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b)0,5x+2/3x+2/3=7/2
1/2x+2/3x+2/3=7/2
x(1/2+2/3)+2/3=7/12
x.7/6+2/3 =7/12
x.7/6 =7/12-2/3
x.7/6 =-1/12
x =-1/12:7/6
x =-1/14
b)3/5-2/15:x=1/2
2/15:x =3/5-1/2
2/15:x =1/10
x =2/15:1/10
x = 4/3
a,=0,9:(4/5.1,25+1:1/3)
=0,9:(1+3)
=0,9:4
=0,225
b,=9,6:6-0,6
=1,6-0,6
=1
c,=7/2.11/4-7/2.5/4
=7/2.(11/4-5/4)
=7/2.3/2
=21/4
a) \(15-3\left(x-1\right)-x=20\)
\(15-3x+3-x=20\)
\(-4x=2\)
\(x=-\frac{1}{2}\)
b) \(6x-2\left(x+3\right)=10\)
\(6x-2x-6=10\)
\(4x=16\)
\(x=4\)
c) \(25:\left(x+4\right)=5\)
\(x+4=25:5=5\)
\(x=1\)
d tự làm nha!
\(C=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{2013}\right)\)
\(C=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{2014}{2013}\)
\(C=\frac{3\cdot4\cdot5\cdot...\cdot2014}{2\cdot3\cdot4\cdot5\cdot...\cdot2013}=\frac{2014}{2}=1007\)
Dấu "." là dấu nhân nhá
\(a)\) \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2S=3+\frac{1}{3^7}\)
\(2S=\frac{3^8+1}{3^7}\)
\(S=\frac{3^8+1}{3^7}.\frac{1}{2}\)
\(S=\frac{3^8+1}{2.3^7}\)
Vậy \(S=\frac{3^8+1}{2.3^7}\)
Chúc bạn học tốt ~
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\)
\(=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{3.7}+\frac{1}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}+0+0+...+0\)
\(=\frac{7}{8}\)
\(C=\left(\dfrac{2}{2}+\dfrac{1}{3}\right)\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}+\dfrac{1}{4}\right)\times...\times\left(\dfrac{2013}{2013}+\dfrac{1}{2013}\right)\)
\(C=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times....\times\dfrac{2014}{2013}\)
\(C=\dfrac{2014}{2}=1007\)
`C = (2/2 + 1/3) xx (3/3 + 1/3) xx (4/4 + 1/4) xx ... xx (2013/2013 + 1/2013)`
`C = 3/2 xx 4/3 xx 5/4 xx ... xx 2014/2013`
`C = 2014/2 = 1007`
Vậy `C = 1007`