\(\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+\frac{15}{98\cdot102}+....+\frac{15}{146\cdot150}\)
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15/90 × 94 + 15/94 × 98 + 15/98 × 102 + ... + 15/146 × 150
= 15/4 × (4/90×94 + 4/94×98 + 4/98×102 + ... + 4/146×150)
= 15/4 × (1/90 - 1/94 + 1/94 - 1/98 + 1/98 - 1/102 + ... + 1/146 - 1/150)
= 15/4 × (1/90 - 1/150)
= 15/4 × 1/30 × (1/3 - 1/5)
= 1/8 × 2/15
= 1/60
=\(\frac{15}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)
=\(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
=\(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)
=\(\frac{15}{4}.\frac{2}{450}\)
=\(\frac{1}{60}\)
\(dungthikkothithoithanks\)
\(\frac{15-\frac{15}{7}-\frac{15}{12}-\frac{15}{98}}{18-\frac{18}{7}-\frac{18}{12}-\frac{18}{98}}\)= \(\frac{15-15\left(\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}{18-18\left(\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}\)=\(\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}\)= \(\frac{15}{18}\)=\(\frac{5}{6}\)
\(\frac{15-\frac{15}{7}-\frac{15}{12}-\frac{15}{98}}{18-\frac{18}{7}-\frac{18}{12}-\frac{18}{98}}=\frac{15\left(1+\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}{18\left(1+\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}=\frac{15}{18}=\frac{5}{6}\)
A=15x(1/7-1/12-1/98)/18(1/7-1/12-1/98)
A=15/18
A=5/6
h nhe!!!
\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}\)
\(A=\frac{15}{18}=\frac{5}{6}\)
2, \(\frac{10}{1.2.3}+\frac{10}{2.3.4}+\frac{10}{3.4.5}+....+\frac{10}{100.101.102}\)
\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{102-100}{100.101.102}\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\frac{2575}{5151}\)
\(=2,499514657\)
\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{18}\right)}=\frac{15}{18}\)
\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}=\frac{15}{18}=\frac{5}{6}\)
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
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 #