tính : 1^2-2^2+3^2-4^2+...-2004^2+2005^2
b) (2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)-2^64
Những câu hỏi liên quan
9 tháng 8 2023 lúc 21:50
bài 6:tính nhanh
7)1^2-2^2+3^2-4^2+....-2004^2+2005^2
8) (2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}
Đọc tiếp
bài 6:tính nhanh
7)1\(^2\)-2\(^2\)+3\(^2\)-4\(^2\)+....-2004\(^2\)+2005\(^2\)
8) (2+1)(2\(^2\)+1)(2\(^4\)+1)(2\(^8\)+1)(2\(^{16}\)+1)(2\(^{32}\)+1)-2\(^{64}\)
7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)
\(A=-\left(1+2+3+...+2004\right)+2005^2\)
\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)
\(A=-1002.2005+2005^2\)
\(A=2005\left(2005-1002\right)=2005.1003=2011015\)
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8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{64}-1\right)-2^{64}\)
\(B=-1\)
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A = 12 – 22 + 32 – 42 + … – 20042 + 20052
b/ B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
giúp nhe ai nhanh mình tk
A = 12 – 22 + 32 – 42 + … – 20042 + 20052
A = 1 + (32 – 22) + (52 – 42)+ …+ ( 20052 – 20042)
A = 1 + (3 + 2)(3 – 2) + (5 + 4 )(5 – 4) + … + (2005 + 2004)(2005 – 2004)
A = 1 + 2 + 3 + 4 + 5 + … + 2004 + 2005
A = ( 1 + 2002 ). 2005 : 2 = 2011015
b/ B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = (22 - 1) (22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = ( 24 – 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = …
B =(232 - 1)(232 + 1) – 264
B = 264 – 1 – 264
B = - 1
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xin lỗi nha chỗ câu a mình lộn
chỗ (1+2002)x2005:2=2011015 là sai nha
(1+2005)x2005:2= 2011015 là đúng nha
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/ A = 12 – 22 + 32 – 42 + … – 20042 + 20052
A = 1 + (32 – 22) + (52 – 42)+ …+ ( 20052 – 20042)
A = 1 + (3 + 2)(3 – 2) + (5 + 4 )(5 – 4) + … + (2005 + 2004)(2005 – 2004)
A = 1 + 2 + 3 + 4 + 5 + … + 2004 + 2005
A = ( 1 + 2005 ). 2005 : 2 = 2011015
b/ B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = (22 - 1) (22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = ( 24 – 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = …
B =(232 - 1)(232 + 1) – 264
B = 264 – 1 – 264
B = - 1
đây là bài hoàn chình nè
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Tính:
a) A= -1^2+2^2-3^2+4^2-...-99^2+100^2
b) B= 1-2^2+3^2-4^2+...-2004^2+2005^2
c) C= (2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)-2^64
THANKS
a,A=-(12-22+32-42+...+992-1002)
=-[(1-2)(1+2)+(3-4)(3+4)+...+(99-100)(99+100)]
=-[(-1).3+(-1).7+...+(-1).199]
=-[(-1).(3+7+...+199]
=\(\frac{\left(199+3\right).50}{2}=5050\)
b, tương tự a
c) C=1(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)-264
=(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)-264
=(22-1)(22+1)(24+1)(28+1)(216+1)(232+1)-264
=(24-1)(24+1)(28+1)(216+1)(232+1)-264
=(28-1)(28+1)(216+1)(232+1)-264
=(216-1)(216+1)(232+1)-264
=(232-1)(232+1)-264
=264-1-264
=-1
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1)_ Tìm phân số, biết rằng nếu lấy 3 / 2 trừ đi phân số đó rồi cộng với 5 / 7 thì được kết quả là 11 / 142)_ Tính hợp lý :a. 17 / 9 + 19 / 13 + 14 / 6 + 7 / 13 + 10 / 6 + 1 / 9b. 2005 x 2007 - 1 / 2004 + 2005 x 2006 3)_ Tính tổng :A 1 / 2 + 1 / 6 + 1 / 12 + . . . + 1 / 42B 2 / 1 x 2 + 2 / 2 x 3 + 2 / 3 x 4 + . . . + 2 / 8 x 9C 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1 / 644)_ Tìm y :a. y x 5 / 6 1 + 1 / 2b. 15 / 28 : y 1 / 7 x 4c. 42 / 25 : y / 5 1 : 5 / 6
Đọc tiếp
1)_ Tìm phân số, biết rằng nếu lấy 3 / 2 trừ đi phân số đó rồi cộng với 5 / 7 thì được kết quả là 11 / 14
2)_ Tính hợp lý :
a. 17 / 9 + 19 / 13 + 14 / 6 + 7 / 13 + 10 / 6 + 1 / 9
b. 2005 x 2007 - 1 / 2004 + 2005 x 2006
3)_ Tính tổng :
A = 1 / 2 + 1 / 6 + 1 / 12 + . . . + 1 / 42
B = 2 / 1 x 2 + 2 / 2 x 3 + 2 / 3 x 4 + . . . + 2 / 8 x 9
C = 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1 / 64
4)_ Tìm y :
a. y x 5 / 6 = 1 + 1 / 2
b. 15 / 28 : y = 1 / 7 x 4
c. 42 / 25 : y / 5 = 1 : 5 / 6
a/b nhân 4 cộng 1/6 = 17/6 số phải tìm là bao nhiêu
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Mỗi bài mình làm một dạng thôi nhé!
