7^3
Các phân số 3/4 ; 7/7 ; 3/2 ; 4/3 được xếp theo thứ tự tăng dần là: (ghi cách giải)
A.3/4 ; 7/7 ; 3/2 ; 4/3
B.7/7 ; 4/3 ; 3/4 ; 3/2
C.3/2 ; 4/3 ; 7/7 ; 3/4
D.3/4 ; 7/7 ; 4/3 ; 3/2
2/5+3/7=...
1/7+3/7+4/7=(1/7+3/7)+....=1/7+(3/7+...)=(1/7+4/7)+...
\(\dfrac{2}{5}+\dfrac{3}{7}=\dfrac{14}{35}+\dfrac{15}{35}=\dfrac{29}{35}\)
\(\dfrac{1}{7}+\dfrac{3}{7}+\dfrac{4}{7}=\left(\dfrac{1}{7}+\dfrac{3}{7}\right)+\dfrac{4}{7}=\dfrac{1}{7}+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)=\left(\dfrac{1}{7}+\dfrac{4}{7}\right)+\dfrac{3}{7}\)
Tính:
A=1+7+7^2 +7^3+..+7^2007
B=1+4+4^2+4^3+...+4^100
C=1+3^2+3^4+3^6+3^8+...+3^100
D=7+7^3+7^5+7^7+7^9+...+7^99
E=2+2^3+2^5+2^7+2^9+...+2^9009
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(7A-A=\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)
\(6A=7^{2008}-1\)
\(A=\frac{7^{2008}-1}{6}\)
Tương tự, \(B=\frac{4^{101}-1}{3},C=\frac{3^{101}-1}{2}\).
\(D=7+7^3+7^5+7^7+...+7^{99}\)
\(7^2.D=7^3+7^5+7^7+7^9+...+7^{101}\)
\(\left(7^2-1\right)D=\left(7^3+7^5+7^7+7^9+...+7^{101}\right)-\left(7+7^3+7^5+7^7+...+7^{99}\right)\)
\(48D=7^{101}-7\)
\(D=\frac{7^{101}-7}{48}\)
Tương tự, \(E=\frac{2^{9011}-2}{3}\)
so sánh 3/7(3/7)^19 và [(-3/7)^5]^4so sánh 3/7(3/7)^19 và [(-3/7)^5]^4
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C = 1 + 7 + 7 mu 2 + 7 mu 3 + ...+ 7 mu 101 chia het cho 8.
B= 1 + 3 + 3 mu 2 + 3 mu 3 + ....+ 3 mu 2000
D = 7 - 7 mu 4 + 7 mu 7 - 7 mu 10 - ... - 7 mu 298 + 7 mu 301
3+7+3+7+3+7+3+7+3+7+3+7+3+7 =
nhanh nhé ! ai gửi đầu tiên mình tick cho !
3+7+3+7+3+7+3+7+3+7+3+7+3+7 = 3.7+7.7=21+49=70
3+7+3+7+3+7+3+7+3+7+3+7+3+7
=10+10+10+10+10+10+10
=10x7
=70
ai k mình mình k lại
Cho \(A=\dfrac{1+7+7^2+7^3+...+7^{11}}{1+7+7^2+7^3+...+7^{10}}\) \(B=\dfrac{1+3+3^2+3^3+...+3^{11}}{1+3+3^2+3^3+...+3^{10}}\)
So sánh A và B
Ta có
A = \(\dfrac{1+7+7^2+7^3+...+7^{11}}{1+7+7^2+7^3+...+7^{10}}\)
Đặt C = 1 + 7 + 72 + 73+...+711
7C = 7 + 72 + 73 + ... + 711 + 712
=> 6C = 712 - 1
C = \(\dfrac{7^{12}-1}{6}\)
Đặt D = 1 + 7 + 72 + 73+...+710
7D = 7 + 72 + 73 + ... + 710 + 711
=> 6D = \(7^{11}-1\)
D = \(\dfrac{7^{11}-1}{6}\)
=> A = \(\dfrac{\dfrac{7^{12}-1}{6}}{\dfrac{7^{11}-1}{6}}\)
A = \(\dfrac{7^{12}-1}{6}\) : \(\dfrac{7^{11}-1}{6}\)
A = \(\dfrac{7^{12}-1}{6}.\dfrac{6}{7^{11}-1}\)
A = \(\dfrac{7^{12}-1}{7^{11}-1}\) = 7, 000000003
Lại có:
B = \(\dfrac{1+3+3^2+3^3+...+3^{11}}{1+3+3^2+3^3+...+3^{10}}\)\
Đặt H = \(1+3+3^2+3^3+...+3^{11}\)
3H = \(3+3^2+3^3+...+3^{12}\)
=> 2H = \(3^{12}-1\)
H = \(\dfrac{3^{12}-1}{2}\)
Đặt Q = \(1+3+3^2+3^3+...+3^{10}\)
3Q = \(3+3^2+3^3+...+3^{10}+3^{11}\)
=> 2Q = \(3^{11}-1\)
Q = \(\dfrac{3^{11}-1}{2}\)
=> B = \(\dfrac{\dfrac{3^{12}-1}{2}}{\dfrac{3^{11}-1}{2}}\)
B = \(\dfrac{3^{12}-1}{2}:\dfrac{3^{11}-1}{2}\)
B = \(\dfrac{3^{12}-1}{2}.\dfrac{2}{3^{11}-1}\)
B = \(\dfrac{3^{12}-1}{3^{11}-1}\)
B = 3, 00001129
Vì 7, 000000003 > 3, 00001129
=> A > B
Vậy A > B
Bài 6 : Tính giá trị các biểu thức .
a. A = -5/7 + 7/-5 + 4/7 + 7/4 .
b. B = 2/-5 + -3/7 + -7/10 + 3/-8 .
c. C = -5/7 + 2/-7 + 4/-9 + 4/9 .
d. D = ( 3 - 3/4 + 2/3 ) - ( 2 + 4/3 - 3/2 ) - ( 1 - 7/3 - 9/2 ) .
(3-1/4+2/3) - (5-1/3-6/5) - (6-7/4+3/2) (6-2/3+1/2) - (5+5/3-3/2)-(3-7/3+5/2)
(5/3-3/7+9)-(2+5/7-2/3)+(8/7-4/3-10) (8-9/4+2/7)-(-6-3/7+5/4)-(3+2/4-9/7 )
mọi người ơi giúp mik với ạ
1. các phân số được sắp xếp theo thứ tự tăng dần là ....
A. 2/3 ; 7/8 ; 7/7 ; 4/3 B. 2/3 ; 4/3 ; 7/7 ; 7/8 C. 2/3 ; 7/7 ; 7/8 ; 4/3 D. 7/8 ; 7/7 ; 2/3 ; 4/3
2. trong các số 254 ; 731 ; 335 ; 948 số nào chia hết cho 2 và 2 ?