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1) `(x-3)^4 >=0`
`2.(x-3)^4>=0`
`2.(x-3)^4-11 >=-11`
`=> A_(min)=-11 <=> x-3=0<=>x=3`
2) `|5-x|>=0`
`-|5-x|<=0`
`-3-|5-x|<=-3`
`=> B_(max)=-3 <=>x=5`.
Bài 1:
Ta có: \(\left(x-3\right)^4\ge0\forall x\)
\(\Leftrightarrow2\left(x-3\right)^4\ge0\forall x\)
\(\Leftrightarrow2\left(x-3\right)^4-11\ge-11\forall x\)
Dấu '=' xảy ra khi x=3
\(D=2015-5\left|x-386\right|-5\left|x-389\right|\)
\(D=2015-5\left(\left|x-386\right|+\left|389-x\right|\right)\)
\(D\le2015-5\left|x-386+389-x\right|\)
\(D\le2015-15=2000\)
Dấu "=" xảy ra khi: \(386\le x\le389\)
\(M=2016-\left|x-2015\right|-\left|x-1975\right|-\left|x-1945\right|\)
\(M=2016-\left(\left|x-2015\right|+\left|x-1975\right|+\left|x-1945\right|\right)\)
Đặt: \(L=\left|x-2015\right|+\left|x-1975\right|+\left|x-1945\right|\)
\(L=\left|x-2015\right|+\left|1945-x\right|+\left|x-1975\right|\)
\(L\ge\left|x-2015+1945-x\right|+\left|x-1975\right|\)
\(L\ge70+\left|x-1975\right|\ge70\)
Suy ra: \(M-L\le2016-70=1946\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}1945\le x\le2015\\x=1975\end{cases}}\Leftrightarrow x=1975\)
\(A=-\left|x-3.5\right|+0.5\le0.5\forall x\)
Dấu '=' xảy ra khi x=3,5
\(B=-\left|1.4-x\right|-2\le-2\forall x\)
Dấu '=' xảy ra khi x=1,4
\(C=-\left|x+\frac{4}{7}\right|+\frac{12}{19}\)
Ta có: \(\left|x+\frac{4}{7}\right|\ge0\)nên \(-\left|x+\frac{4}{7}\right|\le0\)
\(\Rightarrow C=-\left|x+\frac{4}{7}\right|+\frac{12}{19}\le\frac{12}{19}\)
\(\Rightarrow C_{max}=\frac{12}{19}\)
(Dấu "="\(\Leftrightarrow x=\frac{-4}{7}\))
\(D=\left|x-\frac{5}{7}\right|+\frac{2}{3}\)
Vì \(\left|x-\frac{5}{7}\right|\ge0\)nên \(D=\left|x-\frac{5}{7}\right|+\frac{2}{3}\ge\frac{2}{3}\)
\(\Rightarrow D_{min}=\frac{2}{3}\)
(Dấu "="\(\Leftrightarrow x=\frac{5}{7}\))
c) Ta có: \(\left|5x-2\right|\ge0\forall x\)
\(\left|3y+12\right|\ge0\forall y\)
Do đó: \(\left|5x-2\right|+\left|3y+12\right|\ge0\forall x,y\)
\(\Leftrightarrow-\left|5x-2\right|-\left|3y+12\right|\le0\forall x,y\)
\(\Leftrightarrow-\left|5x-2\right|-\left|3y+12\right|+4\le4\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}5x-2=0\\3y+12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\3y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)
bạn làm bài nào đây ạ? 4 - |5x-2| - |3y + 12| mà đâu phải −|5x−2|−|3y+12|+4
a. ta có \(\left|x-386\right|+\left|x-389\right|=\left|x-386\right|+\left|389-x\right|\ge\left|x-386+389-x\right|=3\)
\(\Rightarrow D\le2015-5\times3=2000\)
b. ta có \(\left|x-30\right|+\left|x-4\right|=\left|30-x\right|+\left|x-4\right|\ge\left|30-x+x-4\right|=26\)
\(\Rightarrow E\le\frac{51350}{26}=1975\)