Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
1: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà \(17^{19}+1>17^{18}+1\)
nên 17A>17B
hay A>B
2: \(C=\dfrac{98^{99}+98^{10}+1-98^{10}}{98^{89}+1}=98^{10}+\dfrac{1-98^{10}}{98^{89}+1}\)
\(D=\dfrac{98^{98}+98^{10}+1-98^{10}}{98^{88}+1}=98^{10}+\dfrac{1-98^{10}}{98^{88}+1}\)
mà \(98^{89}+1>98^{88}+1\)
nên C>D
a)=0 vì 24-42=0 số nào nhân vs 0 cũng =0
b) = 100+(98-97)+(96-95)+....+(2-1)
=100+1+1+....+1(có 46 số 1 )
=100+46
=146
a)(217 + 154).(319 - 217).(24 - 42) = 0
b)100+98+96+...+4+2-97-95-...-3-1
= 100 + (98 - 97) + (96 -95) + .... + (4 - 3) + (2 - 1)
= 100 + 1 + 1 + .... + 1 + 1 (98 : 2 = 49 số 1)
= 100 + 49
= 149
a,\(\left(2^{17}+15^4\right).\left(3^{19}-2^{19}\right).\left(4^2-2^4\right)=\left(2^{17}+15^4\right).\left(3^{19}-2^{19}\right).\left(16-16\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).0=0\)
b,\(100+98+96+...+4+2-97-95-....-3-1\)
\(=100+98-97+96-95+......+4-3+2-1\)
\(=100+\left(98-97\right)+\left(96-95\right)+.....+\left(4-3\right)+\left(2-1\right)\)
\(=100+49\times1=100+49=149\)
\(A=98.42-\left\{50.\left[\left(18-2^3\right):2+3^2\right]\right\}\)
\(=98.42-\left\{50.\left[\left(18-8\right):2+9\right]\right\}\)
\(=98.42-\left[50\left(10:2+9\right)\right]\)
\(=98.42-\left(50.14\right)\)
\(=4116-700=3416\)
\(B=-80-\left[-130-\left(12-4\right)^2\right]+2008^0\)
\(=-80-\left(-130-8^2\right)+1\)
\(=-80-\left(-130-64\right)+1\)
\(=-80+130+64+1\)
\(=115\)
\(C=1024:2^4+140:\left(38+2^5\right)-7^{23}:7^{21}\)
\(=1024:16+140:\left(38+32\right)-7^2\)
\(=64+140:70-49\)
\(=64+2-49=17\)
\(D=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left(2^4-4^2\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left(16-16\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).0\)
\(=0\)
\(E=100+98+96+....+4+2-97-95-....-3-1\)
\(=100+\left(98-97\right)+\left(96-95\right)+.....+\left(2-1\right)+\left(1-0\right)\)
\(=100+1+1+...+1+1\)
Vì lập được 49 cặp nên sẽ có 49 số 1
\(\Rightarrow E=100+1.49=100+49=149\)
b)Tận cùng=5 hoặc 0 nhưng mình ngại viết lắm,thông cảm nha
Giải:
a) A=1718+1/1719+1
17A=1719+17/1719+1
17A=1719+1+16/1719+1
17A=1+16/1719+1
Tương tự:
B=1717+1/1718+1
17B=1718+17/1718+1
17B=1718+1+16/1718+1
17B=1+16/1718+1
Vì 16/1719+1<16/1718+1 nên 17A<17B
⇒A<B
b) A=108-2/108+2
A=108+2-4/108+2
A=1+-4/108+2
Tương tự:
B=108/108+4
B=108+4-4/108+1
B=1+-4/108+1
Vì -4/108+2>-4/108+1 nên A>B
c)A=2010+1/2010-1
A=2010-1+2/2010-1
A=1+2/2010-1
Tương tự:
B=2010-1/2010-3
B=2010-3+2/2010-3
B=1+2/2010-3
Vì 2/2010-3>2/2010-1 nên B>A
⇒A<B
Chúc bạn học tốt!
17A=1719+1+16/1719+1
17A=1+16/1719+1
phần in nghiêng mình không hiểu lắm, bn giải thích cho mình được ko?
\(A=\dfrac{17^{100}+17^{96}+17^{92}+....+17^4+1}{17^{102}+17^{100}+17^{98}+....+17^2+1}\)
Gọi \(17^{100}+17^{96}+17^{92}+....+17^4+1\) là B
\(B=17^{100}+17^{96}+17^{92}+....+17^4+1\\ 17^4\cdot B=17^{104}+17^{100}+17^{96}+......+17^8+17^4\\ 17^4\cdot B-B=\left(17^{104}+17^{100}+17^{96}+......+17^8+17^4\right)-\left(17^{100}+17^{96}+17^{92}+....+17^4+1\right)\\ B\cdot\left(17^4-1\right)=17^{104}-1\\ B=\dfrac{17^{104}-1}{17^4-1}\)
Gọi \(17^{102}+17^{100}+17^{98}+....+17^2+1\) là C
\(C=17^{102}+17^{100}+17^{98}+....+17^2+1\\ C\cdot17^2=17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\\ C\cdot17^2-C=\left(17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\right)-\left(17^{102}+17^{100}+17^{98}+....+17^2+1\right)\\ C\cdot\left(17^2-1\right)=17^{104}-1\\ C=\dfrac{17^{104}-1}{17^2-1}\)
=>
\(A=B:C\\ A=\dfrac{17^{104}-1}{17^4-1}:\dfrac{17^{104}-1}{17^2-1}\\ A=\dfrac{17^2-1}{17^4-1}\)
cảm ơn bạn