Tìm GTNN của:
\(\dfrac{6\left|y+5\right|+14}{2\left|y+5\right|+14}\)
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bạn''đẹp trai không bao giờ sai'' ơi,đừng buồn nhé.bạn ấy đúng là không biết suy nghĩ
Tìm GTNN nha.
mk nhầm
Mashiro Shiina giúp mk vs
\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)
\(y_{min}=-3\) khi \(x=1\)
\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)
\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)
\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)
\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)
Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$
$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$
$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$
Bài 2:
$18: \frac{x\times 0,4+0,32}{x}+5=14$
$18: \frac{x\times 0,4+0,32}{x}=14-5=9$
$\frac{x\times 0,4+0,32}{x}=18:9=2$
$x\times 0,4+0,32=2\times x$
$2\times x-x\times 0,4=0,32$
$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$
\(\left(\frac{-1}{25}\right)^{14}:\left(2x-1\right)^2=\left(\frac{1}{5}\right)^{26}\)
=> (2x-1)2 = \(\left(\frac{-1}{25}\right)^{14}:\left(\frac{1}{5}\right)^{26}\)
=> ( 2x - 1 )2 = \(\left(\frac{-1}{25}\right)^{14}:\left(\frac{1}{25}\right)^{13}\)
=> ( 2x - 1 )2 = \(\left[\left(\frac{-1}{25}\right)^{13}.\left(\frac{-1}{25}\right)\right]:\left(\frac{1}{25}\right)^{13}\)
=> ( 2x - 1 )2 = \(\frac{1}{25}\)
=> ( 2x - 1 )^2 = \(\left(\frac{1}{5}\right)^2\)
=> \(\hept{\begin{cases}2x-1=\frac{1}{5}\\2x-1=\frac{-1}{5}\end{cases}}\)
=> \(\hept{\begin{cases}2x=\frac{1}{5}+1\\2x=\frac{-1}{5}+1\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{3}{5}\\x=\frac{2}{5}\end{cases}}\)
Vậy x = 3/5 hay x = 2/5
Bài 2:
x=13 nên x+1=14
\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)
\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)
=14-x=1
x=13 nên x+1=14
f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14
=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14
=14-x=1
Ta có : \(5\left|6y-8\right|\ge0\)
\(\Rightarrow5\left|6y-8\right|+35\ge35\\ \Rightarrow\dfrac{14}{5\left|6y-8\right|+35}\le\dfrac{14}{35}\\ \Rightarrow\dfrac{6}{5}-\dfrac{14}{5\left|6y-8\right|+35}\ge\dfrac{28}{35}\)
Min P = \(\dfrac{28}{35}\)khi y= \(\dfrac{4}{3}\)
Ta có: \(\left(x+y-2\right)^2+7\ge7\Rightarrow\dfrac{14}{\left|y-1\right|+\left|y-3\right|}\ge7\)
\(\Rightarrow\left|y-1\right|+\left|y-3\right|\le2\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left|y-1\right|=0\\\left|y-3\right|=2\end{matrix}\right.\\\left\{{}\begin{matrix}\left|y-1\right|=2\\\left|y-3\right|=0\end{matrix}\right.\\\left|y-1\right|=\left|y-3\right|=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1\\y=3\\y=2\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)
Có gì đâu nkok?
\(A=\dfrac{6\left|y+5\right|+14}{2\left|y+5\right|+14}=\dfrac{6\left|y+5\right|+42}{2\left|y+5\right|+14}-\dfrac{28}{2\left|y+5\right|+14}\)
\(A=\dfrac{3\left(2\left|y+5\right|+14\right)}{2\left|y+5\right|+14}-\dfrac{28}{2\left|y+5\right|+14}\)
\(A=3-\dfrac{28}{2\left|y+5\right|+14}\ge3-\dfrac{28}{14}=1\)
Dấu "=" khi x=-5
You là girl or boy?
You hk lp mấy?