Bài 4:
\(b,\dfrac{\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}}{1+\dfrac{x^3}{1-x^3}}\)
\(=\dfrac{\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}}{\dfrac{1-x^3}{1-x^3}+\dfrac{x^3}{1-x^3}}\)
\(=\dfrac{\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}}{\dfrac{1-x^3+x^3}{1-x^3}}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{\left(1-x\right)\left(1+x+x^2\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{-1}{\left(x-1\right)\left(1+x+x^2\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}.\left[-\left(x-1\right)\left(x^2+x+1\right)\right]\)
\(=\dfrac{-4x\left(x^2+x+1\right)}{x+1}\)
Bài 1:
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
a, ĐKXĐ của phân thức là :
\(\left\{{}\begin{matrix}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
b, \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-4}{x-2}\)
c,\(A=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow-4x+16=3x-6\)
\(\Leftrightarrow-4x-3x=-6-16\)
\(\Leftrightarrow-7x=-22\Leftrightarrow x=\dfrac{22}{7}\)
d, \(A=\dfrac{x-4}{x-2}=\dfrac{x-2}{x-2}-\dfrac{2}{x-2}=1-\dfrac{2}{x-2}\)
Để A là số nguyên
\(\Leftrightarrow\dfrac{2}{x-2}\) là số nguyên
\(\Leftrightarrow2⋮x-2\Leftrightarrow x-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Bạn tự kẻ bảng tính nha
VD : x - 2 = 1 ⇒ x = 3
e,\(x^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
⇔ \(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) ( x = - 3 ko t/m ĐKXĐ )
Thay x = 3vaof bt A ,có :
\(\dfrac{3-4}{3-2}=\dfrac{-1}{1}=-1\)
Vậy ........