[]{x^2-6x+9+5}\)

\(\Leftrightarrow B=5-\sqrt[]{\left(x-3\right)^2+5}\left(1\right)\)

Ta có : \(\left(x-3\right)^2\ge0,\forall x\in R\)

\(\Leftrightarrow\left(x-3\right)^2+5\ge5,\forall x\in R\)

\(\Leftrightarrow\sqrt[]{\left(x-3\right)^2+5}\ge\sqrt[]{5},\forall x\in R\)

\(\Leftrightarrow-\sqrt[]{\left(x-3\right)^2+5}\le-\sqrt[]{5},\forall x\in R\)

\(\Leftrightarrow B=5-\sqrt[]{\left(x-3\right)^2+5}\le5-\sqrt[]{5},\forall x\in R\)

Dấu "=" xả ra khi và chỉ khi \(x-3=0\Leftrightarrow x=3\)

Vậy \(GTLN\left(B\right)=5-\sqrt[]{5}\left(khi.x=3\right)\)

Có: \(C=\frac{1}{\sqrt{x^2-4x+5}}\)

\(\Leftrightarrow C=\frac{1}{\sqrt{\left(x-2\right)^2+1}}\)\(\le1\)

Vậy C_min=1 \(\Leftrightarrow x=2\)

Có: \(B=5-\sqrt{x^2-6x+14}\)

\(\Leftrightarrow B=5-\sqrt{\left(x-3\right)^2+5}\) \(\le5-\sqrt{5}\)

Vậy \(B_{min}=5-\sqrt{5}\Leftrightarrow x=3\)

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3

\(P=\frac{x^2-6x+14}{x^2-6x+12}=\frac{x^2-6x+12+2}{x^2-6x+12}=\frac{x^2-6x+12}{x^2-6x+12}+\frac{2}{x^2-6x+12}=1+\frac{2}{x^2-6x+12}\)

\(=1+\frac{2}{\left(x^2-6x+9\right)+3}=1+\frac{2}{\left(x^2-2.x.3+3^2\right)+3}=1+\frac{2}{\left(x-3\right)^2+3}\)

P lớn nhất \(\Leftrightarrow\) \(\frac{2}{\left(x-3\right)^2+3}\) lớn nhất \(\Leftrightarrow\left(x-3\right)^2+3\) nhỏ nhất

Ta có: \(\) \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2+3\ge3\)

\(\Rightarrow\frac{2}{\left(x-3\right)^2+3}\le\frac{2}{3}\)

Do đó GTLN của \(\frac{2}{\left(x-3\right)^2+3}\) là 2/3

=> GTLN của \(P=1+\frac{2}{3}=\frac{5}{3}\)

Dấu "=" xảy ra <=> x=3

a) \(3-x^2+5x=-\left(x^2-5x-3\right)\)

\(=-\left(x^2-2x.\frac{5}{2}+\frac{10}{4}-\frac{22}{4}\right)\)

\(=-\left(x-\frac{5}{2}\right)^2+\frac{22}{4}\)

\(=-\left(x-\frac{5}{2}\right)^2+\frac{11}{2}\)

Mà: \(\left(x-\frac{5}{2}\right)^2\ge0\)\(\Leftrightarrow-\left(x-\frac{5}{2}\right)^2\le0\)

\(\Leftrightarrow-\left(x-\frac{5}{2}\right)^2+\frac{11}{2}\le\frac{11}{2}\)

\(\Leftrightarrow3-x^2+5x\le\frac{11}{2}\)

Dấu = xảy ra khi: \(\left(x-\frac{5}{2}\right)^2=0\)

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

\(\Leftrightarrow x=\frac{5}{2}\)(T/m)

Vậy GTLN của 3 - x² + 5x là \(\frac{11}{2}\)khi x = \(\frac{5}{2}\).

b) \(12-6x^2-6x=-6\left(x^2+x-2\right)\)

\(=-6\left(x^2+2x.\frac{1}{2}+\frac{1}{4}-\frac{9}{4}\right)\)

\(=-6\left(x+\frac{1}{2}\right)^2+\frac{27}{2}\)

Mà: \(\left(x+\frac{1}{2}\right)^2\ge0\)\(\Leftrightarrow-6\left(x+\frac{1}{2}\right)^2\le0\)

\(\Leftrightarrow-6\left(x+\frac{1}{2}\right)^2+\frac{27}{2}\le\frac{27}{2}\)\(\Leftrightarrow12-6x^2-6x\le\frac{27}{2}\)

Dấu = xảy ra khi: \(\left(x+\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\frac{1}{2}=0\)\(\Leftrightarrow x=-\frac{1}{2}\)(T/m)

Vậy GTLN của 12 - 6x² - 6x là \(\frac{27}{2}\)khi x = \(-\frac{1}{2}\).