1,\(f\left(x\right)=3x^2-2x-7\)

\(=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{22}{3}\)

\(=2\left(x-\dfrac{1}{3}\right)^2-\dfrac{22}{3}\ge-\dfrac{22}{3}\forall x\)

Vậy GTNN của biểu thức là \(-\dfrac{22}{3}\) khi \(x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)

\(b,f\left(x\right)=5x^2+7x=5\left(x^2+\dfrac{7}{5}x+\dfrac{49}{100}\right)-\dfrac{49}{20}\)\(=5\left(x+\dfrac{7}{10}\right)^2-\dfrac{49}{20}\ge-\dfrac{49}{20}\forall x\)

Vậy Giá trị nhỏ nhất của biểu thức là \(-\dfrac{49}{20}\) khi \(x+\dfrac{7}{10}=0\Rightarrow x=-\dfrac{7}{10}\)

\(c,f\left(x\right)=-5x^2+9x-2=-5\left(x^2-\dfrac{9}{5}x+\dfrac{81}{100}\right)+\dfrac{41}{20}\)\(=-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{41}{20}\le\dfrac{41}{20}\forall x\)

Vậy GTLN của biểu thức là \(\dfrac{41}{20}\) khi \(x-\dfrac{9}{10}=0\Rightarrow x=\dfrac{9}{10}\)

\(d,f\left(x\right)=-7x^2+3x=-7\left(x^2-\dfrac{3}{7}x+\dfrac{9}{196}\right)+\dfrac{9}{28}\)\(=-7\left(x-\dfrac{3}{14}\right)^2+\dfrac{9}{28}\le\dfrac{9}{28}\forall x\)

Vậy GTLN của biểu thức là \(\dfrac{9}{28}\) khi \(x-\dfrac{3}{14}=0\Rightarrow x=\dfrac{3}{14}\)

1/ \(f\left(x\right)=3x^2-2x-7\)

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

\(=3\left(x^2-\dfrac{2}{3}+\dfrac{1}{9}-\dfrac{64}{9}\right)\)

\(=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{64}{3}\)

Ta có: \(3\left(x-\dfrac{1}{3}\right)^2\ge0\forall x\Rightarrow3\left(x-\dfrac{1}{3}\right)^2-\dfrac{64}{3}\ge-\dfrac{64}{3}\forall x\)

Dấu "=" xảy ra khi \(x-\dfrac{1}{3}=0\) hay \(x=\dfrac{1}{3}\)

Vậy MIN_f(x) = \(-\dfrac{64}{3}\) khi x = \(\dfrac{1}{3}\).

2/ \(f\left(x\right)=5x^2+7x\)

\(=5\left(x^2+\dfrac{7}{5}x\right)=5\left(x^2+\dfrac{7}{5}x+\dfrac{49}{100}-\dfrac{49}{100}\right)\)

\(=5\left(x+\dfrac{7}{10}\right)^2-\dfrac{49}{20}\)

Ta có: \(5\left(x+\dfrac{7}{10}\right)^2\ge0\forall x\Rightarrow5\left(x+\dfrac{7}{10}\right)^2-\dfrac{49}{20}\ge-\dfrac{49}{20}\forall x\)

Dấu "=" xảy ra khi \(x+\dfrac{7}{10}=0\) hay \(x=-\dfrac{7}{10}\)

Vậy MIN_f(x) = \(-\dfrac{49}{20}\) khi x = \(-\dfrac{7}{10}\).

1/ \(f\left(x\right)=-5x^2+9x-2\)

\(=-5\left(x^2-\dfrac{9}{5}x+\dfrac{2}{5}\right)\)

\(=-5\left(x^2-\dfrac{9}{5}x+\dfrac{81}{100}-\dfrac{41}{100}\right)\)

\(=-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{41}{20}\)

Ta có: \(-5\left(x-\dfrac{9}{10}\right)^2\le0\forall x\Rightarrow-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{41}{20}\le\dfrac{41}{20}\forall x\)

Dấu "=" xảy ra khi \(x-\dfrac{9}{10}=0\) hay \(x=\dfrac{9}{10}\)

Vậy MAX_f(x) = \(\dfrac{41}{20}\) khi x = \(\dfrac{9}{10}\)

2/ \(f\left(x\right)=-7x^2+3x=-7\left(x^2-\dfrac{3}{7}x+\dfrac{9}{196}\right)+\dfrac{9}{28}\)

