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a. ta có \(2019.\left(-2\right)< 0\)
b. \(\left(-2018\right).\left(-2019\right)>0\)
c. \(\left(-1\right).\left(-2\right)\left(-3\right)..\left(-2020\right)>0\)
Bài 1: Tính hợp lý (nếu có thể)
a) 5.(-8).(-2).(-3)\(=\left(-2.5\right).\left(\left(-3\right).\left(-8\right)\right)=-10.24=-240\)
c) 147.333+233.(-147)\(=147\left(333-233\right)=147.100=14700\)
b) (-125).8.(-2).5.19\(=\left(-125.8\right).\left(-2.5\right).19=-1000.\left(-10\right).19=190\text{ }000\)
d) (-115).27+33.(-115)\(=-115.\left(27+33\right)=-115.60=-6900\)
Bài 2: Tìm số nguyên x, biết:
a) 2x+19=15\(\Leftrightarrow2x=15-19=-4\Leftrightarrow x=-2\)
c) 24-(x-3)^3=-3\(\Leftrightarrow\left(x-3\right)^3=27=3^3\Leftrightarrow x-3=3\Leftrightarrow x=6\)
\(...=2022+2020+\left(-2019+2016-2018+2015-2017+2014\right)+...+\left(6-3+5-2+4-1\right)\)
\(=2022+2020+\left(-3-3-3\right)+\left(-3-3-3\right)+...+\left(-3-3-3\right)+\left(-3-2-1\right)\)
\(=2022+2020+\left(-9\right)+\left(-9\right)+...\left(-9\right)+\left(-6\right)\)
\(=2022+2020+\left(-9\right).\left[\left(2019-9\right):6+1\right].\left[\left(2019+6\right)\right]:2+\left(-6\right)\)
\(=2022+2020+\left(-9\right).336.2025:2+\left(-6\right)\)
\(=2022+2020-3061800-6\)
\(=-3057764\)
Ta có: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)
a, \(1^3+2^3+3^3+4^3=\left(1+2+3+4\right)^2=10^2\)
b, \(1^3+2^3+3^3+4^3+5^3=\left(1+2+3+4+5\right)^2=15^2\)
c,
\(3^6:3^2+2^3\cdot2^2-2\\ =3^{6-2}+2^{3+2-1}\\ =3^4+2^4\)
\(a,1^3+2^3+3^3+4^3.\)
\(=\left(1+2+3+4\right)^2.\)
\(=10^2=100.\)
\(b,1^3+2^3+3^3+4^3+5^3.\)
\(=\left(1+2+3+4+5\right)^2.\)
\(=15^2=225.\)
(2 phần a, b thì áp dụng công thức: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2.\))
a)3^3-2=3
b)625/25=25=5^2
c)8^3+2=8^5
d)5^4+3/5^2=5^7:5^2=5^7-2=5^5
C = -1 + 3 - 32 + ... + 32009 - 32010
=> 3C = - 3 + 32 - 33 + ... + 32010 - 32011
=> 3C + C = ( - 3 + 32 - 33 + ... + 32010 - 32011 ) - ( -1 + 3 - 32 + ... + 32009 - 32010 )
=> 4C = -32010 - 1
=> \(C=\frac{-3^{2010} - 1}{4}\)
\(C=\left(3+3^3+3^5+...+3^{2019}\right)-\left(1+3^2+3^4+...+3^{2020}\right)\)
\(C=\left(1+3+3^2+3^3+...+3^{2020}\right)-2\left(1+3^2+3^4+...+3^{2020}\right)\)
\(A=1+3+3^2+...+3^{2020}\)
\(3A=3+3^2+3^3+...+3^{2021}\)
\(2A=3A-A=3^{2021}-1\Rightarrow A=\frac{3^{2021}-1}{2}\)
\(B=1+3^2+3^4+...+3^{2020}\)
\(3^2B=9B=3^2+3^4+3^6+...+3^{2022}\)
\(8B=9B-B=3^{2022}-1\Rightarrow B=\frac{3^{2022}-1}{8}\)
\(\Rightarrow C=A-2B=\frac{3^{2021}-1}{2}-\frac{2\left(3^{2022}-1\right)}{8}=\frac{2.3^{2021}-3^{2022}+1}{4}\)