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a)0,5-|x-3,5|
Vì |x-3,5|\(\ge0\)
Do đó 0,5-|x-3,5|\(\ge0,5\)
Dấu = xảy ra khi x-3,5=0
x=3,5
Vậy Max A=0,5 khi x=3,5
Mỏi cổ quá khi đọc đề bài của bn nên mk làm câu a thôi
Vậy
c) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}\)
\(=\frac{1.2.3.4...2014}{2.3.4.5...2015}=\frac{\left(1.2.3.4...2014\right)}{\left(2.3.4.5...2014\right).2015}=\frac{1}{2015}\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\left(\frac{1}{2013}-\frac{1}{2015}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)
\(C=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)
Bài 1:
c/ \(\left(2017-\dfrac{5}{181}+\dfrac{1}{50}\right)-\left(4+\dfrac{3}{181}-\dfrac{3}{50}\right)-\left(1-\dfrac{8}{181}+\dfrac{3}{50}\right)\)
\(=2017-\dfrac{5}{181}+\dfrac{1}{50}-4-\dfrac{3}{181}+\dfrac{3}{50}-1+\dfrac{8}{181}-\dfrac{3}{50}\)
\(=2012+\dfrac{1}{50}=2012,02\)
d/ \(1-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}-...-\dfrac{1}{99\cdot100}\)
\(=1-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)
\(=1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=1-\left(1-\dfrac{1}{100}\right)=1-1+\dfrac{1}{100}=\dfrac{1}{100}\)
Có 45 tam giác.
Còn 2 câu còn lại đề là j z, chú phải viết rõ thì chụy mới chỉ cho mà biết đk chứ!!!!
a) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{5}\)
\(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{5}+\dfrac{3}{2}\)
\(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{17}{10}\)
\(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{17}{10}:2\)
\(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{17}{20}\)
\(\circledast\)TH1: \(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{17}{20}\)
\(\dfrac{1}{2}x=\dfrac{17}{20}+\dfrac{1}{3}\)
\(\dfrac{1}{2}x=\dfrac{71}{60}\)
\(x=\dfrac{71}{60}:\dfrac{1}{2}\)
\(x=\dfrac{71}{30}\)
\(\circledast\)TH2: \(\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{17}{20}\)
\(\dfrac{1}{2}x=-\dfrac{17}{20}+\dfrac{1}{3}\)
\(\dfrac{1}{2}x=-\dfrac{31}{60}\)
\(x=-\dfrac{31}{60}:\dfrac{1}{2}\)
\(x=-\dfrac{31}{30}\)
Vậy \(x\in\left\{\dfrac{71}{30};-\dfrac{31}{30}\right\}\).
b) \(\dfrac{3}{4}-2\left|2x-\dfrac{2}{3}\right|=2\)
\(2\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)
\(2\left|2x-\dfrac{2}{3}\right|=-\dfrac{5}{4}\)
\(\left|2x-\dfrac{2}{3}\right|=-\dfrac{5}{4}:2\)
\(\left|2x-\dfrac{2}{3}\right|=-\dfrac{5}{8}\) (vô lí)
Vậy \(x\in\varnothing \).
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\)
\(3x=0+1\)
\(3x=1\)
\(x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\)
\(-\dfrac{1}{2}x=0-5\)
\(-\dfrac{1}{2}x=-5\)
\(x=-5:\left(-\dfrac{1}{2}\right)\)
\(x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
BẠN nào nhanh MÌnh tick cho