\(=-7\left(x-\dfrac{3}{14}\right)^2+\dfrac{9}{28}\)

Ta có: \(-7\left(x-\dfrac{3}{14}\right)^2\le0\forall x\Rightarrow-7\left(x-\dfrac{3}{14}\right)^2+\dfrac{9}{28}\le\dfrac{9}{28}\forall x\)

Dấu "=" xảy ra khi \(x-\dfrac{3}{14}=0\) hay x = \(\dfrac{3}{14}\)

Vậy MAX_f(x) = \(\dfrac{9}{28}\) khi x = \(\dfrac{3}{14}\).

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

\(f\left(x\right)=\left|x-2015\right|+\left|x+2016\right|\)

a) Ta có: \(\left|x\right|=\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)

+) Với \(x=\frac{1}{2}\): 

\(f\left(\frac{1}{2}\right)=\left|\frac{1}{2}-2015\right|+\left|\frac{1}{2}+2016\right|=2\)

+) Với \(x=-\frac{1}{2}\)

\(f\left(-\frac{1}{2}\right)=\left|-\frac{1}{2}-2015\right|+\left|-\frac{1}{2}+2016\right|=0\)

c) Áp dụng BĐT |x| + |y| \(\ge\)|x + y|, ta được:

\(f\left(x\right)=\left|x-2015\right|+\left|x+2016\right|=\left|2015-x\right|+\left|x+2016\right|\)

\(\ge\left|\left(2015-x\right)+\left(x+2016\right)\right|=\left|4031\right|=4031\)

(Dấu "="\(\Leftrightarrow\left(2015-x\right)\left(x+2016\right)\ge0\)

TH1: \(\hept{\begin{cases}2015-x\ge0\\x+2016\ge0\end{cases}}\Leftrightarrow-2016\le x\le2015\)

TH2: \(\hept{\begin{cases}2015-x\le0\\x+2016\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2015\\x\le-2016\end{cases}}\left(L\right)\))

Vậy \(f\left(x\right)_{min}=4031\Leftrightarrow-2016\le x\le2015\)

a/ Ta có \(f\left(-x\right)=\left|-x-2014\right|-\left|-x+2014\right|\)

Mà \(\left|-x-2014\right|\le\left|-x\right|+\left|-2014\right|\)(BĐT về giá trị tuyệt đối)

\(\left|-x+2014\right|\le\left|-x\right|+\left|2014\right|\)(BĐT về giá trị tuyệt đối)

=>\(\left|-x-2014\right|-\left|-x+2014\right|\le\left(\left|-x\right|+\left|-2014\right|\right)-\left(\left|x\right|+\left|2014\right|\right)\)

=> \(\left|-x-2014\right|-\left|-x+2014\right|\le\left(x+2014\right)-\left(x+2014\right)\)

=> \(\left|-x-2014\right|-\left|-x+2014\right|\le0\)(1)

và \(f\left(x\right)=\left|x-2014\right|-\left|x+2014\right|\)

Mà \(\left|x-2014\right|\le\left|x\right|+\left|-2014\right|\)(BĐT về giá trị tuyệt đối)

\(\left|x+2014\right|\le\left|x\right|+\left|2014\right|\)(BĐT về giá trị tuyệt đối)

=> \(\left|x-2014\right|-\left|x+2014\right|\le\left(\left|x\right|+\left|-2014\right|\right)-\left(\left|x\right|+\left|2014\right|\right)\)

=> \(\left|x-2014\right|-\left|x+2014\right|\le\left(x+2014\right)-\left(x+2014\right)\)

=> \(\left|x-2014\right|-\left|x+2014\right|\le0\)(2)

Từ (1) và (2) => \(\left|-x-2014\right|-\left|-x+2014\right|=\left|x-2014\right|-\left|x+2014\right|\)

=> \(f\left(x\right)=f\left(-x\right)\)(đpcm)

b/ + Ta có \(\left|x-2014\right|\ge0\)với mọi giá trị của x

\(\left|x+2014\right|\ge0\)với mọi giá trị của x

=> \(\left|x-2014\right|-\left|x+2014\right|\ge0\)với mọi giá trị của x

=> GTNN của f (x) = 0.

và \(\left|x-2014\right|-\left|x+2014\right|\le0\)(cm câu a)

=> GTLN của f (x) = 0.