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Câu 9: Tính bằng cách thuận tiện nhất
102 + 7 +243 + 98
= (102 + 98) + (7 + 243)
= 200 + 250
= 450
Bài 1
5kg 35g = 5035kg
1 tạ 50kg < 150 yến
9600 giây = 200 phút
10 thế kỷ = 10000 năm
Bài 2
102 + 7 + 243 + 98
= (102+98)+(243+7)
= 200 + 250
= 450
a: =1/3x7/5=7/15
b: =11/9x1/2=11/18
c: =1/3x3/4=1/4
d: =1/5x4/5=4/25
e: =2/3x1/4=2/12=1/6
f: =1/2x1/3=1/6
g: =1/3x1/2=1/6
= ( 2380 + 9 x 480) : 25 - 3500 : 25
= ( 2380 + 4320 - 3500 ) : 25
= 3200 : 25
= 128
a: 1-1/2=1/2
b: 1-3/10=7/10
c:3-1/3=8/3
d: 4-1/9=35/9
e: 8-2/3=22/3
f: 5-2/5=23/5
a: bằng 1 phần 2
b: bằng 7 phần 10
c: bằng 2 phần 3
d:bằng 35 phần 9
e: bằng 22 phần 3
g: bằng 23 phần 5
chúc bn học tốt
Tổng hai phân số ban đầu là
7/9+1/5=35/45+9/45=44/45
a: \(\dfrac{5}{3}\cdot\dfrac{4}{5}=\dfrac{4}{3}\)
b: \(\dfrac{2}{3}\cdot\dfrac{3}{5}=\dfrac{2}{5}\)
c: \(\dfrac{7}{6}\cdot\dfrac{6}{5}=\dfrac{7}{5}\)
d: \(\dfrac{8}{5}\cdot\dfrac{5}{7}=\dfrac{8}{7}\)
e: \(\dfrac{7}{6}\cdot\dfrac{6}{7}=\dfrac{7}{7}=1\)
g: \(\dfrac{9}{5}\cdot\dfrac{5}{2}=\dfrac{9}{2}\)
h: \(\dfrac{4}{3}\cdot\dfrac{3}{5}=\dfrac{4}{5}\)
\(a,\dfrac{13}{4}\times\dfrac{2}{3}\times\dfrac{4}{13}\times\dfrac{3}{2}=\left(\dfrac{13}{4}\times\dfrac{4}{3}\right)\times\left(\dfrac{2}{3}\times\dfrac{3}{2}\right)=1\times1=1\)
\(b,\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{18}{21}+\dfrac{3}{21}\right)+\left(\dfrac{19}{32}+\dfrac{13}{32}\right)=1+1+1=3\\ c,\dfrac{4}{7}\times\dfrac{5}{6}+\dfrac{4}{7}\times\dfrac{1}{6}=\dfrac{4}{7}\times\left(\dfrac{5}{6}+\dfrac{1}{6}\right)=\dfrac{4}{7}\times1=\dfrac{4}{7}\\ d,\dfrac{3}{5}\times\dfrac{7}{9}-\dfrac{3}{5}\times\dfrac{2}{9}=\dfrac{3}{5}\times\left(\dfrac{7}{9}-\dfrac{2}{9}\right)=\dfrac{3}{5}\times\dfrac{5}{9}=\dfrac{1}{3}\)
\(e,\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{3}{12}=\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{1}{4}=\dfrac{1}{4}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)=\dfrac{1}{4}\times1=\dfrac{1}{4}\\ f,\dfrac{7}{9}\times\dfrac{8}{5}\times\dfrac{3}{5}=\dfrac{56}{75}\\ g,\dfrac{2}{5}\times\dfrac{3}{4}+\dfrac{3}{4}\times\dfrac{3}{5}=\dfrac{3}{4}\times\left(\dfrac{2}{5}+\dfrac{3}{5}\right)=\dfrac{3}{4}\times1=\dfrac{3}{4}\\ h,\dfrac{2006}{2005}\times\dfrac{3}{4}-\dfrac{3}{4}\times\dfrac{1}{2005}=\dfrac{3}{4}\times\left(\dfrac{2006}{2005}-\dfrac{1}{2005}\right)=\dfrac{3}{4}\times1=\dfrac{3}{4}\)
a: \(=\dfrac{13}{4}\cdot\dfrac{4}{13}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}=1\cdot1=1\)
b: \(=\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{6}{7}+\dfrac{1}{7}+\dfrac{19}{32}+\dfrac{13}{32}=1+1+1=3\)
c: \(=\dfrac{4}{7}\left(\dfrac{5}{6}+\dfrac{1}{6}\right)=\dfrac{4}{7}\)
d: \(=\dfrac{3}{5}\left(\dfrac{7}{9}-\dfrac{2}{9}\right)=\dfrac{3}{5}\cdot\dfrac{5}{9}=\dfrac{1}{3}\)
e: \(=\dfrac{1}{4}\left(\dfrac{5}{9}+\dfrac{4}{9}\right)=\dfrac{1}{4}\)
A=\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
A= \(\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+\frac{1}{8\times9}+\frac{1}{9\times10}\)
A=\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{10}\)
\(A=\frac{3}{20}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{4}-\frac{1}{10}=\frac{3}{20}\)