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Đặt $a=\dfrac{1}{3589};b=\dfrac{1}{297}$
$=>A=a(7+b)-(4-a)2b-7a-3ab$
$=>A=7a+ab-8b+2ab-7a-3ab$
$=>A=-8b=\dfrac{-8}{297}$
\(A=\dfrac{1}{3589}.7\dfrac{1}{297}-3\dfrac{3588}{3589}.\dfrac{2}{297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=\dfrac{1}{3589}.(7+\dfrac{1}{297})-(3+1-\dfrac{1}{3589}).\dfrac{2}{297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=\dfrac{1}{3589}.7+\dfrac{1}{3589}.\dfrac{1}{297}-\dfrac{6}{297}-\dfrac{2}{297}+\dfrac{2}{3589.297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=\dfrac{7}{3589}+\dfrac{1}{3589.297}-\dfrac{8}{297}+\dfrac{2}{3589.297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=0-\dfrac{8}{297}\)
\(A=-\dfrac{8}{297}\)
a: \(A=\dfrac{1}{\left(3-1\right)\left(3+1\right)}+\dfrac{1}{\left(5-1\right)\left(5+1\right)}+...+\dfrac{1}{\left(99-1\right)\left(99+1\right)}\)
\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{49}{100}=\dfrac{49}{200}\)
a)Để biểu thức vô nghĩa thì \(\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\Leftrightarrow x\in\left\{-2;1\right\}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne1\end{matrix}\right.\Leftrightarrow x\notin\left\{-2;1\right\}\)
b) Ta có: \(\dfrac{5x-2}{12}-\dfrac{2x^2+1}{8}=\dfrac{x-3}{6}+\dfrac{1-x^2}{4}\)
\(\Leftrightarrow\dfrac{2\left(5x-2\right)}{24}-\dfrac{3\left(2x^2+1\right)}{24}=\dfrac{4\left(x-3\right)}{24}+\dfrac{6\left(1-x^2\right)}{24}\)
\(\Leftrightarrow10x-4-6x^2-3=4x-12+6-6x^2\)
\(\Leftrightarrow-6x^2+10x-7+6x^2-4x+6=0\)
\(\Leftrightarrow6x-1=0\)
\(\Leftrightarrow6x=1\)
\(\Leftrightarrow x=\dfrac{1}{6}\)
Vậy: \(S=\left\{\dfrac{1}{6}\right\}\)
a: Ta có: |x+4|=1
=>x+4=1 hoặc x+4=-1
=>x=-3(loại) hoặc x=-5
Khi x=-5 thì \(A=\dfrac{\left(-5\right)^2-5}{3\left(-5+3\right)}=\dfrac{20}{3\cdot\left(-2\right)}=\dfrac{-10}{3}\)
b: \(B=\dfrac{x-1+x+1-3+x}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x-3}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{x+1}\)
