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\(\frac{1}{2015}x=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2014}\right)\left(1-\frac{1}{2015}\right)\)
\(\frac{1}{2015}x=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times.....\times\frac{2013}{2014}\times\frac{2014}{2015}\)
\(\frac{1}{2015}x=\frac{1}{2015}\)
\(x=1\)
Chúc bạn học tốt
a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)
\(=\dfrac{1}{2016}\)
b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)
\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)
hay x=10
Vậy: x=10
B2:
a)3x+2=4
3x=4-2
3x=2
x=2/3
b)3(x-1)-5=-20
3(x-1)=-20+5
x-1=-15/3
x-1=-5
x=-5+1
x=-4
c)(x-1)(x+2)=0
nên x-1=0 hoặc x+2=0
x=0+1 x=0-2
x=1 x=-2
d)(x+1)(2x-5)=0
nên x+1=0 hoặc 2x-5=0
x=0-1 2x=0+5
x=1 x=5/2
còn b1 thì cậu đăng câu khác đi, t lười làm
bài 1: a) 1+(-2)+3+(-4)+5+(-6)+....+2015+(-2016)
=[1+(-2)]+[3+(-4)]+[5+(-6)]+....+[2015+(-2016)]
=(-1)+(-1)+(-1)+...+(-1) (có 1008 số -1)
=(-1).1008
=-1008
\(A=1+2^2+2^3+2^4+...+2^{2015}\)
\(2A=2+2^3+2^4+2^5+...+2^{2016}\)
\(2A-A=\left(2+2^3+2^4+2^5+...+2^{2016}\right)-\left(1+2^2+2^3+2^4+...+2^{2015}\right)\)
\(A=2^{2016}-1\)
Đề sai rồi