\(\left(1+\frac{2}{1}\right)x\left(1+\frac{2}{2}\right)x\left(1+\frac{2}{3}\right)x\left(1+\frac{2}{4}\right)x...x\left(1+\frac{2}{2016}\right)x\left(1+\frac{2}{2017}\right)\)
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TN
14 tháng 1 2017
\(VT=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+3\right)\left(x+4\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+3}-\frac{1}{x+4}\)
\(=\frac{1}{x}-\frac{1}{x+4}=\frac{x+4-x}{x\left(x+4\right)}=\frac{4}{x\left(x+4\right)}\)
\(\Rightarrow\frac{4}{x\left(x+4\right)}=\frac{m}{x\left(x+4\right)}=VP\Rightarrow m=4\)
23 tháng 3 2017
Ta có:
\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)
\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)
\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)
\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)
\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)
b/ Thế vô rồi tính nhé
23 tháng 3 2017
Đoạn gần cuối thay y-x= 1 luôn
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)
\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\) giờ mới thay không biết đã tối giản chưa
M
4 tháng 12 2015
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+\right)\left(x+3\right)}+...+\frac{1}{\left(x+2015\right)\left(x+2016\right)}=\frac{1}{x+2016}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+2015}-\frac{1}{x+2016}=\frac{1}{x+2016}\)
\(\frac{1}{x}-\frac{1}{x+2016}=\frac{1}{x+2016}\)
\(\frac{1}{x}-\frac{1}{x+2016}-\frac{1}{x+2016}=0\)
\(\frac{1}{x}-\frac{2x}{x+2016}=0\)
\(\frac{x+2016}{x\left(x+2016\right)}-\frac{2x}{x\left(x+2016\right)}=0\)
\(\frac{x+2016-2x}{x\left(x+2016\right)}=0\Leftrightarrow2016-x=0\Leftrightarrow x=2016\)
22 tháng 6 2019
Phần a vs phần b tính toán thông thường thôi mà bạn, vs 1 h/s lớp 7 thì ít nhất phải làm được chứ?? :((
a) \(x-\frac{4}{5}=\frac{7}{10}-\frac{3}{4}\)
\(\Leftrightarrow x-\frac{4}{5}=\frac{-1}{20}\)
\(\Leftrightarrow x=\frac{-1}{20}+\frac{4}{5}=\frac{15}{20}=\frac{3}{4}\)
b) \(2\frac{1}{3}-x=\frac{-5}{9}+2x\)
\(\Leftrightarrow2\frac{1}{3}-\frac{-5}{9}=2x+x\)
\(\Leftrightarrow3x=\frac{7}{3}+\frac{5}{9}\)
\(\Leftrightarrow3x=\frac{26}{9}\)
\(\Leftrightarrow x=\frac{26}{9}:3=\frac{26}{27}\)
d) .............................. ( Đề bài)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}\)\(-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Leftrightarrow-\frac{1}{x+3}=\frac{1}{2010}\)
\(\Leftrightarrow\frac{1}{-\left(x+3\right)}=\frac{1}{2010}\)\(\Leftrightarrow-\left(x+3\right)=2010\)
\(\Leftrightarrow-x-3=2010\) \(\Leftrightarrow-x=2010+3=2013\)
\(\Leftrightarrow x=-2013\)
Bạn tự kết luận nha!
NT
22 tháng 6 2019
c)
\(\frac{x+3}{2016}+\frac{x+2}{2017}=\frac{x+1}{2018}+\frac{x}{2019}\\ \Leftrightarrow\frac{x+3}{2016}+1+\frac{x+2}{2017}+1=\frac{x+1}{2018}+1+\frac{x}{2019}+1\\ \Leftrightarrow\frac{x+2019}{2016}+\frac{x+2019}{2017}-\frac{x+2019}{2018}-\frac{x+2019}{2019}=0\\ \Leftrightarrow\left(x+2019\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\\ \Rightarrow x-2019=0\\ \Rightarrow x=2019\)
\(=\frac{3}{1}.\frac{4}{2}.\frac{5}{3}...\frac{2018}{2016}.\frac{2019}{2017}\\ =\frac{3.4.5...2018.2019}{1.2.3...2016.2017}\\ =\frac{2018.2019}{2}=1009.2019\)