Số?
a) \(\dfrac{3}{5}=\dfrac{3\times4}{5\times4}=\dfrac{?}{?}\) \(\dfrac{2}{7}=\dfrac{2\times?}{7\times3}=\dfrac{?}{?}\)
b) \(\dfrac{9}{12}=\dfrac{9:3}{12:3}=\dfrac{?}{?}\) \(\dfrac{18}{24}=\dfrac{18:6}{24:?}=\dfrac{?}{?}\)
Quy đồng mẫu số các phân số (theo mẫu).
Mẫu: \(\dfrac{2}{3};\dfrac{3}{4}\) và \(\dfrac{7}{12}\) \(\dfrac{2}{3}=\dfrac{2\times4}{3\times4}=\dfrac{8}{12};\dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\) |
a) \(\dfrac{3}{5};\dfrac{4}{7}\) và \(\dfrac{9}{35}\)
b) \(\dfrac{5}{6};\dfrac{7}{9}\) và \(\dfrac{19}{54}\)
a) \(\dfrac{3}{5}=\dfrac{21}{35};\dfrac{4}{7}=\dfrac{20}{35}\)
b) \(\dfrac{5}{6}=\dfrac{45}{54};\dfrac{7}{9}=\dfrac{42}{54}\)
Tính
a, 4\(\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
b, \(\dfrac{4^6\times9^5+6^9\times120}{-8^4\times3^{12}+6^{11}}\)
c, \(\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{12}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{10}}\)
d,\(\dfrac{30\times4^7\times3^{29}-5\times14^5\times2^{12}}{54\times6^{14}\times9^7-12\times8^5\times7^5}\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
\(\dfrac{4^{10}\times9^6+3^{12}\times8^5}{6^{13}\times4-2^{16}\times3^{12}}\)
\(\dfrac{2^4\times2^6}{\left(2^5\right)^2}-\dfrac{2^5\times15^3}{6^3\times10^2}\)
\(\dfrac{\left(-2\right)^{10}\times3^{31}+2^{40}\times\left(-3\right)^6}{\left(-2\right)^{11}\times\left(-3\right)^{31}+2^{41}\times3^6}\)
giải chi tiết giúp mình nhé
chứng minh
M=\(\dfrac{3}{1^2\times2^2}+\dfrac{5}{2^2\times3^2}+\dfrac{7}{3^2\times4^2}+.......+\dfrac{19}{9^2\times10^2}< 1\)
\(M=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(M=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(M=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(M=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
Bài 1:
\(a,\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...............+\dfrac{1}{1280}\)
\(b,\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+..............+\dfrac{1}{59049}\)
\(c,\dfrac{1}{2}\times3+\dfrac{1}{3}\times4+\dfrac{1}{4}\times5+\dfrac{1}{5}\times6\)
\(\dfrac{7}{12}\times\dfrac{2}{3}-\dfrac{5}{3}\times\dfrac{7}{12}+\dfrac{7}{12}\times3\)
Tính:
a) \(\dfrac{7}{2}\times\dfrac{1}{6}\) b) \(\dfrac{8}{11}\times4\) c) \(\dfrac{8}{9}:\dfrac{2}{5}\) d) \(\dfrac{5}{8}:7\)
a) $\frac{7}{2} \times \frac{1}{6} = \frac{7}{{12}}$
b) $\frac{8}{{11}} \times 4 = \frac{{32}}{{11}}$
c) $\frac{8}{9}:\frac{2}{5} = \frac{8}{9} \times \frac{5}{2} = \frac{{40}}{{18}} = \frac{{20}}{9}$
d) $\frac{5}{8}:7 = \frac{5}{8} \times \frac{1}{7} = \frac{5}{{56}}$
Tính giá trị của mỗi phân số sau:
\(E=\dfrac{11\times3^{29}-\left(3^2\right)^{15}}{2\times3^{14}\times2\times3^{14}}\)
\(G=\dfrac{5\times3^{11}+4\times3^{12}}{3^9\times5^2-3^0\times2^3}\)
\(H=\dfrac{\left(3\times4\times2^{16}\right)^2}{11\times2^{13}\times4^{11}-16^9}\)
\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)
\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)
Goridano
Tổng của hai số đầu là
\(\dfrac{5}{12}\times2=\dfrac{10}{12}\left(1\right)\)
Tổng của 3 số đầu là:
\(\dfrac{19}{36}\times3=\dfrac{19}{12}\left(2\right)\)
Tổng của 4 số là:
\(\dfrac{143}{249}\times4=\dfrac{143}{60}\)
Từ (1) và (2), ta thấy số thứ 3 là: \(\dfrac{19}{12}-\dfrac{10}{12}=\dfrac{48}{60}=\dfrac{4}{5}\)
Từ (2) và (3), ta thấy số cuối là:
\(\left(\dfrac{3}{4}+\dfrac{4}{5}\right)\div2=\dfrac{31}{40}\)
Số đầu là:
\(\dfrac{31}{40}-\dfrac{1}{10}=\dfrac{20}{40}=\dfrac{1}{2}\)
Theo (1), số thứ 2 là:
\(\dfrac{10}{12}-\dfrac{1}{2}=\dfrac{4}{12}=\dfrac{1}{3}\)
Đáp số : \(\dfrac{1}{2};\dfrac{1}{3};\dfrac{3}{4};\dfrac{4}{5}\)