So sánh ( bằng cách nhanh nhất)
a)\(\frac{87}{39}và\frac{2015}{2017}\)
b)\(\frac{n}{n+1}và\frac{n+1}{n+3}\)
c) \(\frac{n}{n+3}va\frac{n-1}{n+4}\)
So sánh các phân số sau :
a) \(\frac{n}{n+5}và\frac{n+9}{n+14}\)
b) \(\frac{n+1}{n+2}và\frac{n+3}{n+4}\)
c) \(\frac{n+9}{n}va\frac{n+11}{2}\)
d) \(\frac{n+12}{n+4}va\frac{n}{n-4}\)
AI NHANH NHẤT MÌNH TÍCH CHO!!!!!!!!!!
so sánh biểu thức P với \(\frac{1}{2}\)biết
\(P=\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+...+\frac{2017}{2015!+2016!+2017!}\)(với n!=1.2.3...n)
\(P=\frac{3}{1!\left(1+2\right)+3!}+\frac{4}{2!\left(1+3\right)+4!}+...+\frac{2017}{2015!\left(1+2016\right)+2017!}\)
\(P=\frac{3}{3\left(1!+2!\right)}+\frac{4}{4\left(2!+3!\right)}+...+\frac{2017}{2017\left(2015!+2016!\right)}\)
\(P=\frac{1}{1!+2!}+\frac{1}{2!+3!}+...+\frac{1}{2015!+2016!}\)
Ta có \(a!>\sqrt{a}\)\(\left(a\inℕ;a>1\right)\) do đó :
\(P>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2015}+\sqrt{2016}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\)
\(\frac{\sqrt{2016}-\sqrt{2015}}{\left(\sqrt{2016}+\sqrt{2015}\right)\left(\sqrt{2016}-\sqrt{2015}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2016}\)
\(-\sqrt{2015}=\sqrt{2016}-1=\frac{1}{2}+\left(\sqrt{2016}-\frac{3}{2}\right)=\frac{1}{2}+\left(\sqrt{2016}-\sqrt{\frac{9}{4}}\right)>\frac{1}{2}\)
Vậy \(P>\frac{1}{2}\)
Chúc bạn học tốt ~
PS : tự nghĩ bừa thui nhé :))
1) CMR : A=(n+2015)(n+2016) + n2 + n chia hết cho 2 với n ϵ N
2) So sánh :
P = \(\frac{2013}{2014^{2013}}+\frac{2014}{2015^{2014}}+\frac{2015}{2016^{2015}}+\frac{2016}{2017^{2016}}\) và
Q = \(\frac{2014}{2017^{2016}}+\frac{2013}{2016^{2015}}+\frac{2016}{2015^{2014}}+\frac{2015}{2014^{2013}}\)
A = (n + 2015)(n + 2016) + n2 + n
= (n + 2015)(n + 2015 + 1) + n(n + 1)
Tích 2 số tự nhiên liên tiếp luôn chia hết cho 2
=> (n + 2015)(n + 2015 + 1) chia hết cho 2
n(n + 1) chia hết cho 2
=> (n + 2015)(n + 2015 + 1) + n(n + 1) chia hết cho 2
=> A chia hết cho 2 với mọi n \(\in\) N (đpcm)
Biết n!=1.2.3...n \(\left(n\inℕ^∗;n\ge2\right)\)và \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+......+\frac{2014}{2015!}\)
Hãy so sánh A với 1
Ta có \(A=\frac{1}{2!}+\frac{2}{3!}+...+\frac{2014}{2015!}\)
=> \(A=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2015-1}{2015!}\)
=> \(A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2015}{2015!}-\frac{1}{2015!}\)
=> \(A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2014!}-\frac{1}{2015!}\)
=> \(A=1-\frac{1}{2015!}< 1\)
so sánh
a\(\frac{n}{n+1}\)và \(\frac{n+2}{n+3}\)
b \(\frac{n}{n+3}\)và \(\frac{n-1}{n+4}\)
c \(\frac{n}{2n+1}\)và\(\frac{3n+1}{6n+3}\)
a). n/n+1 < n+2/n+3
b). n/n+3 > n−1/n+4
c). n/2n+1 < 3n+1/6n+3
k mk nha
\(\frac{n}{n+1}< 1\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+1+2}=\frac{n+2}{n+3}\)
=>n/n+1<n+2/n+3
vậy........
