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\(\frac{2.3+6.9+10.15+14.21}{2.5+6.15+10.25+14.35}\)
Rút gọn:
\(=\frac{3}{5}+\frac{3}{5}+\frac{3}{5}+\frac{3}{5}\)
\(=\frac{12}{5}\)
\(\frac{2.3+6.9+10.15+14.21}{2.5+6.15+10.25+14.35}=\frac{2.3+6.3.3+10.5.3+14.7.3}{2.5+6.3.5+10.5.5+14.7.5}\)
\(=\frac{3\left(2.1+6.3+10.5+14.7\right)}{5\left(2.1+6.3+10.5+14.7\right)}=\frac{3}{5}\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{35.37}\)
\(C=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{35.37}\right)\)
\(C=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{35}-\frac{1}{37}\right)\)
\(C=\frac{1}{2}.\left(1-\frac{1}{37}\right)\)
\(C=\frac{1}{2}.\frac{36}{37}\)
\(C=\frac{18}{37}\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{35.37}\)
\(C=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{35}-\frac{1}{37}\right)\)
\(C=\frac{1}{2}\cdot\left(1-\frac{1}{37}\right)\)
\(C=\frac{1}{2}\cdot\frac{36}{37}=\frac{18}{37}\)
Vay C = \(\frac{18}{37}\)
BÀI 1:
\(N=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{17.20}\)
\(N=2.\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{17.20}\right)\)
\(N=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(N=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(N=2.\frac{9}{20}\)
\(N=\frac{9}{10}\)
BÀI 2:
\(C=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+...+99.100.3\)
\(3B=1.2\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(3B=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+...+98.99.100\right)\)
\(3B=99.100.101\)
\(3B=999900\)
\(\Rightarrow B=999900:3\)
\(B=333300\)
CHÚC BN HỌC TỐT!!!!
\(a,A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+...+\frac{3}{90}\)
\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=3.\left(1-\frac{1}{10}\right)\)
\(A=3.\frac{9}{10}=\frac{27}{10}\)
\(b,B=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)
\(B.\frac{3}{2}=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{17}\)
\(B=\frac{15}{34}:\frac{3}{2}=\frac{5}{17}\)
tA CÓ :
\(\frac{5.6+5.7}{5.8+20}=\frac{5\left(6+7\right)}{5.\left(8+4\right)}=\frac{5.13}{5.12}=\frac{13}{12}\)
\(\frac{8.9-4.15}{12.7-180}=\frac{3.4\left(2.3-5\right)}{12\left(7-15\right)}=\frac{12\left(6-5\right)}{12\left(7-15\right)}=\frac{6-5}{7-15}=\frac{1}{-8}\)
c.\(=3\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\right)\)
\(=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=3\left(1-\frac{1}{101}\right)\)
\(=\frac{300}{101}\)
Bài 1
a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk mình nha!!
Câu 2:
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)
\(=\frac{3\cdot100}{2}\)
\(=\frac{300}{2}=150\)
A có tổng cộng 49 số hạng, nhóm 2 số hạng liên tiếp với nhau được:
\(A=\left(\frac{1}{1.3}-\frac{2}{3.5}\right)+\left(\frac{3}{5.7}-\frac{4}{7.9}\right)+...+\left(\frac{47}{93.95}-\frac{48}{95.97}\right)+\frac{49}{97.99}\)
\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{93.97}+\frac{49}{97.99}\)=> \(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{93.97}+\frac{196}{97.99}=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{97}+\frac{196}{97.99}\)
=> \(4A=1-\frac{1}{97}+\frac{196}{97.99}=\frac{96}{97}+\frac{196}{97.99}=\frac{9700}{97.99}=\frac{100}{99}>1\)
\(4A>1=>A>\frac{1}{4}\)
\(C=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{45.47}\)
\(\Rightarrow C=\frac{6}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\)
\(\Rightarrow C=3.\left(\frac{1}{3}-\frac{1}{47}\right)\)
\(\Rightarrow C=3.\frac{44}{141}\)
\(\Rightarrow C=\frac{44}{47}\)
\(C=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{45.47}=3.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{45.47}\right)=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\\ \)
\(=3.\left(\frac{1}{3}-\frac{1}{47}\right)=\frac{3.44}{141}=\frac{44}{47}\)
tử : đặt 2.5 chung
mẫu : đặt 5.7 chung
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