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a) Cách 1 : Cách 2
1 + 3 +5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 1 + 3 +5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
=(1 + 19) + (3 + 17) +.... + (9 + 11) Áp dụng công thức tính dãy số ta có :
= 20 + 20 + ... + 20 \(\frac{\left[\left(19-1\right):2+1\right].\left(19+1\right)}{2}=\frac{10.20}{2}=10.10=100\)
= 20 x 5 = 100
b) giống bài a nhưng cách 1 làm dài lắm , mình sẽ làm cách 2
áp dụng công thức tính dãy số ta có:
\(\frac{\left[\left(200-4\right):4+1\right].\left(200+4\right)}{2}=\frac{50.204}{2}=50.102=5100\)
a, 3/5 - 1/15 x 10/13
= 3/5 - 2/39
= 117/195 - 10/195
= 107/195
b, (3/5 - 3/20) x 4/5
= (12/20 - 3/20) x 4/5
= 9/20 x 4/5
= 9/25
Chúc bạn học tốt
a) i don't know
b) \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{28.29}\)
\(5.\left(1-\frac{1}{2}\right)+5.\left(\frac{1}{2}-\frac{1}{3}\right)+5.\left(\frac{1}{3}-\frac{1}{4}\right)+...+5.\left(\frac{1}{28}-\frac{1}{29}\right)\)
\(5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{28}-\frac{1}{29}\right)\)
\(5.\left(1-\frac{1}{29}\right)\)
\(5.\frac{28}{29}\)
\(\frac{140}{29}\)
c ) \(\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{28.30}\)
\(2.\left(\frac{1}{2}-\frac{1}{4}\right)+2.\left(\frac{1}{4}-\frac{1}{6}\right)+...+2.\left(\frac{1}{28}-\frac{1}{30}\right)\)
\(2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{28}-\frac{1}{30}\right)\)
\(2.\left(\frac{1}{2}-\frac{1}{30}\right)\)
\(2.\frac{7}{15}=\frac{14}{15}\)
d) đề là thế này nè : 4/6 + 4/12 + 4/20 + 4/30 + 4/42 + 4/56
4/2.3 + 4/3.4 + 4/4.5 + 4/5.6 + 4/6.7 + 4/7.8
4.(1/2-1/3) + 4.(1/3-1/4 ) + 4.(1/4-1/5) + 4.(1/5-1/6 ) + 4.(1/6 - 1/7 ) + 4.(1/7-1/8 )
4.(1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8)
4.(1/2-1/8)
4.1/16
1/4
\(a,4\frac{1}{2}< 4\frac{3}{4}\)
\(b,2\frac{4}{5}< 3\frac{1}{4}\)
\(c,7\frac{2}{9}>5\frac{2}{9}\)
\(d,13\frac{5}{6}< 13\frac{6}{7}\)
Nao Tomori
\(a,4\frac{1}{2}....4\frac{3}{4}\Rightarrow4\frac{1}{2}=\frac{13}{2};4\frac{3}{4}=\frac{19}{4}\)
\(=4\frac{1}{2}< 4\frac{3}{4}\)
\(b,2\frac{4}{5}....3\frac{1}{4}\Rightarrow2\frac{4}{5}=\frac{14}{5};3\frac{1}{4}=\frac{13}{12}\)
\(=2\frac{4}{5}>3\frac{1}{4}\)
\(c,7\frac{2}{9}....5\frac{2}{9}\Rightarrow7\frac{2}{9}=\frac{65}{9};5\frac{2}{9}=\frac{42}{9}\)
\(=7\frac{2}{9}>5\frac{2}{9}\)
\(d,13\frac{5}{6}....13\frac{6}{7}\Rightarrow13\frac{5}{6}=\frac{83}{6};13\frac{6}{7}=\frac{97}{7}\)
\(=13\frac{5}{6}< 13\frac{6}{7}\)
P/s: Quy đồng là bước trung gian nên mk ko ghi bước quy đồng nha
1.
