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a) ĐKXĐ: x∉{0;-5;5}
Ta có: \(M=\left(\frac{x}{x+5}-\frac{5}{5-x}+\frac{10x}{x^2-25}\right)\cdot\left(1-\frac{5}{x}\right)\)
\(=\left(\frac{x}{x+5}+\frac{5}{x-5}+\frac{10x}{x^2-25}\right)\cdot\frac{x-5}{x}\)
\(=\left(\frac{x\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\frac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\frac{10x}{\left(x-5\right)\left(x+5\right)}\right)\cdot\frac{x-5}{x}\)
\(=\frac{\left(x+5\right)^2}{\left(x+5\right)\left(x-5\right)}\cdot\frac{x-5}{x}\)
\(=\frac{x+5}{x-5}\cdot\frac{x-5}{x}=\frac{x+5}{x}\)
b) Đặt \(M=\frac{1}{20}x+1\)
⇒\(\frac{x+5}{x}=\frac{x}{20}+1\)
⇒\(\frac{20\left(x+5\right)}{20x}-\frac{x^2}{20x}-\frac{20x}{20x}=0\)
\(\Leftrightarrow20x+100-x^2-20x=0\)
⇔\(100-x^2=0\)
⇔\(\left(10-x\right)\left(10+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\10+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=-10\left(tm\right)\end{matrix}\right.\)
Vậy: x∈{10;-10}
c) Để M là số nguyên thì x+5⋮x
mà x⋮x
nên 5⋮x
⇔x∈Ư(5)
⇔x∈{1;-1;5;-5}
mà x∉{5;-5}
nên x∈{1;-1}
Vậy: Khi x∈{1;-1} thì M là số nguyên
a) Ta có: \(N=\left(\frac{x+3}{x-3}+\frac{18}{9-x^2}+\frac{x-3}{x+3}\right):\left(1-\frac{x+1}{x+3}\right)\)
\(=\left(\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\frac{18}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{x+3}{x+3}-\frac{x+1}{x+3}\right)\)
\(=\frac{x^2+6x+9-18-\left(x^2-6x+9\right)}{\left(x-3\right)\left(x+3\right)}:\frac{2}{x+3}\)
\(=\frac{x^2+6x-9-x^2+6x-9}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{2}\)
\(=\frac{12x-18}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{2}\)
\(=\frac{12x-18}{x-3}\cdot\frac{1}{2}\)
\(=\frac{12x-18}{2x-6}\)
b)
ĐKXĐ: \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
Đặt \(N=-\frac{1}{2}\)
\(\Leftrightarrow\frac{12x-18}{2x-6}=-\frac{1}{2}\)
\(\Leftrightarrow12x-18=\frac{6-2x}{2}\)
\(\Leftrightarrow12x-18=3-x\)
\(\Leftrightarrow12x-18-3+x=0\)
\(\Leftrightarrow13x-21=0\)
\(\Leftrightarrow13x=21\)
hay \(x=\frac{21}{13}\)(tm)
Vậy: Khi \(N=-\frac{1}{2}\) thì \(x=\frac{21}{13}\)
c) Để N<0 thì 12x-18 và 2x-6 khác dấu
*Trường hợp 1:
\(\left\{{}\begin{matrix}12x-18>0\\2x-6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x>18\\2x< 6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\frac{3}{2}\\x< 3\end{matrix}\right.\)\(\Leftrightarrow\frac{3}{2}< x< 3\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}12x-18< 0\\2x-6>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x< 18\\2x>6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \frac{3}{2}\\x>3\end{matrix}\right.\)(vô lý)
Vậy: Khi N<0 thì \(\frac{3}{2}< x< 3\)
a) Rút gọn:
\(M=\frac{x^2}{\left(x+y\right).\left(1-y\right)}-\frac{y^2}{\left(x+y\right).\left(x+1\right)}-\frac{x^2y^2}{\left(1+x\right).\left(1-y\right)}\)
\(M=\frac{x^2}{\left(x+y\right).\left(1-y\right)}-\frac{y^2}{\left(x+y\right).\left(x+1\right)}-\frac{x^2y^2}{\left(x+1\right).\left(1-y\right)}\)
\(M=\frac{x^2.\left(x+1\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}-\frac{y^2.\left(1-y\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}-\frac{x^2y^2.\left(x+y\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}\)
\(M=\frac{x^2.\left(x+1\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}+\frac{-y^2.\left(1-y\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}+\frac{-x^2y^2.\left(x+y\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}\)
\(M=\frac{x^2.\left(x+1\right)-y^2.\left(1-y\right)-x^2y^2.\left(x+y\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}\)
\(M=x^2-y^2-x^2y^2.\)
Chúc bạn học tốt!