Tìm số m,n để : \(\frac{1}{x\left(x-1\right)}=\frac{m}{x-1}+\frac{n}{x}\)
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a) \(đkxđ\)\(\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne1\end{matrix}\right.\)
K viết lại đề nhé
\(M=\left(\frac{x+2\sqrt{x}-7}{x-9}+\frac{1-\sqrt{x}}{\sqrt{x}-3}\right):\left(\frac{1}{\sqrt{x}+3}-\frac{1}{\sqrt{x}-1}\right)\)
\(\Leftrightarrow\frac{x+2\sqrt{x}-7+\left(1-\sqrt{x}\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}-1-1\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\frac{x+2\sqrt{x}-7+\sqrt{x}+3-x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\left(\sqrt{x}+3\right)\left(1-\sqrt{x}\right)}{4}\)
\(\Leftrightarrow\frac{4}{\sqrt{x}-3}.\frac{1-\sqrt{x}}{4}\)
\(\Leftrightarrow\frac{1-\sqrt{x}}{\sqrt{x}-3}\)
b+c bạn tự làm tiếp nha
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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