\(\dfrac{2^4.5^2.7}{2^3.5.7^2.11}\);\(\dfrac{2^3.3^3.5^3.7.8}{3.2^4.5^3.14}\)
\(\dfrac{4^2.5.11}{44.20}\);\(\dfrac{13.15.16}{18.65.7};\dfrac{7.2.8.5^2}{14.2.5}\)
\(\dfrac{2^3.3^3.5}{3.2^3.5^3}\)
a) \(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=\dfrac{2^3\cdot5\cdot10\cdot7}{2^3\cdot5\cdot7\cdot77}=\dfrac{10}{77}\)
\(\dfrac{2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot2^4\cdot5^3\cdot14}=\dfrac{2^3\cdot3\cdot5^3\cdot7\cdot3^2\cdot8}{3\cdot2^3\cdot2\cdot5^3\cdot14}=\dfrac{7\cdot3^2\cdot8}{2\cdot14}=\dfrac{63\cdot8}{2\cdot14}=18=\dfrac{1386}{77}\)
rut gon phan so: (-2)^3.3^3.5^3.7.8 / 3.5^3.2^4.42
(-2)^3.3^3.5^3.7.8/3.5^3.2^4.42
=(-2)^2.(-2).3^3.5^3.7.8/3.5^3.2^2.2^2.7.6
=4.(-2).3^3.5^3.7.8/3.5^3.4.4.7.6
=-1.3^2.8/6
=-72/6=-12
\(\frac{\left(-2\right)^3.3^3.5^3.7.2^3}{3.5^3.2^4.2.3.7}\)=\(\frac{\left(-2\right)^2.\left(-2\right).3^3.5^3.7.2^3}{3^2.5^3.2^5.7}\)=\(\frac{2^5.\left(-2\right).3}{2^5}\)=\(-6\)
(-2)^3.3^3.5^3.7.8/3.5^3.2^4.42. rút gọn phân số
Rút gọn phân số
\(\dfrac{2^2.3^3.5}{3.2^3.5^3}\)
Giúp em với ạ !
\(\dfrac{2^2\cdot3^3\cdot5}{3\cdot2^3\cdot5^3}=\dfrac{1}{2}\cdot3^2\cdot\dfrac{1}{5^2}=\dfrac{1}{2}\cdot\dfrac{9}{25}=\dfrac{9}{50}\)
Thực hiện phép tính
A, 3.5²-4.2³ B, 3.5²+3÷5⁰-48÷2³ C, 120+40÷4-3.5² D, 5.2³-36÷3²
Giúp mình với vì tối nay mình đi học rồi
\(A=3\cdot5^2-4\cdot2^3\\ =3\cdot25-4\cdot8\\ =75-32\\ =43\\ B=3\cdot5^2+3:5^0-48:2^3\\ =3\cdot25+3-48:8\\ =75+3-6\\ =72\\ C=120+40:4-3\cdot5^2\\ =120+10-75\\ =55\\ D=5\cdot2^3-36:3^2\\ =5\cdot8-36:9\\ =40-4\\ =36\)
Tìm ƯCLN và BCNN của:
a, 3.5^2 và 5^2. 7
b, 2^2. 3.5 ; 3^2 .7 và 3.5.7
Tinh tổng:
a) 2/1.3+2/3.5+2/5.7+.........2/99.101
b) 5/1.3+5/3.5+5/5.7+....................5/99.101
a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
tính tổng :
a.2/1.3+2/3.5+2/5.7+....+2/99.101
b.5/1.3+5/3.5+5/5.7+....+5/99.101
a.2/1.3+2/3.5+2/5.7+................+2/99.101
1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101
1-1/101
100/101
b.5/1.3+5/3.5+5/5.7+............+5/99.101
5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2
5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)
5/2(1-1/3+1/3-1/5+1/5-1/7+........+1/99-1/101)
5/2(1-1/101)
5/2.100/101
250/101
tính tổng :
p= 2/1.3+2/3.5+2.5.7+....................+ 2/49.51
q= 1/1.3+1/3.5+................................... +1/19.21
P = 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/49.51
P = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/49 - 1/51
P = 1 - 1/51
P = 50/51
Q = 1/1.3 + 1/3.5 + ... + 1/19.21
Q = 1/2 .(2/1.3 + 2/3.5 + ... + 2/19.21)
Q = 1/2.(1 - 1/3 + 1/3 - 1/5 + ... + 1/19 - 1/21)
Q = 1/2 . (1 - 1/21)
Q = 1/2. 20/21
Q = 10/21
Ủng hộ mk nha ^_-
\(P=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(P=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(P=1-\frac{1}{51}\)
\(P=\frac{50}{51}\)
\(Q=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)
\(Q=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\right)\)
\(Q=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(Q=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(Q=\frac{1}{2}.\frac{20}{21}\)
\(Q=\frac{10}{21}\)
\(P=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(=1-\frac{1}{51}\)
\(=\frac{50}{51}\)
\(Q=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)
\(2Q=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)
\(2Q=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)
\(2Q=1-\frac{1}{21}\)
\(2Q=\frac{20}{21}\)
\(Q=\frac{20}{21}:2\)
\(Q=\frac{10}{21}\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}\dfrac{2}{7.9}+.........+\dfrac{2}{99.101}\)
\(P=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
Câu 1:
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
= \(\dfrac{1}{3}-\dfrac{1}{101}\)
= \(\dfrac{98}{303}\)
Câu 2 làm tương tự ở câu 1 nhé
