Tính nhanh :
\(33x\left(\frac{3434}{1515}+\frac{3434}{3535}+\frac{3434}{6363}+\frac{3434}{9999}\right)\)
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( 3434/3535 + 3434/6363 + 3434/9999 ) × 55
= ( 34/35 + 34/63 + 34/99 ) × 55
= 34 × ( 1/35 + 1/63 + 1/99 ) × 55
= 34 × 3/55 × 55
= 102/55 × 55
= 102
Mình ko chắc lắm đâu nhá
= \(\frac{12}{15}\) +\(\frac{12}{35}\)+\(\frac{12}{63}\)+\(\frac{12}{99}\)
= 12 x (\(\frac{1}{15}\)+\(\frac{1}{35}\)+\(\frac{1}{63}\)+\(\frac{1}{99}\))
= 12 x ( \(\frac{1}{3x5}\)+\(\frac{1}{5x7}\)+\(\frac{1}{7x9}\)+\(\frac{1}{9x11}\))
= 12 x \(\frac{1}{2}\) x ( \(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))
= 6 x ( \(\frac{1}{3}\) - \(\frac{1}{11}\))
= 6 x \(\frac{8}{33}\)
= \(\frac{48}{33}\)=\(\frac{16}{11}\)
Nhớ tk nha
\(7\cdot x+x:\frac{1}{9}+x=3434\)
\(7\cdot x+x\cdot9+x=3434\)
\(\left(7+9+1\right)x=3434\)
\(17\cdot x=3434\)
\(x=3434:17\)
\(x=202\)
\(7\times x+x:\frac{1}{9}+x=3434\)
\(7\times x+x\times9+x=3434\)
\(\left(7+9\right)\times x+x=3434\)
\(16\times x+x=3434\)
\(17\times x=3434\)
\(x=3434:17\)
\(\Rightarrow x=202\)
1616/1515 + 1616/3535 + 1616/6363 + 1616/9999
= 32/21 + 1616/6363 + 1616/9999
= 16/9 + 1616/9999
= 64/33
Ta có : \(\frac{-3434}{8585}=\frac{x}{5}\)tức là (-3434).5 = 8585x
Vậy \(x=\frac{\left(-3434\right).5}{8585}=-2\).
\(\frac{-3434}{8585}\)=\(\frac{x}{5}\)
=\(\frac{-3434}{8585}=\frac{-3434:1717}{8585:1717}=\frac{-2}{5}\)
Vậy x = -2
học tốt
#Tan
\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)
\(=12\cdot\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=12\cdot\frac{4}{33}\)
\(=\frac{16}{11}\)
\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)
\(=12\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=12\left(\frac{1}{3\cdot5}+\frac{1}{3\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right)\)
\(=12\cdot\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{3\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)
\(=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=6\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=6\cdot\frac{8}{33}\)
\(=\frac{48}{33}\)
Mình giải theo kiểu lớp 6 nhá !
=\(33.\left(\frac{34}{15}+\frac{34}{35}+\frac{34}{63}+\frac{34}{99}\right)\)
=\(33.\left[34.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\right]\)
=\(33.\left[34.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\right]\)
=\(33.\left[34.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).\frac{1}{2}\right]\)
=\(33.\left[34\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).\frac{1}{2}\right]\)
=\(33.\left[34.\left(\frac{1}{3}-\frac{1}{11}\right).\frac{1}{2}\right]\)
=\(33.\left(34.\frac{8}{33}.\frac{1}{2}\right)\)
=\(33.\frac{136}{33}\)
=\(\frac{33.136}{33}\)(*)
=\(136\)
(Bạn có thể bỏ bước có dấu *)