Tìm min A = |3/5-x|+2015, B=|x-11| . |x-14|
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\(a.\dfrac{1}{2}:3+x=\dfrac{14}{5}\)
\(\dfrac{1}{6}+x=\dfrac{14}{5}\)
\(=>x=\dfrac{79}{30}\)
\(b.\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\)
\(\left(\dfrac{8}{5}\cdot\dfrac{4}{7}\right):x=\dfrac{11}{6}\)
\(\dfrac{32}{35}:x=\dfrac{11}{6}\)
\(x=\dfrac{192}{385}\)
\(c.\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\)
\(x:\dfrac{3}{4}=\dfrac{11}{3}-\dfrac{24}{10}\)
\(x:\dfrac{3}{4}=\dfrac{38}{30}\)
\(=>x=\dfrac{19}{20}\)
\(a,\dfrac{1}{2}:3+x=\dfrac{14}{5}\\ \Leftrightarrow x+\dfrac{1}{6}=\dfrac{14}{5}\\ \Leftrightarrow x=\dfrac{79}{30}\\ b,\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\\ \Leftrightarrow x=\dfrac{192}{385}\\ c,\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\\ \Leftrightarrow\dfrac{4}{3}x=\dfrac{19}{15}\\ \Leftrightarrow x=\dfrac{19}{20}\)
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a, X = 273
b, X = 156
c, X = 4
d, X = 505
e, X = \(\frac{37}{245}\)
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mk làm dùm a) nhé, b) tuong tu
a) x = 1 ; -3/2 ; 9
thay vào ta có GTNN = 13
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\(A=2+\frac{21}{\left(x+3y\right)^2}+5\left|x+5\right|+14\)
Ta có:
\(\left(x+3y\right)^2\ge0;\left|x+5\right|\ge0\)
\(\Leftrightarrow\left(x+3y\right)^2+5\left|x+5\right|+14\ge14\)
\(\Leftrightarrow\frac{21}{\left(x+3y\right)^2}+5\left|x+5\right|+14\le\frac{21}{14}=\frac{3}{2}\)
\(\Leftrightarrow A\le\frac{2}{3}+\frac{3}{2}=\frac{13}{6}\)
Dấu '' = '' xảy ra khi:
\(x+5=0\Leftrightarrow x=-5\)
\(x+3y=0\Leftrightarrow y=\frac{-x}{3}=\frac{5}{3}\)
Vậy \(MaxA=\frac{13}{6}\Leftrightarrow x=-5;y=\frac{5}{3}\)
\(A\ge0+2015=2015 \)
Dấu '=' xảy ra khi x=3/5
ta có : B=GTTĐ[(x-11)(x-14)]=GTTĐ(\(x^2-25x+154\))=GTTĐ[\(\left(x-\frac{5}{2}\right)^2-\frac{9}{4}\)]\(\ge-\frac{9}{4}\)
Dấu '=' xảy ra khi x=5/2