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\(B=\frac{1}{x}+\frac{2}{x+1}-\frac{1}{x^2+x}\) ( đkxđ : \(x\ne0;x\ne\pm1\))
<=> \(B=\frac{1}{x}+\frac{2}{x+1}-\frac{1}{x\left(x+1\right)}\)
<=> \(B=\frac{1\left(x+1\right)}{x\left(x+1\right)}+\frac{2x}{x\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
<=> \(B=\frac{1x+1+2x-1}{x\left(x+1\right)}\)
<=> \(B=\frac{3x}{x\left(x+1\right)}\)
\(a,ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)
Sao phân số thứ 2 là \(\frac{1-2}{1+x}\) .Bạn chép đề thật chuẩn mới trả lời đúng nhé
a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)
a, \(P=\left(\frac{x\sqrt{x}}{\sqrt{x}+1}+\frac{x^2}{x\sqrt{x}+1}\right)\left(2-\frac{1}{\sqrt{x}}\right)\)ĐK : \(x\ge0;\sqrt{x}+1>0\)
\(=\left(\frac{x\sqrt{x}\left(x-\sqrt{x}+1\right)+x^2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\left(\frac{2\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\left(\frac{x^2\sqrt{x}-x^2+x\sqrt{x}+x^2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\left(\frac{2\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\left(\frac{x\sqrt{x}\left(x+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\left(\frac{2\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\frac{x\left(x+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
b, \(P=0\Rightarrow\frac{x\left(x+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=0\Leftrightarrow x\left(x+1\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow x=0;x=-1;x=\frac{1}{4}\)Kết hợp với đk vậy \(x=0;x=\frac{1}{4}\)
a) ĐKXĐ \(\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne9\end{matrix}\right.\)
\(A=\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}+\frac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ =\frac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ =\frac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ =\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}-3}\)
b)
\(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}=\sqrt{3}\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-3}-\sqrt{3}=0\\ \Leftrightarrow\frac{\sqrt{x}+2-\sqrt{3}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=0\\ \Leftrightarrow\frac{\sqrt{x}+2-\sqrt{3x}+3\sqrt{3}}{\sqrt{x}-3}=0\\ \Leftrightarrow\sqrt{x}+2-\sqrt{3x}+3\sqrt{3}=0\)
(Bạn thử tìm x đi nha, mk ra số xấu lắm TvT)
c)
\(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}=\frac{\sqrt{x}-3+5}{\sqrt{x}-3}=1+\frac{5}{\sqrt{x}-3}\)
Để A nhận giá trị nguyên thì \(5⋮\sqrt{x}-3\Leftrightarrow\sqrt{x}-3\inƯ\left(5\right)\)
Ta có bảng sau:
\(\sqrt{x}-3\) | 1 | -1 | 5 | -5 |
\(\sqrt{x}\) | 4 | 2 | 8 | -2 |
\(x\) | 16 | 4 | 64 | loại |
Vậy với x=16; x=4 và x=64 thì A nhận giá trị nguyên