Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
Ta có: \(\left|2x-\dfrac{1}{2}\right|\ge0\)
\(\Rightarrow\left|2x-\dfrac{1}{2}\right|-2017\ge-2017\)
\(\Rightarrow A\ge-2017\)
Dấu "=" xảy ra \(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|=0\)
\(\Leftrightarrow2x-\dfrac{1}{2}=0\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy, MinA = -2017 \(\Leftrightarrow x=\dfrac{1}{4}\)
1) Tìm giá trị nhỏ nhất của biểu thức:
A = \(\left|2x-\dfrac{1}{2}\right|\) - 2017
Ta có:
\(\left|2x-\dfrac{1}{2}\right|\) ≥ 0
=> \(\left|2x-\dfrac{1}{2}\right|\) - 2017 ≥ -2017
Dấu " = " xảy ra khi \(2x-\dfrac{1}{2}\) = 0 hay x = \(\dfrac{1}{4}\)
Vậy Min A = -2017 khi x = \(\dfrac{1}{4}\)
Vì \(\left|\left|3x-3\right|+2x+\left(-1\right)^{2016}\right|\ge0\forall x\)
\(\Rightarrow3x+2017^0\ge0\Rightarrow x\ge-\frac{1}{3}\)
Khi đó: \(\left|\left|3x-3\right|+2x+1\right|=3x+1\)
\(\Leftrightarrow\orbr{\begin{cases}\left|3x-3\right|+2x+1=3x+1\\\left|3x-3\right|+2x+1=-3x-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left|3x-3\right|=x\\\left|3x-x\right|=-5x-2\end{cases}}\)
Để |3x - 3| = x => \(x\ge0\)
=> \(\orbr{\begin{cases}3x-3=x\\3x-3=-x\end{cases}\Rightarrow\orbr{\begin{cases}2x=3\\4x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\left(tm\right)\\x=\frac{3}{4}\left(tm\right)\end{cases}}}\)
Để |3x - 3| = - 5x - 2
=> \(-5x-2\ge0\Rightarrow x\le-\frac{2}{5}\)
=> \(\orbr{\begin{cases}3x-3=5x+2\\3x-3=-5x-2\end{cases}\Rightarrow\orbr{\begin{cases}-2x=5\\8x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{5}{2}\left(\text{tm}\right)\\x=\frac{1}{8}\left(\text{loại}\right)\end{cases}}}\)
Vậy \(x\in\left\{\frac{-5}{2};\frac{3}{2};\frac{3}{4}\right\}\)
Có: \(\left(x-7\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-7\right)^2+1\ge1\)
\(\Rightarrow min\left(x-7\right)^2+1=1khi\left(x-7\right)^2=0\)
\(\Rightarrow\left(x-7\right)^2=0^2\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
Vậy GTNN của (x-7)2+1 là 1 tại x=7
Có:\(\left(5x-3\right)^{2018}=\left[\left(5x-3\right)^2\right]^{1009}\)
\(Co:\left(5x-3\right)^2\ge0\)
\(\Rightarrow\left[\left(5x-3\right)^2\right]^{1009}\ge0\)
\(\Rightarrow\left(5x-3\right)^{2018}\ge0\)
\(\Rightarrow\left(5x-3\right)^{2018}-2017\ge-2017\)
\(\Rightarrow min\left(5x-3\right)^{2018}-2017=-2017khi\left(5x-3\right)^2=0\)
\(\Rightarrow5x-3=0\)
\(\Rightarrow x=\frac{3}{5}\)
Vậy GTNN của (5x-3)2018 -2017 là -2017 khi \(x=\frac{3}{5}\)
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
a/ Do \(\left(y-2017\right)^{2014}\ge0\) \(\forall y\Rightarrow A\ge-2017\)
\(\Rightarrow A_{min}=-2017\) khi \(y-2017=0\Rightarrow y=2017\)
b/ \(\left|3y-6045\right|^{2011}\le\left(x-1\right)^{2017}-x\left(x-1\right)^{2017}\)
\(\Leftrightarrow\left|3y-6045\right|^{2011}\le\left(1-x\right)\left(x-1\right)^{2017}\)
\(\Leftrightarrow\left|3y-6045\right|^{2011}\le-\left(x-1\right)\left(x-1\right)^{2017}\)
\(\Leftrightarrow\left|3y-6045\right|^{2011}\le-\left(x-1\right)^{2018}\) (1)
Mà \(\left\{{}\begin{matrix}\left|3y-6045\right|^{2011}\ge0\\-\left(x-1\right)^{2018}\le0\end{matrix}\right.\)
Nên (1) xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}\left|3y-6045\right|^{2011}=0\\-\left(x-1\right)^{2018}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3y-6045=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=2015\end{matrix}\right.\)
\(\Rightarrow B=3.1^2-2.1.2015+6042=2015\)