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31 tháng 5 2018
2) \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
=\(\left(9.2\right)^8-\left(18^4\right)^2-1^2\)
=\(18^8-18^8-1^2\)
\(=0-1^2\)
\(=-1^2=1\)
31 tháng 5 2018
1. a,
(a+b)3 + (a-b) 3 - 2a3 = a3 + 3ab2+ 3a2b + b3+ a3 - 3a2b + 3ab2- b3 - 2a3
= 6ab2
b, 98 . 28 - ( 184 -1)(184 + 1) = ( 9.2)8 - ( 188 - 1) ( hằng đẳng thức)
= 188 - 188 + 1 = 1
15 tháng 8 2017
a. 134^2 - 68.134 + 34^2 = ( 134 - 34 ) ^2 = 100^2 = 10000
b. 9^8.2^8 - ( 18^4 - 1 )(18^4 + 1 ) = 18^8 - 18^8 + 1 = 1
c. 100^2 - 99^2 + 98^2 - 97^2 + ... + 2^2 - 1
=( 100 - 99 )( 100 + 99 ) + ( 98 - 97 )( 98 + 97 ) + ... + ( 2 - 1 )( 2 + 1 )
= 100 + 99 + 98 + 97 + ... + 2 + 1
=( 100 + 1 ).100:2 = 5050
AH
Akai Haruma
Giáo viên
26 tháng 7 2018
Lời giải:
\(A=2018^2-2017.2019=2018^2-(2018-1)(2018+1)\)
\(=2018^2-(2018^2-1^2)=1\)
\(B=9^8.2^8-(18^4-1)(18^4+1)\)
\(=(9.2)^8-[(18^4)^2-1^2]\)
\(=18^8-(18^8-1)=1\)
\(C=163^2+74.163+37^2=163^2+2.37.163+37^2\)
\(=(163+37)^2=200^2=40000\)
\(D=\frac{2018^3-1}{2018^2+2019}=\frac{(2018-1)(2018^2+2018+1)}{2018^2+2019}\)
\(=\frac{2017(2018^2+2019)}{2018^2+2019}=2017\)
AH
Akai Haruma
Giáo viên
26 tháng 7 2018
Sử dụng công thức \((a-b)(a+b)=a^2-b^2\)
\(E=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^8-1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^{16}-1)(2^{16}+1)-2^{32}\)
\(=(2^{32}-1)-2^{32}=-1\)
16 tháng 8 2018
a) \(\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-x^2+2xy-y^2=4xy\)
16 tháng 8 2018
b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\\ =6ab^2\)
13 tháng 6 2018
a) \(127^2+146.127+73^2=127^2+2.73.127+73^2=\left(127+73\right)^2=40000\)b) \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)=18^8-\left(18^8-1\right)=1\)
c) \(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=100+99+98+97+...+2+1\)
\(=\dfrac{100\left(100+1\right)}{2}=5050\)
13 tháng 6 2018
d) \(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\) \(=20^2-19^2+18^2-17^2+16^2-15^2+...+4^2-3^2+2^2-1^2\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+...+\left(2-1\right)\left(2+1\right)\)\(=20+19+18+17+...+2+1\)
\(=\dfrac{20\left(20+1\right)}{2}=210\)
e) \(\dfrac{780^2-220^2}{125^2+150.125+75^2}\)
\(=\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560.1000}{200}=2800\)
15 tháng 8 2017
Bài 1:
a,\(127^2+146.127+73^2=127^2+2.127.73+73^2\)\(=\left(127+73\right)^2=200^2=40000\)
b,\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(18^8-\left(18^8-1\right)=1\)
\(c,100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=199+195+...+3\)
áp dụng công thức Gauss ta đc đáp án là:10100
d, mk khỏi ghi đề dài dòng:
\(\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560000}{40000}=14\)Bài 2:
\(A=\left(2-1\right)\left(2+1\right)\)\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)Cứ tiếp tục ta đc \(A=2^{32}-1< B=2^{32}\)
\(\left(3-1\right)C=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^2+16\right)\)giải như câu a đc:\(\left(3-1\right)C=3^{32}-1\)
\(\Rightarrow C=\dfrac{3^{32}-1}{3-1}=\dfrac{3^{32}-1}{2}< D=3^{32}-1\)
MV
21 tháng 8 2017
1c,
\(=100^2-99^2+98^2-97^2+...+2^2-1^2\\ =\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\\ =\left(100+99\right)\cdot1+\left(98+97\right)\cdot1+...+\left(2+1\right)\cdot1\\ =100+99+98+97+...+2+1\\ =\dfrac{100\cdot101}{2}=5050\)
HH
19 tháng 6 2018
Giải:
1) \(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\dfrac{55}{24}\)
\(=\dfrac{-19}{8}\)
2) \(-1,75-\left(\dfrac{-1}{9}-2\dfrac{1}{18}\right)\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+2\dfrac{1}{18}\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+\dfrac{37}{18}\)
\(=\dfrac{5}{12}\)
3) \(-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{10}\right)\)
\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)
\(=-\dfrac{67}{120}\)
4) \(\dfrac{2}{5}+\left(-\dfrac{4}{3}\right)+\left(-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{1}{2}\)
\(=-\dfrac{43}{30}\)
5) \(\dfrac{3}{12}-\left(\dfrac{6}{15}-\dfrac{3}{10}\right)\)
\(=\dfrac{3}{12}-\dfrac{6}{15}+\dfrac{3}{10}\)
\(=\dfrac{3}{20}\)
6) \(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\)
\(=8\dfrac{5}{11}+3\dfrac{5}{8}-3\dfrac{5}{11}\)
\(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}\)
\(=8+\dfrac{5}{8}\)
\(=\dfrac{69}{8}\)
7) \(-\dfrac{1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}.13\dfrac{9}{11}-\dfrac{1}{4}.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}\left(13\dfrac{9}{11}+6\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+\dfrac{9}{11}+6+\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+6+1\right)\)
\(=-\dfrac{1}{4}.20=-5\)
8) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)
\(=\dfrac{4}{9}\left(-7\right)+6\dfrac{5}{9}\left(-7\right)\)
\(=-7\left(\dfrac{4}{9}+6\dfrac{5}{9}\right)\)
\(=-7\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)\)
\(=-7\left(6+1\right)\)
\(=-7.7=-49\)
Vậy ...
6 tháng 8 2020
1.
a/ \(A=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left[\left(x+y\right)^2-2xy\right]\)
\(=2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-3\left[\left(x+y\right)^2-2xy\right]\)
\(=2.1.\left[1^2-3xy\right]-3\left[1^2-2xy\right]\)
\(=2-6xy-3+6xy\)
\(=-1\)
Vậy...
2.
a. \(127^2+146.127+73^2\)
\(=127^2+2.73.127+73^2\)
\(=\left(127+73\right)^2=200^2=40000\)
b. \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=18^8-18^8+1\)
\(=1\)
\(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=\left(9\cdot2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)
\(=18^8-\left(18^8-1\right)\)
\(=18^8-18^8+1\)
\(=1\)