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TÌM TRƯỚC KHI HỎI
a)Ta có: \(2015=2016-1;2017=2016+1\)
\(\Rightarrow A=2015\cdot2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1< 2016^2=B\)
b)Ta có:
\(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1< 2^{32}=D\)
a)Ta có:A=2015.2017=(2016-1)(2016+1)=20162-1<B=20162
b)Ta có:C=(2+1)(22+1)(24+1)(28+1)(216+1)
=(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)=(22-1)(22+1)(24+1)(28+1)(216+1)
=(24-1)(24+1)(28+1)(216+1)=(28-1)(28+1)(216+1)=(216-1)(216+1)=232-1
=>C<D=232
1) Tính nhanh kết quả của các biểu thức sau:
a) A = 533 + 106.47 + 472
\(=53^2+2.53.47+47^2\)
\(=\left(53+47\right)^2\)
\(=100^2\)
\(=10000\)
b) B = 54 . 34 - (152 - 1)(152 + 1)
\(=15^4-\left(15^4-1\right)\)
\(=15^4-15^4+1\)
\(=1\)
c) C = 502 - 492 + 482 - 472 + ... + 22 - 12
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=99+95+...+3\)
Số số hạng là: (99 - 3) : 4 + 1 = 25
Vậy giá trị của biểu thức là: (99 + 3) . \(\dfrac{25}{2}\) = 1275.
a, Ta co : A = 1999 * 2001
= ( 2000 - 1 ) *( 2000 + 1 )
= \(2000^2-1\)
Vây A < B
cậu ơi tối mình về mình làm tiếp cho bây giờ mình phải đi hok .
a, \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1< 2000^2=B\)
Vậy A<B
b, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}=B\)
Vậy A<B
a, \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)\)
\(=2000^2-1< 2000^2\)
\(\Rightarrow A< B\)
b, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}\)
\(\Rightarrow A< B\)
iiiiiiiiiiiiiiiiiiiiiiiiii !
Bài 2:
a: \(A=1999\cdot2001\)
\(=\left(2000-1\right)\left(2000+1\right)\)
\(=2000^2-1< 2000^2=B\)
Do đó: B lớn hơn
b: \(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1< 2^{16}=D\)
Do đó: D lớn hơn
Tham khảo:
Câu hỏi của Phương Anh Nguyễn Thị - Toán lớp 8 | Học trực tuyến
Bài 1:
a) \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)
\(\Rightarrow x^3-3x^2+3x-1+2^3-x^3+3x^2+6x=17\)
\(\Rightarrow9x+7=17\)
\(\Rightarrow9x=17-7=10\)
\(\Rightarrow x=\dfrac{10}{9}\)
b) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Rightarrow x^3+2^3-x^3+2x=15\)
\(\Rightarrow8+2x=15\)
\(\Rightarrow2x=15-8=7\)
\(\Rightarrow x=\dfrac{7}{2}\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Rightarrow x^3-3x^2.3+3x.3^2-3^3-x^3+3^3+9\left(x^2+2x+1\right)=15\)
\(\Rightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Rightarrow45x+9=15\)
\(\Rightarrow45x=6\)
\(\Rightarrow x=\dfrac{6}{45}=\dfrac{2}{15}\)
d) \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Rightarrow x\left(x^2-5^2\right)-x^3-2^3=3\)
\(\Rightarrow x^3-25x-x^3-8=3\)
\(\Rightarrow-25x-8=3\)
\(\Rightarrow-25x=3+8=11\)
\(\Rightarrow x=-\dfrac{11}{25}\)
Bài 2:
a) Ta có:
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\)
\(B=2^{16}-1\)
Vì 216 - 1 < 216
=> B < A
b) Ta có:
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{128}-1\right)\)
Vì 1/2( 3128 - 1) < 3128 - 1
=> A < B
a) Ta có: A = 3n + 2 + 2014b2
= 3n + 3 + 2013b2 + b2 - 1
= 3(n + 1 + 671b2) + (b - 1)(b + 1)
Vì b là số nguyên tố khác 3 nên b có dạng 6m - 1, 6m + 1 (m ∈ N*)
*Với b = 6m - 1 thì (b - 1)(b + 1) = (6m - 2)6m ⋮ 3
*Với b = 6m + 1 thì (b - 1)(b + 1) = 6m(6m + 2) ⋮ 3
Do đó: (b - 1)(b + 1) ⋮ 3 với mọi b là số nguyên tố khác 3.
Suy ra A = 3(n + 1 + 671b2) + (b - 1)(b + 1) ⋮ 3
Vậy A là hợp số với mọi b là số nguyên tố khác 3 và n ∈ N.
Nhiều đọc là đã cảm thấy nản như thế ko ai giúp đâu bạn đăng từng bài 1 thôi ngủ đi ko ai làm đâu :"))))
Có
a) A= 2015. 2017 = ( 2016 - 1)(2016 + 1)
= 20162 - 1 < 20162 = B
=> A < B ( 20162 - 1 < 20162 )
b) C = (2+1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
= ( 2 - 1)(2+1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
= ( 24 - 1)(24 + 1)(28 + 1)(216 + 1)
= (28 -1)(28 + 1)(216 + 1)
= ( 216 - 1)( 216 + 1)
= 232 - 1 > 223 = D
Vậy C > D ( 232 - 1 < 223 )