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\(C=\dfrac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}=\dfrac{\left|x-2017\right|+2019-1}{\left|x-2017\right|+2019}=1-\dfrac{1}{\left|x-2017\right|+2019}\)
Vì \(\left|x-2017\right|\ge0\Rightarrow\left|x-2017\right|+2019\ge2019\Rightarrow\dfrac{1}{\left|x-2017\right|+2019}\le\dfrac{1}{2019}\)
\(\Rightarrow C=1-\dfrac{1}{\left|x-2017\right|+2019}\ge1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
Dấu "=" xảy ra <=> \(\left|x-2017\right|=0\Leftrightarrow x=2017\)
Vậy \(A_{Min}=\dfrac{2018}{2019}\) khi x = 2017
Có: \(|x-1|\ge0\)
\(|x-2|\ge0\)
.................
\(|x-2019|\ge0\)
=> \(A\ge0\)
Vậy giá trị nhỏ nhất của A là 0
\(A=\left|x-2017\right|+\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\)
\(\Rightarrow A=\left|x-2017\right|+\left|x-2018\right|+\left|2019-x\right|+\left|2020-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2017\right|+\left|x-2018\right|+\left|2019-x\right|+\left|2020-x\right|\ge\left|x-2017+x-2018+2019-x+2020-x\right|\)
\(\Rightarrow A\ge\left|4\right|\)
\(\Rightarrow A\ge4.\)
Dấu '' = '' xảy ra khi:
\(\left(x-2017\right).\left(x-2018\right).\left(2019-x\right).\left(2020-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2017\ge0\\x-2018\ge0\\2019-x\ge0\\2020-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2017\le0\\x-2018\le0\\2019-x\le0\\2020-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2017\\x\ge2018\\x\le2019\\x\le2020\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2017\\x\le2018\\x\ge2019\\x\ge2020\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2018\le x\le2019\\x\in\varnothing\end{matrix}\right.\)
Vậy \(MIN_A=4\) khi \(2018\le x\le2019.\)
Chúc bạn học tốt!
a,Ta có:
\(\left|4x-\frac{7}{3}\right|\ge0\Rightarrow\left|4x-\frac{7}{3}\right|+2004\ge2004\)
Dấu "=" xảy ra \(\Leftrightarrow\left|4x-\frac{7}{3}\right|=0\Leftrightarrow4x-\frac{7}{3}=0\Leftrightarrow4x=\frac{7}{3}\Leftrightarrow x=\frac{7}{12}\)
b,Ta có:
\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=\left|x-1\right|+\left|x-2\right|+\left|3-x\right|+\left|4-x\right|\ge x-1+x-2+3-x+4-x=4\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\begin{cases}x-1\ge0\\x-2\ge0\\3-x\ge0\\4-x\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge1\\x\ge2\\x\le3\\x\le4\end{cases}\)\(\Leftrightarrow2\le x\le3\)
Câu C sai đề
A=\(\left|4x-\frac{7}{3}\right|+2004\ge2004\)
Dấu "=" xảy ra khi: x=7/12
Vậy GTNN của A là 2004 tại x=7/12
phần A, B bạn làm như bạn nguyễn quang trung còn C,D làm theo mình:
\(C=\frac{2017}{2018}-\left|x-\frac{3}{5}\right|\)
vì \(\left|x-\frac{3}{5}\right|\ge0\forall x\)
nên \(\frac{2017}{2018}-\left|x-\frac{3}{5}\right|\le\frac{2017}{2018}\forall x\)
vậy \(MaxC=\frac{2017}{2018}\Leftrightarrow x=\frac{3}{5}\)
\(D=\left|x-2\right|+\left|y+1\right|+3\)
\(\left|x-2\right|\ge0;\left|y+1\right|\ge0\forall x\)
nên \(\left|x-2\right|+\left|y+1\right|+3\ge3\forall x\)
vậy \(MinA=3\Leftrightarrow x=2;y=-1\)
a ) Ta có : A = \(\left|x+\frac{1}{2}\right|\ge0\forall x\)
Vậy Amin = 0 , khi x = \(-\frac{1}{2}\)
b) \(B=\left|\frac{3}{7}-x\right|+\frac{1}{9}\)
Mà : \(\left|\frac{3}{7}-x\right|\ge0\forall x\)
Nên : \(B=\left|\frac{3}{7}-x\right|+\frac{1}{9}\ge\frac{1}{9}\forall x\)
Vậy Bmin = \(\frac{1}{9}\) kh x = \(\frac{3}{7}\)
\(A=\left|2018-x\right|+\left|x-2017\right|\ge2018-x+x-2017=1\)
dấu = xãy ra khi \(\left(2018-x\right)\left(x-2017\right)\ge0\Leftrightarrow2017\le x\le2018\)
vậy \(A_{min}=1\) khi \(2017\le x\le2018\)
\(B=\left|x-1\right|+\left|2019-x\right|+\left|x-1999\right|\ge x-1+2019-x+\left|x-1999\right|\)
\(B\ge\left|x-1999\right|+2020\ge2020\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}x-1\ge0\\2019-x\ge0\\x-1999=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1\le x\le2019\\x=1999\end{matrix}\right.\Rightarrow x=1999\)
vậy \(B_{min}=2020\) khi x=1999
\(A=\left|2018-x\right|+\left|2017-x\right|\)
\(A=\left|2018-x\right|+\left|x-2017\right|\)
Áp dụng BĐT:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow A\ge\left|2018-x+x-2017\right|\)
\(\Rightarrow A\ge1\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2018-x\ge0\Rightarrow x\le2018\\x-2017\ge0\Rightarrow x\ge2017\end{matrix}\right.\\\left\{{}\begin{matrix}2018-x< 0\Rightarrow x< 2018\\x-2017< 0\Rightarrow x< 2017\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow2017\le x\le2018\)
B tương tự