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=3/4*4/5*...*99/100

=3/100

11 tháng 3 2022

\(\left(\dfrac{-5}{24}+0,75+\dfrac{7}{12}\right):\left(-2\dfrac{1}{4}\right)\\ =\left(\dfrac{-5}{24}+\dfrac{3}{4}+\dfrac{7}{12}\right):\dfrac{-9}{4}\\ =\dfrac{9}{8}.\dfrac{-4}{9}\\ =\dfrac{-1}{2}\)

28 tháng 2 2018

\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)

\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)

\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)

2.

\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)

\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)

3.

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)

\(=\frac{1}{100}\)

4.

\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)

\(=\frac{1}{2}\)

28 tháng 2 2018

mình chỉ làm được câu 3 thôi

có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)

\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)

\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)

\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)

\(=\frac{-1}{100}\)

Bài 3: 

b: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)+10=0\)

\(\Leftrightarrow6x^2+12-6x^2+12x-6=0\)

hay \(x=-\dfrac{1}{2}\)

Bài 2: 

a: \(x^3+3x^2+3x+1=\left(x+1\right)^3\)

b: \(m^3+9m^2n+27mn^2+27n^3=\left(m+3n\right)^3\)

17 tháng 1 2018

Đặt B = \(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\)

\(=\frac{1}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{44.49}\right)\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)

\(=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}\cdot\frac{45}{196}=\frac{9}{196}\)

Đặt C = \(\frac{1-3-5-....-49}{89}\)

\(=\frac{1-\left(3+5+...+49\right)}{89}\)

\(=\frac{1-\frac{\left(49+3\right).24}{2}}{89}\)

\(=\frac{1-624}{89}=\frac{-623}{89}=-7\)

\(\Rightarrow A=B.C=\frac{9}{196}\cdot\left(-7\right)=\frac{-9}{28}\)

16 tháng 1 2018

X có vô số giá trị!

\(=\left[\dfrac{9-3}{162}\cdot\dfrac{81}{17}+\dfrac{35}{34}\right]:\left(\dfrac{3}{5}+\dfrac{7}{102}\right)\cdot\dfrac{102}{5}+2017\)

\(=\left[\dfrac{6}{2}\cdot\dfrac{1}{17}+\dfrac{35}{34}\right]:\dfrac{341}{510}\cdot\dfrac{102}{5}+2017\)

\(=\dfrac{41}{34}\cdot\dfrac{510}{341}\cdot\dfrac{102}{5}+2017=\dfrac{12546}{341}+2017=\dfrac{700343}{341}\)