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a) Ta có : 2 + 4 + 6 + 8 +... + 16 + 18
SSH : (18 - 2) : 2 + 1 = 9(số)
Tổng : (2 + 18).9 : 2 = 90
\(A=\frac{27\cdot45+27\cdot45}{90}=\frac{27\left(45+45\right)}{90}=\frac{27\cdot90}{90}=27\)
b) Ta có : 135.1420 + 45.780.3 = 135.1420 + 135.780 = 135(1420 + 780) = 135.2200 (*)
3 + 6 + 9 + 12 + ... + 24 + 27
Số số hạng : (27 - 3) : 3 + 1 = 9(số)
Tổng : (3 + 27).9 : 2 = 135 (**)
Từ (*) và (**) suy ra
=> \(B=\frac{135\cdot2200}{135}=2200\)
a)
2 + 4 + 6 + ... + 16 + 18
Số số hạng :
( 18 - 2 ) / 2 + 1 = 9
Tổng :
( 18 + 2 ) x 9 / 2 = 90
\(A=\frac{27\cdot45+27\cdot45}{90}\)
\(=\frac{27\left(45+45\right)}{90}\)
\(=\frac{27\cdot90}{90}\)
\(=27\)
b)
3 + 6 + 9 + 12 + ... + 24 + 27
Số số hạng :
( 27 - 3 ) / 3 + 1 = 9
Tổng :
( 27 + 3 ) x 9 /2 = 135
\(B=\frac{135\cdot1420+45\cdot780\cdot3}{135}\)
\(=\frac{135\cdot1420+135\cdot780}{135}\)
\(=\frac{135\left(1420+780\right)}{135}\)
\(=\frac{135\cdot2200}{135}\)
= 2200
Giusp m với. Tìm số A biết. 121:A dư 10, 61 : A dư 10. giải thích cách làm theo lớp 6 b nhé.Thank
a) 378
b) 3
c) 2
d) 2
e) \(\frac{8748}{1715}\)
Mình thấy bài e) bạn có ghi thiếu ko vậy.81^2 x;: hay là cộng trừ vậy?
\(a.\frac{27.45+27.55}{2+4+6+...+14+16+18}=\frac{27.100}{\frac{\left(2+18\right).9}{2}}=30\)
\(b.\frac{26.108-26.12}{32-28+24-20+16-12+8-4}=\frac{26\left(108-12\right)}{\left(32-28\right).4}=\frac{26.96}{4.4}=156\)
\(c.\frac{27.4500+135.550.2}{2+4+6+...+14+16+18}=\frac{270.450+270.550}{\frac{\left(2+8\right).9}{2}}\)
\(d.\frac{48.700-24.45.20}{45-40+35-30+25-20+15-10+5}=\frac{48.700-48.450}{5.5}\)\(=\frac{48\left(700-450\right)}{25}=\frac{48.250}{25}=480\)
#ĐinhBa
B1
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)
\(x-\frac{11}{6}=1:\frac{3}{50}\)
\(x-\frac{11}{6}=\frac{50}{3}\)
\(x=\frac{50}{3}+\frac{11}{6}\)
\(x=\frac{37}{2}\)
b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)
\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)
\(\frac{5}{7}:x=-\frac{4}{15}\)
\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)
\(x=-\frac{75}{28}\)
c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)
\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)
\(\frac{2}{5}.x=-1\)
\(x=-1:\frac{2}{5}\)
\(x=-\frac{5}{2}\)
B2
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)
\(=\frac{7}{6}.24:5-\frac{33}{10}\)
\(=28:5-\frac{33}{10}\)
\(=\frac{28}{5}-\frac{33}{10}\)
\(=\frac{56}{10}-\frac{33}{10}\)
\(=\frac{23}{10}\)
b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)
\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)
\(=\frac{61}{70}+0:\frac{25}{144}\)
\(=\frac{61}{70}+0\)
\(=\frac{61}{70}\)