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A = 2006 x 125 + 1000 / 126 x 2005 - 888
A = (2005 + 1) x 125 + 1000 / (125 + 1) x 2005 - 888
A = 2005 x 125 + 125 + 1000 / 125 x 2005 + 2005 - 888
A = 2005 x 125 + 1125 / 125 x 2005 + 1117
A = 1125 / 1117
8 x 2010 x 125 - 2009 x 437 - 2009 x 563
= ( 8 x 125 ) x 2010 - 2009 x ( 437 + 563 )
= 1000 x 2010 - 2009 x 1000
= 1000 x ( 2010 - 2009 )
= 1000 x 1
= 1000
=))
\(\frac{2010.125+1000}{126.2010-1010}=\frac{10\left(201.125+100\right)}{10\left(201.126-101\right)}=\frac{201.125-101+201}{2011.126-101}\)
\(=\frac{201.126-101}{201.126-101}=1\)
=2010 x 125 + 2010x 126 + 1000+1010
=2010 x 125 + 2010 x 126 + 2010
=2010 x ( 125 + 126 +1 )
=2010 x 252
=506520
\(\frac{2010\cdot2011+1000}{2012\cdot2010-1010}\)
= \(\frac{2010\cdot2011+1000}{\left(2011+1\right)\cdot2010-1010}\)
= \(\frac{2010\cdot2011+1000}{2011\cdot2010+2010-1010}\)
= \(\frac{2010\cdot2011+1000}{2011\cdot2010+1000}\)
= 1
\(\frac{2010.2011+1000}{2012.2010-1010}\)
\(=\frac{2010.2011+2010-1010}{2012.2010-1010}\)
\(=\frac{2010.\left(2011+1\right)-1010}{2012.2010-1010}\)
\(=\frac{2010.2012-1010}{2012.2010-1010}\)
\(=1\)
\(\frac{\left(2005+1\right).125+1000}{\left(125+1\right).2005-888}\)
= \(\frac{2005.125+125+1000}{125.2005+2005-888}\)
= \(\frac{2005.125+1125}{125.2005+1117}\)
= \(\frac{250625+1125}{250625+1117}\)
= \(\frac{125875}{125871}\)
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
\(A=\frac{1}{100}.\left(1-\frac{1}{101}\right)+\frac{1}{100}.\left(\frac{1}{2}-\frac{1}{102}\right)+\frac{1}{100}.\left(\frac{1}{3}-\frac{1}{103}\right)+...+\frac{1}{100}.\left(\frac{1}{25}-\frac{1}{125}\right)\)
\(A=\frac{1}{100}.\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{25}-\frac{1}{125}\right)\)
\(A=\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)\)
\(B=\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}\)
\(B=\frac{1}{25}.\left(1-\frac{1}{26}\right)+\frac{1}{25}.\left(\frac{1}{2}-\frac{1}{27}\right)+\frac{1}{25}.\left(\frac{1}{3}-\frac{1}{28}\right)+...+\frac{1}{25}.\left(\frac{1}{100}-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+...+\frac{1}{100}-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-\frac{1}{28}-...-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1+\frac{1}{2}+...+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-...-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{125}\right)\)\(B=\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)\)
Ta thấy biểu thức trong ngoặc của hai vế A và B giống nhau
Vậy A : B = \(\frac{1}{100}:\frac{1}{25}=\frac{1}{4}\)
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
\(\Rightarrow A=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{24.25}\right)+\left(\frac{1}{101.102}+\frac{1}{102.103}+...+\frac{1}{124.125}\right)\)
\(A=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{24}-\frac{1}{25}\right)+\left(\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+...+\frac{1}{124}-\frac{1}{125}\right)\)
\(A=\left(1-\frac{1}{25}\right)+\left(\frac{1}{101}-\frac{1}{125}\right)\)
\(A=\frac{24}{25}+\frac{24}{12625}\)
Bạn tự tính luôn nha trog máy tính của mình là : 0,961... ( k làm thành phân số được )
lết quả là :
\(\frac{-1}{2009}\)
ai thấy đúng thì tk nha
bằng 1 bạn ạ
bằng 1 bạn ạ