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Bài 1:
a) \(\dfrac{5}{18}\) và \(\dfrac{4}{3}\)
Mẫu số chung: \(18\)
\(\dfrac{4}{3}=\dfrac{4\times6}{3\times6}=\dfrac{24}{18}\)
b) \(\dfrac{7}{5}\) và \(\dfrac{8}{11}\)
Mẫu số chung: \(55\)
\(\dfrac{7}{5}=\dfrac{7\times11}{5\times11}=\dfrac{77}{55}\) ; \(\dfrac{8}{11}=\dfrac{8\times5}{11\times5}=\dfrac{40}{55}\)
c) \(\dfrac{7}{12}\) và \(\dfrac{5}{6}\)
Mẫu số chung: \(12\)
\(\dfrac{5}{6}=\dfrac{5\times2}{6\times2}=\dfrac{10}{12}\)
Bài 2:
a) \(\dfrac{5}{10}\) và \(\dfrac{25}{75}\)
\(\dfrac{5}{10}=\dfrac{5:5}{10:5}=\dfrac{1}{2}\) ; \(\dfrac{25}{75}=\dfrac{25:25}{75:25}=\dfrac{1}{3}\)
Mẫu số chung: 6
\(\dfrac{1}{2}=\dfrac{1\times3}{2\times3}=\dfrac{2}{6}\) ; \(\dfrac{1}{3}=\dfrac{2\times2}{3\times2}=\dfrac{4}{6}\)
b) \(\dfrac{42}{56}\) và \(\dfrac{18}{48}\)
\(\dfrac{42}{56}=\dfrac{42:14}{56:14}=\dfrac{3}{4}\) ; \(\dfrac{18}{48}=\dfrac{18:6}{48:6}=\dfrac{3}{8}\)
Mẫu số chung: 8
\(\dfrac{3}{4}=\dfrac{3\times2}{4\times2}=\dfrac{6}{8}\)
Bài 3:
a) \(\dfrac{12\times5}{4\times9\times5}\) \(=\dfrac{3\times4\times5}{4\times3\times3\times5}=\dfrac{1}{3}\)
b) \(\dfrac{32\times50}{48\times75}\) \(=\dfrac{4\times8\times25\times2}{2\times3\times8\times25\times3}=\dfrac{4}{9}\)
c) \(\dfrac{63\times81}{21\times27}\) \(=\dfrac{7\times9\times3\times3\times9}{3\times7\times3\times9}=9\)
d) \(\dfrac{11\times10}{4\times55}\) \(=\dfrac{11\times5\times2}{2\times2\times11\times5}=\dfrac{1}{2}\)
( x - 1 )3 = 27
( x - 1 )3 = 33
\(\Rightarrow\) x - 1 = 3
x = 4
\(y\times42+y\times57+y=25400\)
\(y\times\left(42+57+1\right)=25400\)
\(y\times100=25400\)
\(y=25400:100\)
\(y=254\)
Theo đề, số học sinh tham quan xếp mỗi xe 30, 40 hoặc 48 em đều vừa đủ nên số học sinh đi tham quan thược \(ƯC\left(30,40,48\right)\)
\(\RightarrowƯC\left(30,40,48\right)=\left\{0;240;480;720;960;1200\right\}\)
Vì nhà trường tổ chức cho khoảng 900 đến 1000 học sinh nên số học sinh đi tham quan là 960 học sinh
Vậy có 960 học sinh đi tham quan
a) Công suất tỏa nhiệt của bếp là:
\(P=I^2.R=2,5^2.80=500\left(W\right)\)
b) Nhiệt lương để đun nước từ 25oC
\(Q_i=m.c.\Delta t=1,5.4200.\left(100-25\right)=472500\left(J\right)\)
Nhiệt lượng bếp tỏa ra là:
\(Q=I^2.R.t=P.t=500.20.60=600000\left(J\right)\)
Hiệu suất của bếp là:
\(H=\dfrac{Q_i}{Q}.100\%=\dfrac{472500}{600000}.100\%=78,75\%\)
\(\left(x\times11\right):3=\left(3372\times5\right):3\)
\(\left(x\times11\right):3=16860:3\)
\(\left(x\times11\right):3=5620\)
\(x\times11=5620\times3\)
\(x\times11=16860\)
\(x=16860:11\)
\(x=\dfrac{16860}{11}\)
\(456.75+134.68-2009\)
\(=34200+9112-2009\)
\(=43312-2009\)
\(=41303\)
\(55.48-110.24+123\)
\(=55.24.2-110.24+123\)
\(=\left(55.2-110\right).24+123\)
\(=0.24+123\)
\(=0\)