\(a,\frac{1}{7}+\frac{3}{4}+\frac{2}{7}+\frac{4}{7}=\frac{1}{7}+\frac{2}{7}+\frac{4}{7}+\frac{3}{4}\)
\(=1+\frac{3}{4}=\frac{7}{4}.\)
\(b,\frac{4}{7}:\frac{2}{5}+\frac{3}{7}:\frac{2}{5}-\frac{7}{9}=\left(\frac{4}{7}+\frac{3}{7}\right):\frac{2}{5}-\frac{7}{9}\)
\(=1:\frac{2}{5}-\frac{7}{9}=\frac{5}{2}-\frac{7}{9}\)
\(=\frac{31}{18}.\)
a)3/4+(1/7+2/7+4/7)
=3/4+1
=7/4
b)2/5:(4/7+3/7)-7/9
=2/5:1-7/9
=2/5-7/9
=-17/45
a) \(\frac{3}{2}-\frac{5}{6}:x=\frac{5}{15}-\frac{3}{15}\)
\(\Leftrightarrow\frac{3}{2}-\frac{5}{6}:x=\frac{2}{15}\)
\(\Leftrightarrow\frac{5}{6}:x=\frac{3}{2}-\frac{2}{15}\)
\(\Leftrightarrow\frac{5}{6}:x=\frac{41}{30}\)
\(\Leftrightarrow x=\frac{5}{6}:\frac{41}{30}\)
\(\Leftrightarrow x=\frac{25}{41}\)
b) \(x-\frac{6}{7}.\frac{14}{8}=\frac{1}{2}-\frac{2}{5}\)
\(\Leftrightarrow x-\frac{3}{2}=\frac{1}{10}\)
\(\Leftrightarrow x=\frac{1}{10}+\frac{3}{2}\)
\(\Leftrightarrow x=\frac{8}{5}\)
c) \(x:\frac{6}{5}+\frac{2}{3}=\frac{7}{3}\)
\(\Leftrightarrow x:\frac{6}{5}=\frac{7}{3}-\frac{2}{3}\)
\(\Leftrightarrow x:\frac{6}{5}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{5}{3}.\frac{6}{5}\)
\(\Leftrightarrow x=2\)
1.
7/15:(1/2-9/10xX)-1/6=13/6
=>7/15:(1/2-9/10xX)=13/6+1/6
=>7/16:(1/2-9/10xX)=7/3
=>1/2-9/10xX=7/16:7/3
=>1/2-9/10xX=3/16
=>9/10xX=1/2-3/16
=>9/10xX=5/16
=>X=5/16:9/10
=>X=25/72
a) \(\frac{1.3+3.5+5.7+7.9}{3.6+9.10+15.14+21.18}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.2.3+3.5.2.3+5.7.2.3+7.9.2.3}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.6+3.5.6+5.7.6+7.9.6}\)
= \(\frac{1.3+3.5+5.7+7.9}{6.\left(1.3+3.5+5.7+7.9\right)}=\frac{1}{6}\)
Dấu "." là dấu nhân cấp 2
b) \(\frac{1.2+2.3+3.4+4.5}{3.6+6.9+9.12+12.15}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.3.3+2.3.3.3+3.4.3.3+4.5.3.3}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.9+2.3.9+3.4.9+4.5.9}\)
= \(\frac{1.2+2.3+3.4+4.5}{9.\left(1.2+2.3+3.4+4.5\right)}=\frac{1}{9}\)
Dấu "." là dấu nhân cấp 2
c) \(\frac{0,3+\frac{3}{7}+\frac{3}{11}}{0,4+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{\frac{3}{10}+\frac{3}{7}+\frac{3}{11}}{\frac{4}{10}+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{3.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}{4.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}=\frac{3}{4}\)
a) Ta thấy \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
b) \(A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)
\(A.B=\frac{1.\left(3.5...99\right).\left(2.4.6...100\right)}{\left(2.4.6...100\right).\left(3.5.7...99\right).101}=\frac{1}{101}\)
c) vì A < b nên A . A < A . B < \(\frac{1}{101}< \frac{1}{100}\)
do đó : A . A < \(\frac{1}{10}.\frac{1}{10}\)suy ra A < \(\frac{1}{10}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\) (DẤU ''.''LÀ DẤU NHÂN NHA P)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=1+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+...+\left(-\frac{1}{13}+\frac{1}{13}\right)-\frac{1}{15}\)
\(A=1-\frac{1}{15}\)
A=\(\frac{14}{15}\)