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a) $\frac{7}{9} \times \frac{{15}}{{28}} \times \frac{9}{7} = (\frac{7}{9} \times \frac{9}{7}) \times \frac{{15}}{{28}} = 1 \times \frac{{15}}{{28}} = \frac{{15}}{{28}}$
b) $\frac{9}{{32}} \times \left( {\frac{2}{3} \times \frac{{14}}{{21}}} \right) = \frac{9}{{32}} \times \left( {\frac{2}{3} \times \frac{2}{3}} \right) = \frac{9}{{32}} \times \frac{4}{9} = \frac{{36}}{{288}} = \frac{1}{8}$
\(y+y\times\frac{1}{3}\div\frac{2}{9}+y\div\frac{2}{7}=252\)
\(3y\times\frac{1}{3}\div\frac{2}{9}\div\frac{2}{7}=252\)
\(3y\times\frac{21}{4}=252\)
\(3y=252\div\frac{21}{4}\)
\(3y=48\)
\(y=48\div3\)
\(y=16\)
Cách giải
X = 8/9 / ( 1+1/3+1/6+1/10+1/15+1/21)
1+1/3=4/3
4/3+1/6=3/2
3/2+1/10=8/5
8/5+1/15=5/3
5/3+1/21=12/7
4/3+3/2=17/6
8/5+5/3=49/15
(17/6+49/15)+17/6=183/30+17/6=286/30
8/9:286/30=2288/270=1140/135=228/27=76/9
X = 76/9
\(a,\frac{1}{2}\times\frac{4}{5}\div\frac{6}{10}\)
\(=\frac{2}{5}\div\frac{6}{10}\)
\(=\frac{2}{5}\times\frac{10}{6}\)
\(=\frac{2}{3}\)
\(b,\frac{24}{35}\div\left(\frac{4}{5}\times\frac{8}{7}\right)\)
\(=\frac{24}{35}\div\frac{32}{35}\)
\(=\frac{24}{35}\times\frac{35}{32}\)
\(=\frac{3}{4}\)
a)\(\frac{2}{3}:\frac{4}{9}+\frac{1}{3}:\frac{4}{9}\)
=\(\frac{4}{9}:\left(\frac{2}{3}+\frac{1}{3}\right)\)
=\(\frac{4}{9}:1=\frac{4}{9}\)
b
a) $\frac{2}{3} - \frac{1}{3} = \frac{{2 - 1}}{3} = \frac{1}{3}$
b) $\frac{7}{{12}} - \frac{5}{{12}} = \frac{{7 - 5}}{{12}} = \frac{2}{{12}} = \frac{1}{6}$
c) $\frac{{17}}{{21}} - \frac{{10}}{{21}} = \frac{{17 - 10}}{{21}} = \frac{7}{{21}} = \frac{1}{3}$
=(1+1+1+1+1+1+1+1)+(1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45)
Đặt A = 1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45
Ta có:
A x 1/2= 1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90
1/6=1/2x3=1/2-1/3
1/12=1/3x4=1/3-1/4
……………………
1/90=1/9x10=1/9-1/10
A x 1/2=1/2-1/3+1/3-1/4+1/4-1/5+…+1/9-1/10
A x 1/2=1/2-1/10=4/10
A=4/10:1/2=4/5
Vậy 4/3+7/6+11/10+16/15+22/21+29/28+37/36+46/45=1+1+1+1+1+1+1+1+4/5=8+4/5=44/5
\(\frac{4}{3}+\frac{7}{6}+\frac{11}{10}+...+\frac{46}{45}\)
\(=1+\frac{1}{3}+1+\frac{1}{6}+1+\frac{1}{10}+...+1+\frac{1}{45}\)
\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)(8 chữ số 1)
\(=8+\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)
Đặt A = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)
=> \(\frac{1}{2}\)A = \(\frac{1}{2}\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)
= \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Vậy A = \(\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)
Do đó biểu thức trên là 8 + \(\frac{4}{5}\) = \(\frac{44}{5}\)
Đáp số: \(\frac{44}{5}\)
a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)
Phần b T2
c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)
lấy (1/3 + 1/15 +1/10 + 1/21 ) + (1/36 + 1/28 + 1/6) + (1/45 + 1/55)
= (4/50 + 3/70) + 2/100
= 7/120 + 2/100
= 9/220
\(\frac{2}{7}:y=\frac{10}{21}.\frac{9}{14}\)
\(\frac{2}{7}:y=\frac{15}{49}\)
\(y=\frac{2}{7}:\frac{15}{49}\)
\(y=\frac{2}{7}.\frac{49}{15}\)
\(y=\frac{14}{15}\)
\(y-\frac{1}{3}=\frac{10}{21}:\frac{15}{28}\)
\(y-\frac{1}{3}=\frac{10}{21}.\frac{28}{15}\)
\(y-\frac{1}{3}=\frac{8}{9}\)
\(y=\frac{8}{9}+\frac{1}{3}\)
\(y=\frac{8}{9}+\frac{3}{9}\)
\(y=\frac{11}{9}\)
a) Ta có : \(y-\frac{1}{3}=\frac{10}{21}\div\frac{15}{28}\)
\(\Rightarrow\) \(y-\frac{1}{3}=\frac{8}{9}\)
\(\Rightarrow\) \(y\) \(=\frac{8}{9}+\frac{1}{3}\)
\(\Rightarrow\) \(y\) \(=\frac{11}{9}\)
Vậy \(y=\frac{11}{9}\)
b) Ta có : \(\frac{2}{7}\div y=\frac{10}{21}\times\frac{9}{14}\)
\(\Rightarrow\) \(\frac{2}{7}\div y=\frac{15}{49}\)
\(\Rightarrow\) \(y=\frac{2}{7}\div\frac{15}{49}\)
\(\Rightarrow\) \(y=\frac{14}{15}\)
Vậy \(y=\frac{14}{15}\)
Cbht !!!