tinh gia tri bieu thuc: A=(1-1/(1+2))(1-1/(1+2+3))...(1-1/(1+2+3+...+2006))
Những câu hỏi liên quan
cho x+y =1 . tinh gia tri cua bieu thuc A=x^3+y^3+3xy
chox-y=1. tinh gia tri cua bieu thuc B=x^3-y^3-3xy
cho x+y=1 . tinh gia tri cua bieu thuc C=x^3+y^3+3xy(x^2+y^2)+6x^2*y^2(x+y)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
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cho bieu thuc a=-1/3+1/3^2-1/3^3+1/3^4-1/3^5+...+1/3^100 tinh gia tri cua bieu thuc b=4/a/+1/3^100
1) Cho bieu thuc A=\(3+\frac{2}{x-1}\). Tinh gia tri cua bieu thuc A khi |2x-3|=1
2) Rut gon bieu thuc B=\(\frac{x}{x-1}\)-\(\frac{x-5}{x+1}\)-\(\frac{3-x}{1-x^2}\)
3) Tim cac gia tri nguyen cua x de bieu thuc \(\frac{B}{A}\)co gia tri nguyen duong
tinh gia tri bieu thuc
A=1+1/2(1+2)+1/3(1+2+3)+....+1/2015(1+2+3+...+2015)
tinh gia tri cua bieu thuc :
a] (18 1/2 + 5 3/8 - 7 5/24) : 16 2/3
b] 2005 x 2007 -1 / 2004 + 2005 x 2006
a) chắc bạn cũng biết
b) \(\frac{2005.2007-1}{2004+2005.2006}=\frac{2005.\left(2006+1\right)-1}{2004+2005.2006}=\frac{2005.2006+2005.1-1}{2004+2005.2006}=\frac{2005.2006+2004}{2004+2005.2006}=1\)
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Tinh gia tri cua bieu thuc : A =1+1/2+1/2^2+1/2^3+...+1/2^2014
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2013}}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2013}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\right)\)
\(A=2-\frac{1}{2^{2014}}\)
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tinh gia tri bieu thuc
A=(1999*1998+1998+1997)*1+1/2:1/1/2-1/1/3
tinh gia tri bieu thuc [1+1/100] * [1+1/99]*.....* [1+1/3] * [1+1/2]
Ta có : \(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right).......\left(1+\frac{1}{3}\right)\left(1+\frac{1}{2}\right)\)
\(=\frac{101}{100}.\frac{100}{99}.\frac{99}{98}......\frac{4}{3}.\frac{3}{2}=\frac{101}{2}\)
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\(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right).....\left(1+\frac{1}{3}\right).\left(1+\frac{1}{2}\right)\)
\(=\frac{101}{100}.\frac{100}{99}.....\frac{4}{3}.\frac{3}{2}=\frac{101}{2}\)
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Đặt \(A=\left[1+\frac{1}{100}\right]\cdot\left[1+\frac{1}{99}\right]\cdot....\cdot\left[1+\frac{1}{3}\right]\cdot\left[1+\frac{1}{2}\right]\)
\(A=\frac{101}{100}\cdot\frac{100}{99}\cdot....\cdot\frac{4}{3}\cdot\frac{3}{2}\)
\(A=\frac{101}{\frac{4}{2}}=\frac{101}{2}\)
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tinh gia tri bieu thuc -1.-1^2.-1^3.-1^4...-1^2010.-1^2011