Bài 1: tìm GTLN:
a) D=1-|2x-3|
b) E=-|10-5x|-14,2
c) P=4-|5x-2|-|3y+12|
d) A= 5-3(2x-1)2
BÀI 2: TÍNH:
A=|x+\(\dfrac{1}{2}\)|-|x+2|+|x-\(\dfrac{3}{4}\)| khi x=\(\dfrac{-1}{2}\)
B= 2x+2xy-y với |x|=2,5;y=\(\dfrac{-3}{4}\)
C=3a-3ab-b; B=\(\dfrac{5a}{3}-\dfrac{3}{b}\)
Mong mn giúp mk vs, mk đang cần gấp
Help me ..........................!
Bài 1
a, \(D=1-\left|2x-3\right|\)
Ta có : \(\left|2x-3\right|\ge0\)
\(\Rightarrow1-\left|2x-3\right|\le1\)
Dấu "=" xảy ra khi \(\left|2x-3\right|=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=3:2=\dfrac{3}{2}\)
\(b,\) Ta có : \(\left|10-5x\right|\ge0\Rightarrow\left|10-5x\right|+14,2\ge14,3\Rightarrow-\left|10-5x\right|-14,2\le-14,2\)
Dấu "=" xảy ra khi \(-\left|10-5x\right|=0\)
\(\Leftrightarrow10-5x=0\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=10:5=2\)
Vậy \(Emax=-14,2\Leftrightarrow x=2\)
\(c,\) Ta có : \(\left|5x-2\right|\ge0\)
\(\left|3y-12\right|\ge0\)
⇒ \(\left|5x-2\right|+\left|3y+12\right|-4\ge-4\)
⇒ \(4-\left|5x-2\right|-\left|3y+12\right|\le4\)
Dấu "=" xảy ra khi \(\left[{}\begin{matrix}\left|5x-2\right|=0\\\left|3y+12\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=2\\3y=-12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)
\(d,\) \(A=5-3\left(2x-1\right)^2\)
Ta có : \(\left(2x-1\right)^2\ge0\)
\(\Rightarrow3.\left(2x-1\right)^2\ge0\)
\(\Rightarrow3.\left(2x-1\right)^2-5\ge-5\)
\(\Rightarrow5-3\left(2x-1\right)^2\le5\)
Dấu "=" xảy ra khi \(\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(Amax=5\Leftrightarrow x=\dfrac{1}{2}\)
Bài 2:
a: \(A=\left|\dfrac{-1}{2}+\dfrac{1}{2}\right|-\left|-\dfrac{1}{2}+2\right|+\left|-\dfrac{1}{2}-\dfrac{3}{4}\right|\)
\(=-\left|\dfrac{3}{2}\right|+\left|-\dfrac{5}{4}\right|\)
=-3/2+5/4
=5/4-6/4=-1/4
b: TH1: x=2,5
\(B=2\cdot2.5+2\cdot2.5\cdot\dfrac{-3}{4}+\dfrac{3}{4}\)
\(=5+5\cdot\dfrac{-3}{4}+\dfrac{3}{4}=\dfrac{23}{4}-\dfrac{15}{4}=2\)
TH2: x=-2,5
\(B=2\cdot\left(-2.5\right)+2\cdot\left(-2.5\right)\cdot\dfrac{-3}{4}+\dfrac{3}{4}\)
\(=-5+\dfrac{15}{4}+\dfrac{3}{4}=-5+\dfrac{9}{2}=-\dfrac{1}{2}\)
