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B=\(\frac{2002.\left(14+1\right)-1}{2001+2002.14}\)=\(\frac{2002.14+2001}{2001+2002.14}\)=1
Giải hộ mình câu này với:
(215/216+3971/3972-1/5)x(-9/35+2/5-1/7)=..........................?
a)\(\frac{8}{13}+\frac{4}{9}+\frac{1}{3}+\frac{5}{13}+3\)
\(=\left(\frac{8}{13}+\frac{5}{13}\right)+\left(\frac{4}{9}+\frac{1}{3}\right)+3\)
\(=1+\left(\frac{4}{9}+\frac{3}{9}\right)+3\)
\(=1+\frac{7}{9}+3\)
\(=1+3+\frac{7}{9}\)
\(=4\frac{7}{9}\)
\(=\frac{43}{9}\)
b)\(\frac{8}{5}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{2}{5}-1\)
\(=\frac{4}{7}\cdot\left(\frac{8}{5}+\frac{2}{5}\right)-1\)
\(=\frac{4}{7}\cdot2-1\)
\(=\frac{8}{7}-1\)
\(=\frac{1}{7}\)
a) \(\frac{8}{13}+\frac{4}{9}+\frac{1}{3}+\frac{5}{13}+3\)
\(=\left(\frac{8}{13}+\frac{5}{13}\right)+\left(\frac{4}{9}+\frac{3}{9}\right)+3\)
\(=1+\frac{7}{9}+3=4\frac{7}{9}\)
b)\(\frac{8}{5}.\frac{4}{7}+\frac{4}{7}.\frac{2}{5}-1\)
\(=\frac{4}{7}.\left(\frac{8}{5}+\frac{2}{5}\right)-1\)
\(=\frac{4}{7}.2-1\)
\(=\frac{8}{7}-\frac{7}{7}=\frac{1}{7}\)
\(3+\frac{2}{5}-\frac{3}{4}\times\frac{4}{5}\)
\(=3+\frac{2}{5}+\frac{3}{5}\)
\(=3+1\)
\(=4\)
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