b)\(\frac{n}{n+3}>\frac{n}{n+4}>\frac{n-1}{n+4}\Rightarrow\frac{n}{n+3}>\frac{n}{n+4}\)
vậy.....
c)\(\frac{n}{2n+1}=\frac{3n}{6n+3}< \frac{3n+1}{6n+3}\)
vậy.......
a) \(\frac{n}{n+1}=1-\frac{1}{n+1};\frac{n+2}{n+3}=1-\frac{1}{n+3}\)
Vì \(\frac{1}{n+1}>\frac{1}{n+3}\)=) \(1-\frac{1}{n+1}< 1-\frac{1}{n+3}\)
=) \(\frac{n}{n+1}< \frac{n+2}{n+3}\)
b) Áp dụng tính chất : Nếu \(\frac{a}{b}< 1\)=) \(\frac{a}{b}< \frac{a+m}{b+m}\)
Ta có : \(\frac{n-1}{n+4}< 1\)=) \(\frac{n-1}{n+4}< \frac{n-1+1}{n+4+1}=\frac{n}{n+5}< \frac{n}{n+3}\)
=) \(\frac{n-1}{n+4}< \frac{n}{n+3}\)
So sánh các phân số sau :
a)\(\frac{n}{n+1}\) và \(\frac{n+1}{n+2}\)
b)\(\frac{n+2016}{n+2017}\)và \(\frac{n+2017}{n+2018}\)
c)\(\frac{a+7}{a+4}\)và \(\frac{a+15}{a+12}\)
d)\(\frac{a-1908}{a-1899}\)và \(\frac{a-2112}{a-2103}\)
Quy đồng: \(\frac{n}{n+1}\)= \(\frac{n\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}\)=\(\frac{n^2.2n}{\left(n+1\right)\left(n+2\right)}\)
\(\frac{n+1}{n+2}\)= \(\frac{\left(n+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+2\right)}\)= \(\frac{n^2+2n+1}{\left(n+1\right)\left(n+2\right)}\)
Vì n2+2n+1 < n2.2n+1 nên...
Vậy...
Ko chắc nha
Nghe nó ko có lý kiểu j j ý
So sánh các phân số sau ( bằng cách hợp lí)
g) \(\frac{n}{n+3}\)Và \(\frac{n+1}{n+2}\)
h) \(\frac{n+1}{n+2}\)và \(\frac{n+3}{n+4}\)
h) Ta có: \(\frac{n+1}{n+2}=1-\frac{1}{n+2}\)
\(\frac{n+3}{n+4}=\frac{1}{n+4}\)
Vì \(n+2< n+4\)\(\Rightarrow\frac{1}{n+2}>\frac{1}{n+4}\)
\(\Rightarrow1-\frac{1}{n+2}< 1-\frac{1}{n+4}\)\(\Rightarrow\frac{n+1}{n+2}< \frac{n+3}{n+4}\)
1.So sánh \(\frac{2016}{2017}+\frac{2017}{2018}\)với \(1\)( không tính kết quả )
2.So sánh: \(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
3. Với n là số nguyên dương hãy so sánh 2 phân số sau: \(\frac{n}{n+8}\)và \(\frac{n-2}{n+9}\)
1. \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)>1
2. A>B
so sánh
a) \(\frac{n}{n+1}và\frac{n+2}{n+3}\)
b) \(\frac{n}{n+3}và\frac{n-1}{n+4}\)