tính tổng
A = 70 .[ 121212/565656 + 121212/727272 + 121212/909090
B = 191/210 + 161/240 +129/272 + 95/306
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=\(70\left(\frac{121212}{565656}+\frac{121212}{727272}+\frac{121212}{909090}\right)\)
=\(70\left(\frac{3.40404}{14.40404}+\frac{121212}{6.121212}+\frac{2.60606}{15.60606}\right)\)
=\(70\left(\frac{3}{14}+\frac{1}{6}+\frac{2}{15}\right)\)
=\(70.\frac{18}{35}=38\)
Ta có \(70\left(\frac{121212}{565656}+\frac{121212}{727272}+\frac{121212}{909090}\right)\)
\(=70\left(\frac{3\cdot40404}{14\cdot40404}+\frac{121212}{121212\cdot6}+\frac{2\cdot60606}{15\cdot60606}\right)\)
\(=70\left(\frac{3}{14}+\frac{1}{6}+\frac{2}{15}\right)\)
\(=70\cdot\frac{18}{35}=36\)
a, 1/210+1/240+1/272+1/306
= 1/14.15 + 1/15.16 + 1/16.17 +1/17.18
= 1/14 - 1/15 + 1/15 - 1/16 +1/16 - 1/17 + 1/17 - 1/18
= 1/14 - 1/18 = 1/63
a) \(\frac{1}{210}+\frac{1}{240}+\frac{1}{272}+\frac{1}{306}\)
=\(\frac{1}{14.15}+\frac{1}{15.16}+\frac{1}{16.17}+\frac{1}{17.18}\)
=\(\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}+\frac{1}{16}-\frac{1}{17}+\frac{1}{17}-\frac{1}{18}\)
=\(\frac{1}{14}-\frac{1}{18}=\frac{1}{63}\)
b)\(\frac{191}{210}+\frac{161}{240}+\frac{129}{272}+\frac{95}{306}\)
=\(\frac{191}{14.15}+\frac{161}{15.16}+\frac{129}{16.17}+\frac{95}{17.18}\)
=\(\frac{191}{14}-\frac{191}{15}+\frac{161}{15}-\frac{161}{16}+\frac{129}{16}-\frac{129}{17}+\frac{95}{17}-\frac{95}{18}\)
tự làm tiếp
Đặt
\(A=210.\left(\frac{111111}{121212}+\frac{111111}{202020}+\frac{111111}{303030}+\frac{111111}{424242}+\frac{111111}{565656}+\frac{111111}{727272}+\frac{111111}{909090}\right)\)
\(=210.\left(\frac{11}{12}+\frac{11}{20}+\frac{11}{30}+\frac{11}{42}+\frac{11}{56}+\frac{11}{72}+\frac{11}{90}\right)\)
\(=210.11\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=2310.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=2310.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2310.\left(\frac{1}{3}-\frac{1}{10}\right)\)
\(=2310.\frac{7}{30}\)
\(=539\)
210.(111111/121212 +111111/202020+11111/303030 + 111111/424242 +111111/565656 +111111/727272 +111111/909090)
= 210 . ( 11/12+11/20+11/42+11/56+11/72+11/90)
= 210.(11.(1/12+1/20+1/42+1/56+1/72+1/90))
=210.(11(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10))
=210.(11.(1/3-1/10))
=210.(11.7/30)
=210.77/30
=593
T I C K GIÙM MIK
a = 36
b=149/63
thanks