Tim GTNN cua bieu thức:A= |2x+ 1/5|+|2x+1/6|+|2x+1/7|
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Ta có: M = \(\frac{x^4+x^2+5}{x^4+2x^2+1}\)
M = \(\frac{\left(x^4+2x^2+1\right)-\left(x^2+1\right)+5}{\left(x^2+1\right)^2}\)
M = \(1-\frac{1}{x^2+1}+5\cdot\frac{1}{\left(x^2+1\right)^2}\)
Đặt \(\frac{1}{x^2+1}=y\)
Khi đó, ta có: M = \(1-y+5y^2=5\left(y^2-\frac{1}{5}y+\frac{1}{100}\right)+\frac{19}{20}=5\left(y-\frac{1}{10}\right)^2+\frac{19}{20}\ge\frac{19}{20}\forall y\)
Dấu "=" xảy ra <=> y - 1/10 = 0 <=> y = 1/10 <=> \(\frac{1}{x^2+1}=\frac{1}{10}\) <=> x2 + 1 = 10
<=> x2 = 9 <=> \(x=\pm3\)
Vậy MinM = 19/20 khi x = 3 hoặc x = -3
a/ \(2x^2+8x+1=2\left(x^2+4x+\frac{1}{2}\right)=2\left(x^2+2.2x+4-4+\frac{1}{2}\right)\)
\(=2\left[\left(x+2\right)^2-\frac{7}{2}\right]=2\left(x+2\right)^2-7\ge-7\)
Vậy Min A = -7 khi x + 2 = 0 => x = 2
b/ \(2x^2+3x+1=2\left(x^2+\frac{3}{2}x+\frac{1}{2}\right)=2\left(x^2+2.\frac{3}{4}.x+\frac{9}{16}-\frac{9}{16}+\frac{1}{2}\right)\)
\(=2\left[\left(x+\frac{3}{4}\right)^2-\frac{1}{16}\right]=2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)
Vậy Min B = -1/8 khi x + 3/4 = 0 => x = -3/4
\(\sqrt{x^2-4x+5}=\sqrt{\left(x-2\right)^2+1}\ge1\)
Đặt \(\sqrt{x^2-4x+5}=a\Rightarrow a\ge1\)
\(M=2\left(x^2-4x+5\right)+\sqrt{x^2-4x+5}-4\)
\(M=2a^2+a-4=2a^2+3a-2a-3-1\)
\(M=a\left(2a+3\right)-\left(2a+3\right)-1\)
\(M=\left(a-1\right)\left(2a+3\right)-1\)
Do \(a\ge1\Rightarrow\left\{{}\begin{matrix}a-1\ge0\\2a+3>0\end{matrix}\right.\) \(\Rightarrow\left(a-1\right)\left(2a+3\right)\ge0\Rightarrow M\ge-1\)
\(\Rightarrow M_{min}=-1\) khi \(a=1\Leftrightarrow x=2\)
a)Ta thấy:
\(\left(2x+\frac{1}{3}\right)^2\ge0\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^2-\frac{5}{6}\ge0-\frac{5}{6}=-\frac{5}{6}\)
\(\Rightarrow A\ge-\frac{5}{6}\)
Dấu "=" <=>x=-1/6
Vậy MinA=-5/6<=>x=-1/6
b)Ta thấy:\(\hept{\begin{cases}\left|2x+3\right|\\\left|y-\frac{1}{2}\right|\end{cases}\ge}0\)
\(\Rightarrow\left|2x-3\right|+\left|y-\frac{1}{2}\right|\ge0\)
\(\Rightarrow\left|2x-3\right|+\left|y-\frac{1}{2}\right|+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}\)
\(\Rightarrow B\ge\frac{3}{4}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|2x-3\right|=0\\\left|y-\frac{1}{2}\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}\)
Vậy...
\(\text{a)Để C đạt GTNN}\)
\(\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\\\left(y-\frac{1}{5}\right)^2\end{cases}\ge0}\)
\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)
\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10\)
\(\Rightarrow C\ge-10\)
\(\text{Vậy minC=-10 khi x=-2;y= }\frac{1}{5}\)
b)\(\text{Để D đạt GTLN}\)
=>(2x-3)2+5 đạt GTNN
Mà (2x-3)2\(\ge\)5
\(\Rightarrow GTLN\)của \(A=\frac{4}{5}\)khi \(x=\frac{3}{2}\)