a)\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-10\right)⋮2\) 2
CMR: vs mọi n thuộc Z thì
a) \(\left(n^2-3n+1\right)\left(n+2\right)-n^3+2⋮5\)
b)\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-10\right)⋮2\)
a: \(=n^3+2n^2-3n^2-6n+n+2-n^3+2\)
\(=-n^2+5n\)
Cái này nếu n=1 thì ko thỏa mãn nha bạn
b: \(=6n^2+30n+n+5-6n^2+30n-10n+50\)
\(=49n+55\)
Nếu n là số lẻ thì 49n+55 chia hết cho 2
Còn nếu n là số chẵn thì 49n+55 ko chia hết cho 2 nha bạn
Chứng minh rằng với mọi n thuộc Z thì :
a) \(\left(n^2+3n-1\right).\left(n+2\right)-n^3+2⋮5\)
b) \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)⋮2\)
c) \(\left(2n-1\right).3-\left(2n-1\right)⋮8\)
d) \(n^2\left(n+1\right)+2n\left(n+1\right)⋮6\)
a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10⋮2\)
d: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
chứng minh rằng với mọi số nguyên n thì
\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-10\right)⋮2\)
Bn có sai ko? Hay đề là tìm n để Biểu thức \(⋮\) 2
Ta có: \(\left(3n+5\right)\left(2n-10\right)=2\left(n-5\right)\left(3n+5\right)\) \(⋮\) 2
=> Theo đề bài phải c/m: \(\left(6n+1\right)\left(n+5\right)\) \(⋮\) 2 (*)
Xét n là số lẻ => \(\left(6n+1\right)\left(n+5\right)\) là số chẳn => Biểu thức \(⋮\) 2
Xét n là số chẳn => \(\left(6n+1\right)\left(n+5\right)\) là số lẻ => \(⋮̸\) 2
=> Để (6n+1)(n+5)−(3n+5)(2n−10) \(⋮\) 2 thì n là số lẻ, n\(\in Z\)
Chứng minh vs mọi n thuộc Z thì:
\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2⋮5\)
\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)⋮2\)
a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2-n^3+2\)
\(=5n^2+5n⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=\left(6n^2+30n+n+5\right)-\left(6n^2-3n+10n-5\right)\)
\(=6n^2+31n+5-6n^2-7n+5\)
\(=24n+10⋮2\)
CMR: với mọi số tự nhiên n thì:
a)\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\) chia hết cho 5
b)\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)chia hết cho 2
a, Ta có: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
\(\Rightarrowđpcm\)
b, \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+31n+5-6n^2-7n+5\)
\(=24n+10=2\left(12n+5\right)⋮2\)
\(\Rightarrowđpcm\)
a)
= n3 + 2n2 + 3n2 + 6n - n - 2 + 2
= 5n2 + 5n
= 5(n2 + n ) chia hết cho 5
b)
= 2(12n +5) chia hết cho 2
a/ Chứng minh ới mọi số nguyên \(n\)thì: \(\left(n^2-3n+1\right)\left(n+2\right)-n^3+2\)chia hết cho 5
b/ Chứng minh với mọi số nguyên \(n\)thì: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-10\right)\)chia hết cho 2
Chứng minh rằng với mọi giá trị nguyên của n ta luôn có:
a) \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2⋮5\)
b) \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)⋮2\)
a,
\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\\ =\left(n^2+3n-1\right)n+\left(n^2+3n-1\right)2-n^3+2\\ =n^3+3n^2-n+2n^2+6n-2-n^3+2\\ =5n^2+5n\\ =5\cdot\left(n^2+n\right)⋮5\\ \RightarrowĐpcm\)
b,
\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\\ =\left(6n+1\right)n+\left(6n+1\right)5-\left(3n+5\right)2n-\left(3n+5\right)\\ =6n^2+n+30n+5-6n^2-10n-3n-5\\ =18n⋮2\\ \RightarrowĐpcm\)
a) \(lim\frac{\left(-2\right)^n+3^n}{\left(-2\right)^{n+1}+3^{n+1}}\)
b) \(lim\frac{\left(2n-1\right)\left(n+1\right)\left(3n+4\right)}{\left(5-6n\right)^3}\)
c) \(lim\left(\sqrt{n^2+5n+1}-\sqrt{n^2-2}\right)\)
d) \(lim\frac{5\cdot3^n-6^{n+1}}{4\cdot2^n+6^n}\)
e) \(lim\left(-2n^3-3n^2+5n-2020\right)\)
a/ \(=lim\frac{\left(-\frac{2}{3}\right)^n+1}{-2.\left(-\frac{2}{3}\right)^n+3}=\frac{1}{3}\)
b/ \(=lim\frac{\left(2-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(3+\frac{4}{n}\right)}{\left(\frac{5}{n}-6\right)^3}=\frac{2.1.3}{\left(-6\right)^3}=-\frac{1}{36}\)
c/ \(=lim\frac{5n+3}{\sqrt{n^2+5n+1}+\sqrt{n^2-2}}=\frac{5+\frac{3}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{2}{n}}}=\frac{5}{1+1}=\frac{5}{2}\)
d/ \(=lim\frac{5.\left(\frac{1}{2}\right)^n-6}{4.\left(\frac{1}{3}\right)^n+1}=\frac{-6}{1}=-6\)
e/ \(=-n^3\left(2+\frac{3}{n}-\frac{5}{n^2}+\frac{2020}{n^3}\right)=-\infty.2=-\infty\)
a; lim\(\frac{\sqrt{6n^4+n+1}}{2n^2+1}\)
b; lim \(\frac{\left(n+1\right)\left(2n+1\right)^2\left(3n+1\right)^3}{n^2\left(n+2\right)^2\left(1-3n\right)^2}\)