tinh :
A = (-1).\(\left(1\right)^2\).\(\left(-1\right)^3\)..........\(\left(-1\right)^{2010}\).\(\left(-1\right)^{2011}\)
\(A=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)....\left(1-\frac{2011}{2010}\right)\)
Suy ra : A = ( 1 - 1 / 2010 ) . ( 1 - 2 / 2010 ) .... 0 . ( 1 - 2011 / 2010 ) = 0
Suy ra A = 0
A = 1. ( 1/2010 + 2/2010 ) - ( 3/2010 + 4/2010 ) - ... - ( 2010/2010 + 2011/2010 )
= 1/2010 - 2011/2010
= -2010/2010
1.tính tổng
a. A=\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{20}\right)\)
b. B=\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right).....\left(1-\frac{2011}{2010}\right)\)
a)\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)
\(A=\frac{1.2.3...19}{2.3.4...20}\)
\(A=\frac{1}{20}\)
\(1^2-2^2+3^2-4^2+...+2009^2-2010^2+2011^2\)
\(=1-\left(2-3\right)\left(2+3\right)-\left(4-5\right)\left(4+5\right)-....-\left(2010-2011\right)\left(2010+2011\right)\)
\(=1-\left(-1\right).5-\left(-1\right).9-...-\left(-1\right).4021\)
\(=1+5+9+...+4021\)
\(=\frac{\left[\left(4021-1\right):4+1\right]\left(4021+1\right)}{2}\)
\(=2023066\)
thì làm sao???Hỏi xong rồi tự trả lời thì có ích gì
(✿◠‿◠)(๛ČℌUƔÊŇ♥Ť❍Ą́Ňツ)
Ê nhóc đừng có nghĩ lung tung
Tính tich : \(A=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)......\left(1-\frac{2011}{2010}\right)\)
trong dãy tích A sẽ có phân số \(1-\frac{2010}{2010}=1-1=0\)
=>A=0
\(A=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right).........\left(1-\frac{2010}{2010}\right)\left(1-\frac{2011}{2010}\right)\)
\(A=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right)......0.\left(1-\frac{2011}{2010}\right)\)
A = 0
Tính A= \(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)....\left(1-\frac{2011}{2010}\right)\)
\(A=\frac{ }{ }sdadsad\text{đ}\text{s}gh\text{d}fg\text{d}\)sf
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)
\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)
\(x=\frac{1}{2011}:\frac{1}{2011}=1\)
Vậy x=1
\(\frac{1}{2011}.x=\frac{1}{2}.\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)......\left(\frac{2010}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{2}{4}.\left(\frac{4}{6}\right).\left(\frac{6}{8}\right).......\left(\frac{4018}{4020}\right).\left(\frac{4020}{4022}\right)\)
\(\frac{1}{2011}.x=\frac{2.4.6.8.....4018.4020}{4.6.8.10.....4020.4022}\)
\(\frac{1}{2011}.x=\frac{2}{4022}\)
\(\Rightarrow\)\(x=\frac{2}{4022}:\frac{1}{2011}=1\)
Ai thấy đún thì ủng hộ mink nha !!!
Thanks you very much !!
Chúc các bạn luôn học giỏi !!!
Chứng minh rằng:
\(\dfrac{1}{3\left(\sqrt{2}+1\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+\dfrac{1}{7\left(\sqrt{4}+\sqrt{3}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)
\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)
\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)
Tính giá trị biểu thức: \(\left(-1\right).\left(-1\right)^2.\left(-1^3\right).\left(-1\right)^4.....\left(-1\right)^{2010}.\left(-1\right)^{2011}\)
Vì số mũ lẻ
=> (-1). (-1)2. (-1)3. (-1)4.... (-1)2011= -1
= (-1)1+2+3+4+......+2011
=(-1)2023066
= 1
Bn nào đúng ạ? Bài này có nhiều sách giải ra kết quả khác nhau :(
cho \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\). hãy tính giá trị biểu thức sau: \(A=f\left(\dfrac{1}{2012}\right)+f\left(\dfrac{2}{2012}\right)+...+f\left(\dfrac{2010}{2012}\right)+f\left(\dfrac{2011}{2012}\right)\)
Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)