1. Ta có: \(f\left(x\right)=9x^2-12x+1=\left(3x\right)^2-2.3x.2+2^2-3\)

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

Vì \(\left(3x-2\right)^2\ge0\) với mọi x \(\Rightarrow\left(3x-2\right)^2-3\ge-3\) hay \(f\left(x\right)\ge-3\)

Dấu ''='' xảy ra \(\Leftrightarrow\left(3x-2\right)^2=0\Rightarrow3x-2=0\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)

Vậy min f(x) =-3 khi \(x=\dfrac{2}{3}\)

2. Ta có: \(f\left(x\right)=2x^2-7x+5=2.\left(x^2-3,5x\right)+5=2.\left(x^2-2.x.1,75+1,75^2\right)-2.1,75^2+5\)

\(=2.\left(x-1,75\right)^2-1,125\)

Vì \(2.\left(x-1,75\right)^2\ge0\Rightarrow2.\left(x-1,75\right)^2-1,125\ge-1,125\Rightarrow f\left(x\right)\ge-1,125\)

Dấu ''='' xảy ra \(\Leftrightarrow2.\left(x-1,75\right)^2=0\Rightarrow x-1,75=0\Rightarrow x=1,75\)

Vậy min f(x)=-1,125 khi x=1,75

3.\(3x^2-10x=3.\left(x^2-\dfrac{10}{3}x\right)=3.\left(x^2-2.x.\dfrac{5}{3}\right)\)

\(=3.\left[x^2-2.x.\dfrac{5}{3}+\left(\dfrac{5}{3}\right)^2\right]-3.\left(\dfrac{5}{3}\right)^2\)

\(=3.\left(x-\dfrac{5}{3}\right)^2-\dfrac{25}{3}\)

Vì \(3.\left(x-\dfrac{5}{3}\right)^2\ge0\Rightarrow3.\left(x-\dfrac{5}{3}\right)^2-\dfrac{25}{3}\ge-\dfrac{25}{3}\Rightarrow f\left(x\right)\ge-\dfrac{25}{3}\)

Dấu ''='' xảy ra \(\Leftrightarrow3.\left(x-\dfrac{5}{3}\right)^2=0\Rightarrow x-\dfrac{5}{3}=0\Rightarrow x=\dfrac{5}{3}\)

Vậy min f(x)=\(-\dfrac{25}{3}\) khi \(x=\dfrac{5}{3}\)

a) có nghĩa khi \(x-1\ne0\Rightarrow x\ne1\)

b)\(f\left(7\right)=\frac{7+2}{7-1}=\frac{9}{6}\)

c)\(f\left(x\right)=\frac{x+2}{x-1}=\frac{1}{4}\Leftrightarrow x+2=4x-4\)

\(\Leftrightarrow-3x=-6\Leftrightarrow x=2\)

e)\(f\left(x\right)>1\Rightarrow\frac{x+2}{x-1}-1>0\)

\(\Rightarrow\frac{3}{x-1}>0\) thấy 3>0 nên x-1>0 =>x>1

Bài 2:

a)\(P=9-2\left|x-3\right|\)

Thấy: \(\left|x-3\right|\ge0\)\(\Rightarrow2\left|x-3\right|\ge0\)

\(\Rightarrow-2\left|x-3\right|\le0\)

\(\Rightarrow9-2\left|x-3\right|\le9\)

Khi x=3

b)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(Q=\left|x-2\right|+\left|x-8\right|\)

\(=\left|x-2\right|+\left|8-x\right|\)

\(\ge\left|x-2+8-x\right|=6\)

Khi \(2\le x\le8\)

bài 1 bựa nớ ta làm rồi chj nựa

f(x)= 2x²-7x+1

\(=2\left(x^2-\frac{7x}{2}+\frac{1}{2}\right)\)

\(=2\left(x^2-\frac{7x}{2}+\frac{49}{16}\right)-\frac{41}{8}\)

\(=2\left(x-\frac{7}{4}\right)^2-\frac{41}{8}\ge0-\frac{41}{8}=-\frac{41}{8}\)

Dấu = khi \(2\left(x-\frac{7}{4}\right)^2=0\Leftrightarrow x-\frac{7}{4}\Leftrightarrow x=\frac{7}{4}\)

Vậy...