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\(I=-3+\left|\frac{1}{2}-x\right|\)
Vì \(\left|\frac{1}{2}-x\right|\ge0\)
\(\Rightarrow-3+\left|\frac{1}{2}-x\right|\ge-3\)
Dấu = xảy ra khi \(\frac{1}{2}-x=0\Rightarrow x=\frac{1}{2}\)
Vậy Min I = -3 khi x=1/2
2 . | 1/2 . x - 1/3 | - 3/2 = 1/4
2 . | 1/2 . x - 1/3 | = 1/4 + 3/2
2 . | 1/2 . x - 1/3 | = 7/4
| 1/2 . x - 1/3 | = 7/4 : 2
| 1/2 . x - 1/3 | = 7/8
1/2 . x - 1/3 = 1/8 1/2 . x - 1/3 = -1/8
1/2 . x = 1/8 + 1/3 1/2 . x = -1/8 + 1/3
1/2 . x = 11/24 1/2 . x = 5/24
x = 11/24 : 1/2 x = 5/24 : 1/2
x = 11/24 . 1/2 x = 5/24 . 2/1
x = 11/48 x = 5/12
2 )
2 . | 1/2 . x - 1/3 | - 3/2 = 1/4
2 . | 1/2 . x - 1/3 | = 1/4 + 3/2
2 . | 1/2 . x - 1/3 | = 7/4
| 1/2 . x - 1/3 | = 7/4 : 2
| 1/2 . x - 1/3 | = 7/8
1/2 . x = 7/8 + 1/3
1/2 . x = 29/24
1/2 . x = 29/24 1/2 . x = -29/24
x = 29/24 : 1/2 x = -29/24 : 1/2
x = 29/24 . 2/1 x= -29/24 . 2/1
x = 29/12 x = -29/12
Một trong hai bài có một bài đúng mk cũng chả bk
a,|x+1/2|=2/5
\(\Rightarrow\)\(\orbr{\begin{cases}\frac{x+1}{2}\\\frac{x+1}{2}\end{cases}}\)=+-2/5
x+1/2=2/5\(\Rightarrow\)x+1=4/5\(\Rightarrow\)x=9/5
x+1/2=-2/5\(\Rightarrow\)x+1=-4/5\(\Rightarrow\)x=1/5
Vậy x\(\in\){1/5;9/5}
a: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2-10+3}{15}=\dfrac{-5}{15}=\dfrac{-1}{3}\)
b: \(=\left(6+\dfrac{1}{8}-\dfrac{1}{2}\right)\cdot4=\dfrac{48+1-4}{8}\cdot4=\dfrac{45}{2}\)
c: \(=\dfrac{1}{4}\cdot4-2\cdot\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)
d: \(F=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2008\cdot2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{1004}{2010}=\dfrac{1004}{1005}\)
\(2\left|\dfrac{1}{2}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\left|\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{3}{2}\Rightarrow2\left|\dfrac{1}{2}\right|=\dfrac{7}{4}\)
\(\Leftrightarrow\left|\dfrac{1}{2}\right|=\dfrac{7}{4}:2\Rightarrow\left|\dfrac{1}{2}\right|=\dfrac{7}{8}\)....?
sai đề k bn