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TN
9 tháng 8 2016
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...
TN
25 tháng 7 2017
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 \(x-\dfrac{3}{14}=0\Rightarrow x=\dfrac{3}{14}\)
25 tháng 7 2017
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 MINf(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 MINf(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 MAXf(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 MAXf(x) = \(\dfrac{9}{28}\) khi x = \(\dfrac{3}{14}\).
11 tháng 7 2017
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 \(x=\dfrac{5}{3}\)
LD
7 tháng 5 2019
\(f\left(x\right)=-x^2+3x-1=-\left(x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}\right)+\frac{5}{4}\)
\(=-\left(x-\frac{3}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\)
Vậy \(Min_{f\left(x\right)}=\frac{5}{4}\Leftrightarrow x=\frac{3}{2}\)
VT
2
5 tháng 4 2018
a)
\(f\left(x\right)=3x^2-5x+1\)
\(3f\left(x\right)=9x^2-15x+3\)
\(3f\left(x\right)=\left(9x^2-15x+\frac{25}{4}\right)-\frac{13}{4}\)
\(3f\left(x\right)=\left(3x-\frac{5}{2}\right)^2-\frac{13}{4}\)
Mà \(\left(3x-\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow3f\left(x\right)\ge\frac{-13}{4}\)
\(\Leftrightarrow f\left(x\right)\ge-\frac{13}{12}\)
Dấu '=' xảy ra khi :
\(3x-\frac{5}{2}=0\Leftrightarrow3x=\frac{5}{2}\Leftrightarrow x=\frac{5}{6}\)
5 tháng 4 2018
\(f\left(x\right)=2x^2-9x-3\)
\(2f\left(x\right)=4x^2-18x-6\)
\(2f\left(x\right)=\left(4x^2-18x+\frac{81}{4}\right)-\frac{105}{4}\)
\(2f\left(x\right)=\left(2x-\frac{9}{2}\right)^2-\frac{105}{4}\)
Mà \(\left(2x-\frac{9}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2f\left(x\right)\ge-\frac{105}{4}\)
\(\Leftrightarrow f\left(x\right)\ge-\frac{105}{8}\)
Dấu "=" xảy ra khi :
\(2x-\frac{9}{2}=0\Leftrightarrow2x=\frac{9}{2}\Leftrightarrow x=\frac{9}{4}\)
