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\(\frac{3}{4}+\frac{2}{3}-\frac{1}{6}\)
\(=\frac{3}{4}+\frac{2}{3}\)
\(=\frac{17}{12}\)
\(=\frac{17}{12}-\frac{1}{6}\)
\(=\frac{90}{72}\)
a,=25/6;7/6=25/6x6/7=25/7
b,=7/2x32/6=56/3
c,=17/5-11/10=34/10-11/10=23/10
d,=8/3+11/4=32/12+33/12=65/12
Gọi tổng trên là A
Ta có : \(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{2560}\)
\(2A=2\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{2560}\right)\)
\(2A=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1280}\)
\(2A-A=\left(\frac{2}{5}+\frac{1}{5}+...+\frac{1}{1280}\right)-\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{2560}\right)\)
\(A\left(2-1\right)=\frac{2}{5}-\frac{1}{2560}\)
\(A.1=\frac{1024}{2560}-\frac{1}{2560}\)
\(A=\frac{1023}{2560}\)
Ta có : A = 1/5 + 1/10 + 1/20 + ... + 1/2560
2A = 2 ( 1/5 + 1/10 + ... + 1/2560 )
2A = 2/5 + 1/5 + 1/10 + .. + 1/2560
2A - A = ( 2/5 + 1/5 + ... + 1/1280 ) - ( 1/5 + 1/10 + ... + 1/2560 )
A = 2 - 1 = 2/5 - 1/2560
A.1 = 1024/2560 - 1/2560
A = 1023 = 2560
a, 6/9+5/7+1/3=2/3+5/7+1/3=5/7+1=12/7
b, 17/7+6/5-20/14=17/7+6/5-10/7=6/5+1=11/5
c,2/5x1/4+3/4x2/5=2/5x(1/4+3/4)=2/5x1=2/5
d, 6/11:4/6+5/11:2/3=6/11:2/3+5/11:2/3=(6/11+5/11):2/3=3/2
nha
C1 : \(\frac{7}{11}\):\(\frac{3}{5}\)+\(\frac{4}{11}\):\(\frac{3}{5}\)
=\(\frac{7}{11}\)x\(\frac{5}{3}\)+\(\frac{4}{11}\)x\(\frac{5}{3}\)
=\(\frac{35}{33}\)+\(\frac{20}{33}\)
=\(\frac{55}{33}\)=\(\frac{5}{3}\)
C2: \(\frac{7}{11}\):\(\frac{3}{5}\)+\(\frac{4}{11}\):\(\frac{3}{5}\)
= \(\frac{3}{5}\)x (\(\frac{7}{11}\)+\(\frac{4}{11}\))
=\(\frac{3}{5}\)x 1= \(\frac{3}{5}\)
\(C1:\frac{7}{11}:\frac{3}{5}+\frac{4}{11}:\frac{3}{5}\)
\(=\frac{7}{11}x\frac{5}{3}+\frac{4}{11}x\frac{5}{3}\)
\(=\frac{35}{33}+\frac{20}{33}\)
\(=\frac{55}{33}=\frac{5}{3}\)
\(C2:\frac{7}{11}:\frac{3}{5}+\frac{4}{11}:\frac{3}{5}\)
\(=\frac{7}{11}x\frac{5}{3}+\frac{4}{11}x\frac{5}{3}\)
\(=\frac{5}{3}x\left(\frac{7}{11}+\frac{4}{11}\right)\)
\(=\frac{5}{3}x1=\frac{5}{3}\)
3/5+4/7+6/11=41/35+6/11
=661/385
3/5 + 4/7 + 6/11 = 661/385