\(\frac{6x-3}{3x-1}\)
Những câu hỏi liên quan
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\\ \left(\frac{3x}{1-3n}+\frac{2n}{3x+1}\right):\left(\frac{6x^2+10x}{1-6x+9x^2}\right)\\ \left(\frac{9}{x^3-9n}+\frac{1}{x+3}\right):\left(\frac{x}{3n+9}\right)\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
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Q= \(\left(x^3-1-\frac{7-x^3}{3+x^3}\right).\frac{4}{x^5+3x^2}:\left(\frac{6x^4-24}{x^9+6x^6+9x^3}.\frac{2x}{3x^3+6}\right)\)
1, Thực hiện tính cộng, trừ, nhân, chia các phân thức sau:
a,\(\frac{2x-7}{10x-4}-\frac{3x+5}{4-10x}\)
b,\(\frac{2x+3}{4x^2y^2}:\frac{6x+9}{10x^2y}\)
c,\(\frac{x^2-y^2}{6x^2y^2}:\frac{x+y}{3xy}\)
d,\(\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}\)
a) \(\frac{2x-7}{10x-4}-\frac{3x+5}{4-10x}\)
\(=\frac{2x-7}{10x-4}-\frac{-\left(3x+5\right)}{-\left(4-10x\right)}\)
\(=\frac{2x-7}{10x-4}-\frac{5-3x}{10x-4}\)
\(=\frac{2x-7-\left(5-3x\right)}{10x-4}\)
\(=\frac{2x-7-5+3x}{10x-4}\)
\(=\frac{5x-12}{10x-4}\)
Rút gọn
a) \(\left(\frac{4}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
b) \(\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
c) \(\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}\)
a) (\((\frac{3x}{1-3x}+\frac{2x}{3x+1}):\frac{6x^2+10x}{1-6x+9x^2}\)
1. \(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
2. \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
3. \(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
a,\(\frac{1}{2x-2}-\frac{x-1}{3x^2+6x+3}\)
b,\(\frac{4}{x^2-1}+\frac{1}{x+1}+\frac{-2}{x-1}\)
c,\(\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)
ai jup em vs
Đề bài yêu cầu gì bạn?
\(\frac{6x}{x^2-3x-5}+\frac{3x}{x^2+3-5}=-1\)
Chắc đề là thế này đúng ko bạn: \(\frac{6x}{x^2-3x-5}+\frac{3x}{x^2+3x-5}=-1\)
ĐKXĐ: ...
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{6}{x-\frac{5}{x}-3}+\frac{3}{x-\frac{5}{x}+3}=-1\)
Đặt \(x-\frac{5}{x}=a\) ta được:
\(\frac{6}{a-3}+\frac{3}{a+3}=-1\)
\(\Leftrightarrow6\left(a+3\right)+3\left(a-3\right)=-\left(a-3\right)\left(a+3\right)\)
\(\Leftrightarrow a^2+9a=0\Rightarrow\left[{}\begin{matrix}a=0\\a=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{5}{x}=0\\x-\frac{5}{x}=-9\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=5\\x^2+9x-5=0\end{matrix}\right.\)
\(\left(\frac{X^2+3X}{X^3+3X^2+9X+27}+\frac{3}{X+9}\right):\left(\frac{1}{X-3}-\frac{6X}{X^3-3X^2+9X-27}\right)\)
= \(\left[\frac{x.\left(x+3\right)}{\left(x+3\right).\left(x^2+9\right)}+\frac{3}{x+9}\right]:\left[\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\) ]
\(=\frac{x+3}{x^2-9}.\frac{\left(x-3\right).\left(x^2+9\right)}{x^2+9-6x}\)
= \(\frac{\left(x-3\right).\left(x+3\right)}{\left(x-3\right)^2}\)
= \(\frac{x+3}{x-3}\)
k mik nhé. Plssss~
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tìm điều kiện xác định của biểu thức:
\(a)\frac{6x}{-\sqrt{x+7}}-\frac{3}{-5x-4}+\frac{\sqrt{x}}{-3x+2}\)
\(b)\frac{5-\sqrt{x}}{x+4}+\frac{\sqrt{x-2}-3}{-2x-10}\)
\(c)\frac{\sqrt{6x}}{-x-3}-\frac{4x}{2x+3}\)
\(d)\frac{\sqrt{2x-7}}{3x-4}-\frac{\sqrt{6x}}{x-3}+3x-1\)
a) \(\left\{{}\begin{matrix}x\ge0\\-\sqrt{x+7}< 0\\-5x-4\ne0\\-3x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+7>0\\-5x\ne4\\-3x\ne-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>-7\\x\ne\frac{-4}{5}\\x\ne\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne\frac{2}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x\ge0\\x+4\ne0\\x-2\ge0\\-2x-10\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-4\\x\ge2\\-2x\ne10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne-5\end{matrix}\right.\Leftrightarrow x\ge2\)
c) \(\left\{{}\begin{matrix}x\ge0\\-x-3\ne0\\2x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-3\\x\ne-\frac{3}{2}\end{matrix}\right.\Leftrightarrow x\ge0\)
d) \(\left\{{}\begin{matrix}2x-7\ge0\\x\ge0\\3x-4\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{7}{2}\\x\ge0\\x\ne\frac{4}{3}\\x\ne3\end{matrix}\right.\Leftrightarrow x\ge\frac{7}{2}\)
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