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=\(\frac{15}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)
=\(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
=\(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)
=\(\frac{15}{4}.\frac{2}{450}\)
=\(\frac{1}{60}\)
\(dungthikkothithoithanks\)
\(\frac{15-\frac{15}{7}-\frac{15}{12}-\frac{15}{98}}{18-\frac{18}{7}-\frac{18}{12}-\frac{18}{98}}=\frac{15\left(1+\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}{18\left(1+\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}=\frac{15}{18}=\frac{5}{6}\)
a) \(A=\frac{15^{16}+1}{15^{17}+1}\)và\(B=\frac{15^{15}+1}{15^{16}+1}\)
ta có \(A=\frac{15^{16}}{15^{17}}\)và\(B=\frac{15^{15}}{15^{16}}\)
ta dễ nhận thấy phần cơ số của hai phân số A và B = nhau
mà phần mũ của các lũy thừa phân số A đều lớn hơn phân số B
\(\Rightarrow\frac{15^{16}}{15^{17}}>\frac{15^{15}}{15^{16}}\)
\(\Rightarrow\frac{15^{16}+1}{15^{17}+1}>\frac{15^{15}+1}{15^{16}+1}\)
\(\Rightarrow A>B\)
\(A=\frac{15^{16}+1}{15^{17}+1}vaB=\frac{15^{15}+1}{15^{16}+1}\)
+)Ta thấy\(A=\frac{15^{16}+1}{15^{17}+1}< 1\)
\(\Rightarrow A< \frac{15^{16}+1+14}{15^{17}+1+14}=\frac{15^{16}+15}{15^{17}+15}=\frac{15.\left(15^{15}+1\right)}{15.\left(15^{15}+1\right)}=\frac{15^{15}+1}{15^{16}+1}=B\)
Vậy A<B
b)Đề sai
Chúc bạn học tốt
F=7/4(1/90-1/94+1/94-1/98+...+1/158-1/162)
=7/4(1/90-1/162)
=7/4.2/405
=7/810
Vậy F=7/810
đặt A=1/1.2.3+1/2.3.4+...+1/37.38.39
2A=1/1.2-1/2.3+1/2.3-1/3.4+...+1/37.38-1/38.39
2A=1/1.2-1/38.39
2A=1/2-1/1482
2A=370/741
A=370/741.1/2
A=185/741
HOÀN THÀNH
a) \(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}.........1\frac{1}{99}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}......\frac{100}{99}\)
\(=\frac{\left(2.2\right).\left(3.3\right).\left(4.4\right).\left(5.5\right)....\left(10.10\right)}{\left(1.3\right).\left(2.4\right).\left(3.5\right).\left(4.6\right).....\left(9.11\right)}\)
\(=\frac{\left(2.3.4.5...10\right).\left(23.4.5....10\right)}{\left(1.2.3.4...9\right).\left(3.4.5.6....11\right)}=\frac{10}{1}.\frac{2}{11}=\frac{20}{11}\)
b) \(\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}=\frac{99}{98}-\frac{98}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)-\left(\frac{98}{97}-\frac{1}{97}\right)\)
\(=\frac{98}{98}-\frac{97}{97}=1-1=0\)
\(=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{15.16}\)
\(=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
=\(\frac{1}{8}\)
Tích cho mình nhé cảm ơn
\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)
\(=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=2\left(\frac{4}{16}-\frac{1}{16}\right)\)
\(=2\times\frac{3}{16}\)
\(=\frac{6}{16}=\frac{3}{8}\)
Đặt \(A=\frac{15+\frac{15}{7}-\frac{15}{11}+\frac{15}{2009}-\frac{15}{13}}{\frac{4}{2009}-\frac{4}{13}+\frac{4}{7}-\frac{4}{11}+4}\)
\(=\frac{15\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{2009}-\frac{1}{13}\right)}{4\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{2009}-\frac{1}{13}\right)}\)
\(=\frac{15}{4}\)
Đặt \(B=\frac{5\cdot2010-1996}{14+4\cdot2010}\)
\(=\frac{5\left(1996+4\right)-1996}{14+4\cdot2010}\)
\(=\frac{5\cdot1996+20-1996}{14+4\left(1996+4\right)}\)
\(=\frac{4\cdot1996+20}{4\cdot1996+30}\)
\(\Rightarrow A\cdot B=\frac{4\cdot1996+20}{4\cdot1996+30}\cdot\frac{15}{4}=\frac{15\cdot4\left(1996+5\right)}{4\left(4\cdot1996+30\right)}=\frac{15\left(1996+5\right)}{4\cdot1996+30}=\frac{30015}{8004}\)
mặc dầu ko khoa học lắm nhưng mình thấy cũng được đấy
Ta có: \(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)
= \(15.\left(\frac{1}{90.94}+\frac{1}{94.98}+\frac{1}{98.102}+...+\frac{1}{146+150}\right)\)
= \(15.\left[\frac{1}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+\frac{4}{98.102}+...+\frac{4}{146+150}\right)\right]\)
= \(15.\left[\frac{1}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\right]\)
= \(15.\left[\frac{1}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\right]\)
= \(15.\left(\frac{1}{4}.\frac{1}{225}\right)\)
= \(=\frac{1}{60}\)
Bài làm
\(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)
= \(15.\frac{1}{90.94}+15.\frac{1}{94.98}+15.\frac{1}{98.102}+...+15.\frac{1}{146.150}\)
= \(15.\left(\frac{1}{90.94}+\frac{1}{94.98}+\frac{1}{98.102}+...+\frac{1}{146.150}\right)\)
= \(15.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)
= \(15.\left(\frac{1}{90}-\frac{1}{150}\right)\)
= \(15.\left(\frac{5}{450}-\frac{3}{450}\right)\)
= \(15.\frac{2}{450}\)
= \(\frac{2}{30}\)
# Chúc bạn học tốt #