\(\text{cho A =2016-2014+2012-2010+....-1842.Tinh A}\)
Những câu hỏi liên quan
cho A = 2016 - 2014 + 2012 - 2010 + ......-1842
tính A
(6/2008*2010+6/2010*2012+6/2012*2014+6/2014*2016+6/2016*2018)+3/x=3/2008
Tính tổng A= 2016 - 2014 + 2012 - 2010 +...+ 20 -18 + 16 - 14
a) A = \(\dfrac{\text{4024×2014−2}}{2011+2012×2010}\) mình biết kết quả ý này bằng 2 bạn nào giải giùm rồi xem có đúng kq ko
b) B = \(\dfrac{\text{2012×2013+2014}}{2010−2012×2015}\) ý này bằng 1
\frac{x-10}{2010}+\frac{x-8}{2012}+\frac{x-6}{2014}+\frac{x-4}{2016}+\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2016}{4}+\frac{x-2014}{6}+\frac{x-2012}{8}+\frac{x-2010}{10}
cho a,b khác 0 thỏa mãn a^2014 + b^2014 = a^2013 + b^2013 = a^2012 + b^2012
chứng minh rằng : a^2014 + b^2014 = a^2010 + b^2010
Đề \(\Rightarrow a^{2014}+b^{2014}-2\left(a^{2013}+b^{2013}\right)+a^{2012}+b^{2012}=0\)
\(\Leftrightarrow a^{2012}\left(a^2-2a+1\right)+b^{2012}\left(b^2-2b+1\right)=0\)
\(\Leftrightarrow a^{2012}\left(a-1\right)^2+b^{2012}\left(b-1\right)^2=0\)
\(\Leftrightarrow\left(a=0\text{ hoặc }a=1\right)\text{ và }\left(b=0\text{ hoặc }b=1\right)\)
\(+a=0\text{ hoặc }a=1\text{ thì }a^{2014}=a^{2010}\)
\(+b=0\text{ hoặc }b=1\text{ thì }b^{2014}=b^{2010}\)
Suy ra \(a^{2014}+b^{2014}=a^{2010}+b^{2010}\)
Đúng 0
Bình luận (0)
\(\frac{x+1}{2016}+\frac{x+3}{2014}=\frac{x+5}{2012}+\frac{x+7}{2010}\)
. Ta có: \(\frac{x+1}{2016}+\frac{x+3}{2014}=\frac{x+5}{2012}+\frac{x+7}{2010}\) \(\Leftrightarrow\frac{x+1}{2016}+1+\frac{x+3}{2014}+1=\frac{x+5}{2012}+1\frac{x+7}{2010}+1\)
. \(\Leftrightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}-\frac{x+2017}{2012}-\frac{x+2017}{2010}=0\) \(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\right)\)
\(\Leftrightarrow x+2017=0\) \(\Leftrightarrow x=-2017\)
Đúng 0
Bình luận (0)
\(\frac{x+1}{2016}+\frac{x+3}{2014}=\frac{x+5}{2012}+\frac{x+7}{2010}\)
\(\Rightarrow\left(\frac{x+1}{2016}+1\right)+\left(\frac{x+3}{2014}+1\right)=\left(\frac{x+5}{2012}+1\right)+\left(\frac{x+7}{2010}+1\right)\)
\(\Rightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}=\frac{x+2017}{2012}+\frac{x+2017}{2010}\)
\(\Rightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}-\frac{x+2017}{2012}-\frac{x+2017}{2010}=0\)
\(\Rightarrow\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2017=0\)\(\left(Vì\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\ne0\right)\)
\(\Rightarrow x=0-2017\)
\(\Rightarrow x=-2017\)
Vậy x=-2017
Đúng 0
Bình luận (0)
a) A = \(\dfrac{\text{4024×2014−2}}{2011+2012×2010}\) mình biết kết quả ý này bằng 2 bạn nào giải giùm rồi xem có đúng kq ko
b) B = \(\dfrac{\text{2012×2013+2014}}{2010−2012×2015}\) ý này bằng 1
mình không biết kq =mấy
nhứng mình c/m kq =2 là sai
\(A-2=\dfrac{4024.2014-2}{Khongquantam}-2=\dfrac{4024.2014-2-2.2011-2.2012.2010}{Khongquantam}\)
\(A-2=\dfrac{2\left(2012.2014-2011-2012.2010-1\right)}{Khongquantam}=\dfrac{2\left[2012.\left(2014-2010\right)-2011-1\right]}{Khongquantam}\)
\(A-2=\dfrac{2\left[4.2012-2011-1\right]}{Khongquantam}=\dfrac{2\left[3.2011+3\right]}{Khongquantam}\)
\(A-2=\dfrac{2\left[3.\left(2011+1\right)\right]}{Khongquantam}=\dfrac{2.3.2012}{Khongquantam}\ne0\)\(A-2\ne0\)
\(\Rightarrow A\ne2\Rightarrow kq=2=sai\)
Đúng 0
Bình luận (3)
bài 2: Tính nhanh
a) A = 201^2
b) B= 498^2
c) C= 93. 107
d) D= 2016^2 - 2015. 2017
e) E= 2016^3 - 1
________
2016^2 + 2017
g) G= 2016^2 - 2015^2 + 2014^2 - 2013^2 + 2012^2 - 2011^2 + 2010^2 - 1^2
\(201^2=\left(200+1\right)^2=200^2+2.200.1+1^2=40000+400+1=40401\)
\(498^2=\left(500-2\right)^2=500^2-2.500.2+2^2=250000-2000+4=248004\)
Đúng 1
Bình luận (0)
\(93.107=\left(100-7\right)\left(100+7\right)=100^2-7^2=10000-49=9951\)
\(2016^2-2015.2017=2016^2-\left(2016-1\right)\left(2016+1\right)=2016^2-2016^2+1^2=1\)
Đúng 1
Bình luận (0)