1) \(\left(\frac{3}{2}-\frac{x}{y}\right)+\frac{5}{7}=\frac{11}{14}\)
\(\Rightarrow\frac{x}{y}=\frac{3}{2}-\left(\frac{11}{14}-\frac{5}{7}\right)=\frac{10}{7}\)
2) a)
\(\frac{17}{9}+\frac{19}{13}+\frac{14}{6}+\frac{7}{13}+\frac{10}{6}+\frac{1}{9}\)
\(=\left(\frac{17}{9}+\frac{1}{9}\right)+\left(\frac{19}{13}+\frac{7}{13}\right)+\left(\frac{14}{6}+\frac{10}{6}\right)\)
\(=2+2+4\)
\(=8\)
3) a)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
\(A=\frac{1}{1}-\frac{1}{7}=\frac{6}{7}\)
4) a)
\(y.\frac{5}{6}=1+\frac{1}{2}\)
\(\Rightarrow y.\frac{5}{6}=\frac{3}{2}\)
\(\Rightarrow y=\frac{3}{2}.\frac{6}{5}=\frac{9}{5}\)
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Xem thêm câu trả lời
19 tháng 9 2020 lúc 17:14
Bài 1 : Tính
\(a,A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(b,B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(c,R\left(x\right)=x^4-17x^3+17x^2-17x+20\) với x=16
\(d,S\left(x\right)=x^{10}-13x^9+13x^8-13x^7+...+13x^2-13x+10\) với x=12
a)
$A=(1^2-2^2)+(3^2-4^2)+....+(2003^2-2004^2)+2005^2$
$=(1-2)(1+2)+(3-4)(3+4)+....+(2003-2004)(2003+2004)+2005^2$
$=-(1+2)-(3+4)-...-(2003+2004)+2005^2$
$=-(1+2+3+...+2004)+2005^2=-\frac{2004.2005}{2}+2005^2$
$=2005^2-1002.2005=2005(2005-1002)=2011015$
b)
$B=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{16}-1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{32}-1)(2^{32}+1)-2^{64}$
$=2^{64}-1-2^{64}=-1$
c) Do $x=16$ nên $x-16=0$
$R(x)=x^4-17x^3+17x^2-17x+20$
$=(x^4-16x^3)-(x^3-16x^2)+x^2-16x-x+20$
$=x^3(x-16)-x^2(x-16)+x(x-16)-x+20$
$=x^3.0-x^2.0+x.0-x+20=-x+20=-16+20=4$
d) Do $x=12$ nên $x-12=0$. Khi đó:
$S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+(x^2-12x)-x+10$
$=x^9(x-12)-x^8(x-12)+x^7(x-12)-....+x(x-12)-x+10$
$=(x-12)(x^9-x^8+x^7-....+x)-x+10$
$=0-x+10=-x+10=-12+10=-2$
a) Tính A = 12 - 22 + 32 - 42 + ... - 20042 + 20052
b) Tính B = (2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) - 264
Giúp mình với !
Mình cần gấp !
Ai đúng nhanh mình tick cho !
b) Ta có: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2+1\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(=2^{64}-1-2^{64}=-1\)
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1)1/2+1/22+1/23+....+1/22004
2)1/2+1/4+1/8+1/16+1/32+1/64
3)1/6+1/12+1/20+1/30+1/42+1/56
tính: A=(2+1).(2^2+1).(2^4+1).(2^8+1).(2^16+1)+(2^32+1)-(2^64).
A = (2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) - 264
A = (2 - 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) - 264
A = (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) - 264
A = (24 - 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) - 264
A = (28 - 1)(28 + 1)(216 + 1)(232 + 1) - 264
A = (216 - 1)(216 + 1)(232 + 1) - 264
A = (232 - 1)(232 + 1) - 264
A = 264 - 1 - 264
A = -1
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Tính nhanh
A. 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
B. 2/1*3 + 2/3*5 + 2/5*7 + 2/7*9 + 2/9*11