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\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).......\left(52^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1+1+\dfrac{1}{2}\right)\left(1-1+\dfrac{1}{2}\right)........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right)........\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......\left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
\(a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.\) (1)
ĐK xác định : x≠0 ; y≠0
Đặt ẩn phụ : a = \(\dfrac{1}{x}\) ; b = \(\dfrac{1}{y}\)
Thay vào (1) ta được :
\(\left\{{}\begin{matrix}4a+b=12\\a+b=-3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}3a=15\\a+b=-3\end{matrix}\right.< =>\left\{{}\begin{matrix}a=5\\b=-8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{8}\end{matrix}\right.\)
Vậy S = {(\(\dfrac{1}{5};-\dfrac{1}{8}\))}
\(b.\left\{{}\begin{matrix}5\dfrac{1}{x}+2\dfrac{1}{y}=6\\2\dfrac{1}{x}-\dfrac{1}{y}=3\end{matrix}\right.\) (2)
ĐK xác định : x≠0 ; y≠0
Đặt ẩn phụ : a = 1/x ; b = 1/y
Thay vào (2) ta được : \(\left\{{}\begin{matrix}5a+2b=6\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}5a+2b=6\\4a-2b=6\end{matrix}\right.< =>\left\{{}\begin{matrix}9a=12\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{4}{3}\\b=-\dfrac{1}{3}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-3\end{matrix}\right.\)
Vậy S = {(\(\dfrac{3}{4};-3\) )}
c) \(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.\)
ĐK xác định : x≠0 ; y ≠0
Áp dụng quy tác cộng đại số ta có :
\(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\3\dfrac{1}{x}-3\dfrac{1}{y}=15\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-3\dfrac{1}{y}=-13\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{3}{13}\\x=\dfrac{3}{28}\end{matrix}\right.\)
Vậy S = {(\(\dfrac{3}{28};\dfrac{3}{13}\))}
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\)
ĐK xác định : x≠0 ; y≠0
áp dụng quy tắc cộng đại số ta có :
\(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.< =>\left\{{}\begin{matrix}2\dfrac{1}{x}-8\dfrac{1}{y}=10\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-5\dfrac{1}{y}=9\\\dfrac{1}{x}-4\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{5}{9}\\x=-\dfrac{5}{11}\end{matrix}\right.\)
Vậy S = {(\(-\dfrac{5}{11};-\dfrac{5}{9}\))}
e) ĐK xác định x≠0 ; y≠0
\(\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\6\dfrac{1}{x}-\dfrac{1}{y}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\18\dfrac{1}{x}-3\dfrac{1}{y}=6\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-17\dfrac{1}{x}=-2\\\dfrac{1}{x}-3\dfrac{1}{y}=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=\dfrac{17}{2}\\y=-\dfrac{17}{22}\end{matrix}\right.\)
Vậy S={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}
C=\(\dfrac{x-x^3}{x^2+1}\left(\dfrac{1}{1+2x+x^2}+\dfrac{1}{1-x^2}\right)+\dfrac{1}{1+x}\)
\(=\dfrac{x\left(1-x^2\right)}{x^2+1}\left(\dfrac{1}{\left(1+x\right)^2}+\dfrac{1}{\left(1-x\right)\left(1+x\right)}\right)+\dfrac{1}{1+x}\)
\(=\dfrac{x\left(1-x\right)\left(1+x\right)}{x^2+1}\left(\dfrac{1-x+1+x}{\left(1-x\right)\left(1+x\right)^2}\right)+\dfrac{1}{1+x}\)
\(=\dfrac{x\left(1-x\right)\left(1+x\right).2}{\left(x^2+1\right)\left(1-x\right)\left(1+x^2\right)}+\dfrac{1}{1+x}\)
\(=\dfrac{2x}{\left(x^2+1\right)\left(1+x\right)}+\dfrac{1}{1+x}\)
\(=\dfrac{2x+\left(x^2+1\right)}{\left(x^2+1\right)\left(1+x\right)}\)
\(=\dfrac{2x+x^2+1}{\left(x^2+1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x+1}{\left(x^2+1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x^2+1\right)\left(x +1\right)}\)
\(=\dfrac{x+1}{x^2+1}\)
\(\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}=\dfrac{1}{3}\left(\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{3n}{6\left(3n+2\right)}=\dfrac{n}{6n+4}\)
\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}=\dfrac{1}{4}\left(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{3.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{n\left(n+2\right)}{3\left(2n+1\right)\left(2n+3\right)}\)
\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n^2+2n+1}{n^2\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n\left(n+1\right)+1}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{\left[n\left(n+1\right)+1\right]^2}{n^2\left(n+1\right)^2}}=\dfrac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\dfrac{1}{n\left(n+1\right)}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow C=1+\dfrac{1}{1}-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(\Rightarrow C=2019-\dfrac{1}{2019}\)
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7