CMR với mọi số tự nhiên n lớn hơn hoặc bằng 1 thì:
\(\left(1+\frac{1}{1\times3}\right)\left(1+\frac{1}{2\times4}\right)\left(1+\frac{1}{3\times5}\right).......\left(1+\frac{1}{n\times\left(n+2\right)}\right)< 2\)
CMR
\(1\times3+2\times4+3\times5+\left(n-1\right)\left(n+1\right)=\frac{\left(n-1\right)n\left(2n+1\right)}{6}\)
Tính C=\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+....+\frac{1}{n\times\left(n+1\right)\times\left(n+2\right)}\)
Bạn nào giúp mik nhớ viết cả cách giải cho mik nhé!!!!!!!!!!
tính \(A=\frac{1}{2}\times\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{2015\times2017}\right)\)
\(A=\frac{1}{2}\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)...\left(1+\frac{1}{2015\cdot2017}\right)\)\(A=\frac{1}{2}\left(\frac{1\cdot3+1}{1\cdot3}\right)\left(\frac{2\cdot4+1}{2\cdot4}\right)...\left(\frac{2015\cdot2017+1}{2015\cdot2017}\right)\)
\(A=\frac{1^2}{2}\cdot\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\cdot\cdot\frac{2016^2}{2015\cdot2017}\)
\(A=\frac{1^2\cdot2^2\cdot3^2\cdot\cdot\cdot2016^2}{2\cdot1\cdot3\cdot2\cdot4\cdot\cdot\cdot2015\cdot2017}\)
\(A=\frac{2016}{2017}\)
A = \(\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times....\times\left(1+\frac{1}{5\times7}\right)\)=?
1=3/3=4/4=5/5=...
=> 1+1/1*3=3/1*3=1/1
=> 1+1/2*4=4/2*4=1/2
=>...
Bieu thuc se con lai la 1*1/2*1/3*1/4*1/5
Vay A=1/120
Tinh \(\left(1-\frac{2}{2\times3}\right)\times\left(1-\frac{2}{3\times4}\right)\times\left(1-\frac{2}{4\times5}\right)\times...\times\left(1-\frac{2}{2015\times2016}\right)\)
=\(\left(1+\frac{1}{1\times3}\right)\left(1+\frac{1}{2\times4}\right)\left(1+\frac{1}{3\times5}\right)...\left(1+\frac{1}{2016\times2018}\right)\)
Ai giúp với
= 1.3+1/1.3 . 2.4+1/2.4 . ....... . 2016.2018+1/2016.2018
= 2^2/1.3 . 3^2/2.4 . ....... . 2017^2/2016.2018
= 2.3. ...... . 2017/1.2. ..... . 2016 . 2.3. ..... . 2017/3.4. ...... . 2018
= 2017 . 2/2018
= 2017/1009
Tk mk nha
Tính luôn :
= 4/3 . 9/8 . 16/15 .. . 4068289/2016 . 2018
= 4/3 . 9/8 . 16/15 . .. 2017 x 2017 / 2016. 2018
= 4 . 9 . 16 ... 2017 . 2017 / 3 . 8 . 15 . ...2016 . 2018
= 2 . 2 . 3 . 3 . 4 . 4 ... 2017 . 2017 / 3 . 2 . 4 . 3 . 5 ...2016 . 2018
= ( 2 . 3 . 4 ... 2017 ) . ( 2 . 3 . 4 ... 2017 ) / ( 3 . 4 . 5 ... 2016 ) . ( 2 . 3 . 4 . 5 ...2018 )
= 2 . 2017 / 2018
= 2017 / 1009
Thực hiện phép tính sau: (với n là số tự nhiên, n lớn hơn hoặc bằng 2)
\(D=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{n^2}\right)\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
hỏi dễ hơn đi
\(\left(1-\frac{2}{2\times3}\right)\left(1-\frac{2}{3\times4}\right)\left(1-\frac{2}{4\times5}\right)...\left(1-\frac{2}{99\times100}\right)\)= ?
BÀI 1: CMR với mọi số tự nhiên \(n\ge3\)
\(B=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....+\frac{1}{n^3}< \frac{1}{12}\)
BÀI 2: CMR với mọi số tự nhiên \(n\ge1\)
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n\left(n+2\right)}\right)< 2\)
BÀI 3: CMR với mọi số tự nhiên \(n\ge2\)
\(B=\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)....\left(1-\frac{1}{n\left(n+1\right)}\right)>\frac{1}{3}\)
M.N giúp mk với!!!!!
vì bài dài quá nên mình làm từng bài 1 nhé
1. Ta thấy : \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)
Do đó :
\(B< \frac{1}{2}.\left[\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]< \frac{1}{2}.\frac{1}{6}=\frac{1}{12}\)
2.
Nhận xét : \(1+\frac{1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Do đó :
\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2.3...\left(n+1\right)}{1.2...n}.\frac{2.3...\left(n+1\right)}{3.4...\left(n+2\right)}=\frac{n+1}{1}.\frac{2}{n+2}< 2\)
3.
Nhận xét ; \(1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Do đó : \(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{\left(n-1\right)n\left(n+2\right)}{n\left(n+1\right)}\)
Rút gọn được : B = \(\frac{1}{n}.\frac{n+2}{3}>\frac{1}{3}\)