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

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

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

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

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

\(B=1+\frac{2}{\left(x-3\right)^2+3}\le1+\frac{2}{3}\)

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

Dấu " = " xảy ra \(\Leftrightarrow x=3\)

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

=\(\frac{x^2-6x+9+3+2}{x^2-6x+9+3}\)

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

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

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

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

*Ta có:(x-3)² \(\ge\) 0;với mọi x;cộng 3 vào 2 vế

\(\Rightarrow\)(x-3)²+3 \(\ge\) 0+3;với mọi x

\(\Rightarrow\)(x-3)²+3 \(\ge\) 3;với mọi x

\(\Rightarrow\)\(\frac{2}{\left(x-3\right)^2+3}\) \(\ge\) 0;với mọi x;lấy hai vế cộng cho1​​

\(\Rightarrow\)​\(1+\frac{2}{\left(x-3\right)^2+3}\)​ \(\ge\)1+0;với mọi x

Vậy .................................

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

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: \(x^2\left(x-3\right)-4x+12\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

b: \(2a\left(x+y\right)+x+y=\left(x+y\right)\left(2a+1\right)\)

c: \(6x^2-12x-7x+14\)

\(=6x\left(x-2\right)-7\left(x-2\right)\)

\(=\left(x-2\right)\left(6x-7\right)\)

a) 3x - 10 = 2x + 13

<=> 3x - 2x = 13+10

<=> x = 23

      Vậy x = 23

b) x + 12 = -5 - x

<=> x + x = -5 - 12

<=> 2x = -17

<=> x = -17 : 2

<=> x = -8,5

      Vậy x = -8,5

c) x + 5 = 10 - x

<=> x + x = 10 - 5

<=> 2x = 5

<=> x = 5 : 2

<=> x = 2,5

      Vậy x = 2,5

d) 6x + 2 ^ 3 = 2x - 12

<=> 6x + 8 = 2x - 12

<=> 6x - 2x = -12 - 8

<=> 4x = -20

<=> x = -20 : 4

<=> x = -5

     Vậy x = -5

e) 12 - x = x + 1

<=> -x - x = 1 - 12

<=> -2x = -11

<=> x = -11 : (-2)

<=> x = 5,5

     Vậy x= 5,5

f) 14 - 4x = 3x + 20

<=> -4x - 3x = 20 - 14

<=> -7x = 6

<=> x = 6 : (-7)

<=> x = \(-\frac{6}{7}\)

7/48 - (1/2 x 2 + 1/6 x 4 + 1/8 x 5 + 1/12 x 7 + 1/14 x 8) : x = 0

7/48 - (1 + 2/3 + 5/8 + 7/12 + 4/7) : x = 0 (đã rút gọn)

7/48 - (336/336 + 224/336 + 210/336 + 196/336 + 192/336) : x = 0 (quy đồng)

7/48 - 193/56 : x  = 0

193/56 : x = 0 + 7/48

193/56 : x = 7/48

              x = 193/56 : 7/48

              x = 1158/49

tìm x à                    

|3x-4|+3x+5=0

\(3x\left(x-5\right)-3x^2=3x+8\)

<=>  \(3x^2-15x-3x^2=3x+8\)

<=>  \(18x=-8\)

<=>  \(x=-\frac{4}{9}\)

Vậy....

\(A=-2x^2+5x-8\)

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

\(A=-2\left(x^2-2\cdot x\cdot\frac{5}{4}+\frac{25}{16}+\frac{39}{16}\right)\)

\(A=-2\left[\left(x-\frac{5}{4}\right)^2+\frac{39}{16}\right]\)

\(A=-2\left(x-\frac{5}{4}\right)^2-\frac{39}{6}\le\frac{-39}{6}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{5}{4}\)

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

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

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

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

\(2B=-\left[\left(x-y\right)^2+\left(x+2\right)^2+\left(y+2\right)^2-8\right]\)

\(B=-\frac{\left(x-y\right)^2+\left(x+2\right)^2+\left(y+2\right)^2}{2}+4\le4\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=-2\)

\(C=\frac{3}{4x^2-4x+5}=\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\forall x\)

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

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

\(=1+\frac{2}{\left(x-3\right)^2+3}\le1+\frac{2}{3}=\frac{5}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=3\)