B=x^2-5x-1 Tìm GTNN
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HV
0
NM
0
4 tháng 7 2021
Ta có: \(B=x^2+5x+6\)
\(=x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{1}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{5}{2}\)
Vậy: \(B_{min}=-\dfrac{1}{4}\) khi \(x=-\dfrac{5}{2}\)
H
4 tháng 7 2021
\(B=x^2+5x+6=\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{25}{4}+6=\left(x+\dfrac{5}{2}\right)^2-\dfrac{1}{4}\)
Vì \(\left(x+\dfrac{5}{2}\right)^2\ge0\) nên \(B\ge-\dfrac{1}{4}\)
Vậy GTNN của B là \(-\dfrac{1}{4}\)
Dấu = xảy ra \(\text{⇔}x+\dfrac{5}{2}=0\text{⇔}x=-\dfrac{5}{2}\)
4 tháng 4 2022
d. Áp dụng BĐT Caushy Schwartz ta có:
\(x+y+\dfrac{1}{x}+\dfrac{1}{y}\le x+y+\dfrac{\left(1+1\right)^2}{x+y}=x+y+\dfrac{4}{x+y}\le1+\dfrac{4}{1}=5\)
-Dấu bằng xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
7 tháng 7 2016
\(B=5x^2+x+1\)
\(=>5\left(x^2+\frac{1}{5}x+\frac{1}{5}\right)\)
\(=>5\left(x^2+2.x.\frac{1}{10}+\frac{1}{100}+\frac{19}{100}\right)\)
\(=>5\left(\left(x+\frac{1}{10}\right)^2+\frac{19}{100}\right)\)
\(=>\frac{19}{20}+5\left(x+\frac{1}{10}\right)^2\ge\frac{19}{20}\)
MIN B = \(\frac{19}{20}< =>x+\frac{1}{10}=0=>x=\frac{-1}{10}\)
NV
7 tháng 7 2016
B = 5x2 + x - 1
\(=5\left(x^2+\frac{1}{5}x-\frac{1}{5}\right)=5\left[x^2+2.\frac{1}{10}.x+\left(\frac{1}{10}\right)^2-\left(\frac{1}{10}\right)^2-\frac{1}{5}\right]\)
\(=5\left[\left(x+\frac{1}{10}\right)^2-\frac{21}{100}\right]=5\left(x+\frac{1}{10}\right)^2-\frac{21}{20}\ge-\frac{21}{20}\)
Vậy MinB = -21/20 khi \(x+\frac{1}{10}=0\Rightarrow x=-\frac{1}{10}\)
NT
0
HP
2
NC
18 tháng 8 2020
Bài 1:
a) \(x^2-5x+1=0\)
\(\Leftrightarrow\left(x^2-5x+\frac{25}{4}\right)-\frac{21}{4}=0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2-\frac{\left(\sqrt{21}\right)^2}{2^2}=0\)
\(\Leftrightarrow\left(x-\frac{5+\sqrt{21}}{2}\right)\left(x+\frac{\sqrt{21}-5}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5+\sqrt{21}}{2}=0\\x+\frac{\sqrt{21}-5}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\end{cases}}\)
b) \(3x^2-12x-1=0\)
\(\Leftrightarrow3\left(x^2-4x+4\right)-13=0\)
\(\Leftrightarrow\left(x-2\right)^2-\left(\sqrt{\frac{13}{3}}\right)^2=0\)
\(\Leftrightarrow\left(x-2-\sqrt{\frac{13}{3}}\right)\left(x-2+\sqrt{\frac{13}{3}}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2+\sqrt{\frac{13}{3}}\\x=2-\sqrt{\frac{13}{3}}\end{cases}}\)
NC
18 tháng 8 2020
Bài 2:
a) \(A=\frac{1}{4}x^2-x+1=\left(\frac{1}{2}x-1\right)^2\ge0\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(\frac{1}{2}x-1\right)^2=0\Rightarrow\frac{1}{2}x=1\Rightarrow x=2\)
Vậy Min(A) = 0 khi x = 2
b) \(B=3x^2-4x-2=3\left(x^2-\frac{4}{3}x+\frac{4}{9}\right)-\frac{10}{3}=3\left(x-\frac{2}{3}\right)^2-\frac{10}{3}\ge-\frac{10}{3}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(3\left(x-\frac{2}{3}\right)^2=0\Rightarrow x=\frac{2}{3}\)
Vậy \(Min\left(B\right)=-\frac{10}{3}\Leftrightarrow x=\frac{2}{3}\)
KN
9 tháng 10 2019
a) \(A=5-8x-x^2=-\left(x^2+8x-5\right)\)
\(=-\left(x^2+8x+16-21\right)\)
\(=-\left[\left(x+4\right)^2-21\right]\)
\(=-\left(x+4\right)^2+21\le21\)
Vậy \(A_{max}=21\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
\(B=5x-3x^2=-3\left(x^2-\frac{5}{3}x\right)\)
\(=-3\left(x^2-\frac{5}{3}x+\frac{35}{36}-\frac{25}{36}\right)\)
\(=-3\left[\left(x-\frac{5}{6}\right)^2-\frac{25}{36}\right]\)
\(=-3\left[\left(x-\frac{5}{6}\right)^2\right]+\frac{25}{12}\le\frac{25}{12}\)
Vậy \(B_{min}=\frac{25}{12}\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)
\(B=x^2-5x-1\)
\(=x^2-2.x.\frac{5}{2}+\frac{25}{4}-\frac{29}{4}\)
\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)
Dấu "=" xảy ra khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)