Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)
b: =(-4)+(-4)+...+(-4)
=-4*25=-100
c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)
=10*53
=530
Đặt \(A=1.3.5.7...99\)
\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\)
Ta có:
\(A=1.3.5.7...99\)
\(\Rightarrow A=\dfrac{\left(1.3.5.7...99\right)\left(2.4.6.8...100\right)}{2.4.6.8...100}\)
\(\Rightarrow A=\dfrac{1.2.3.4...100}{2.4.6.8...100}\)
\(\Rightarrow A=\dfrac{1.2.3.4...100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.50\right)}\)
\(\Rightarrow A=\dfrac{\left(1.2.3.4...50\right)\left(51.52.53...100\right)}{\left(1.2.3.4...50\right)\left(2.2.2.2...2\right)}\)
\(\Rightarrow A=\dfrac{51.52.53.54...100}{2.2.2.2...2}\)
\(\Rightarrow A=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}....\dfrac{100}{2}\)
\(\Rightarrow A=B\)
Vậy \(1.3.5.7...99=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\) (Đpcm)
VT: 1.3.5.7....99=\(\dfrac{(1.3.5.7.....99).\left(2.4.6....100\right)}{2.4.6....100}\)
\(=\dfrac{\left(1.3.5.7.....99\right)\left(2.4.6.....100\right)}{1.2.2.2.2.3.....2.50}\)\(=\dfrac{\left(1.2.3.4.....50\right)\left(51.52.53....100\right)}{\left(1.2.3.4......50\right)\left(2.2.2.2.2....2\right)}\)
\(=\dfrac{51.52.53......100}{2.2.2.2.....2}=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}......\dfrac{100}{2}=VP\left(đpcm\right)\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\cdot\dfrac{1}{2}-2\cdot\dfrac{1}{4}-...-2\cdot\dfrac{1}{100}\)
\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\dfrac{1}{1}-\dfrac{1}{2}-...-\dfrac{1}{50}\)
\(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)
\(\Rightarrow A=B\)
tớ giải chi tiết hơn nhá:
A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=(\(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{99}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
A=\(\left(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
A=\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)
Vậy A=B
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)
\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)
\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)
tính chất trên gọi là tính chất bắc cầu, ta so sánh hai phân số với một số (phân số) thứ 3.
a) \(\dfrac{3}{4}+\dfrac{3}{5}-\dfrac{18}{60}\) ( MTC: 60)
= \(\dfrac{3.15}{4.15}+\dfrac{3.12}{5.12}-\dfrac{18}{60}\)
= \(\dfrac{45}{60}+\dfrac{36}{60}-\dfrac{18}{60}\)
= \(\dfrac{45+36-18}{60}\)=\(\dfrac{63}{60}=\dfrac{21}{20}\)
b)\(\dfrac{17}{8}-\dfrac{11}{6}-\dfrac{2}{9}\) (MTC:72)
=\(\dfrac{17.9}{8.9}-\dfrac{11.12}{6.12}-\dfrac{2.8}{9.8}\)
= \(\dfrac{153}{72}-\dfrac{132}{72}-\dfrac{16}{72}\)
=\(\dfrac{153-132-16}{72}\)
=\(\dfrac{5}{72}\)
c)\(\dfrac{23}{29}+\dfrac{5}{11}+\dfrac{17}{11}\) (MTC:319)
= \(\dfrac{23.11}{29.11}+\dfrac{5.29}{11.29}+\dfrac{17.29}{11.29}\)
=\(\dfrac{253}{319}+\dfrac{145}{319}+\dfrac{493}{319}\)
=\(\dfrac{253+145+493}{319}\)=\(\dfrac{891}{319}=\dfrac{81}{29}\)
c) \(\dfrac{20}{45}+\dfrac{14}{35}+\dfrac{32}{44}\)
= \(\dfrac{4}{9}+\dfrac{2}{5}+\dfrac{8}{11}\)(Rút gọn b/thức)(MTC:495)
=\(\dfrac{4.55}{9.55}+\dfrac{2.99}{5.99}+\dfrac{8.45}{11.45}\)
=\(\dfrac{220}{495}+\dfrac{198}{495}+\dfrac{360}{495}\)
=\(\dfrac{220+198+360}{495}\)=\(\dfrac{778}{495}\)
e)\(17\dfrac{25}{27}+3\dfrac{7}{2}\)
= \(\dfrac{484}{27}+\dfrac{13}{2}\) (MTC:54)
=\(\dfrac{484.2}{27.2}+\dfrac{13.27}{2.27}\)
\(=\dfrac{968}{54}+\dfrac{351}{54}\)
=\(\dfrac{968+351}{54}=\dfrac{1319}{54}\)
b)Ta có :
\(A=1.3.5...........99\)
\(\Rightarrow A=\dfrac{\left(1.3.7.9.............99\right)\left(2.4.6.8........100\right)}{2.4.6.8.............100}\)
\(\Rightarrow A=\dfrac{1.2.3.4.............100}{2.4.6.8................100}\)
\(\Rightarrow A=\dfrac{1.2.3.4..................100}{\left(2.1\right)\left(2.2\right)...............\left(2.50\right)}\)
\(\Rightarrow A=\dfrac{51.52.53...........................100}{2.2.2.2.............................2}\)
\(\Rightarrow A=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}.............\dfrac{100}{2}\)
\(\Rightarrow A=D\)
~ Chúc bn học tốt ~
mk hỏi này sao mà 1.2.3.4.....100/(2.1).(2.2)...(2.50)lại =51.52.53..........100/2.2.2........2