công trừ phân thức
\(\frac{y}{xy-5x^2}-\frac{15y-25x}{y^2-25x^2}\)
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Tính
\(\dfrac{y}{xy-5x^2}-\dfrac{15y-25x}{y^2-25x^2}\)
\(=\dfrac{y}{x\left(y-5x\right)}+\dfrac{25x-15y}{\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y\left(y+5x\right)+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2+5xy+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2+30xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
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Tính
\(\dfrac{y}{xy-5x^2}-\dfrac{15y-25x}{y^2-25x^2}\)
\(\dfrac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}=\dfrac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}=\dfrac{y-5x}{x\left(y+5x\right)}\)
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Phép trừ phân thức đại số :
\(\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}\)
\(\frac{1}{x-5x^2}\)\(-\)\(\frac{25x-15}{25x^2-1}\)\(=\)\(\frac{-1}{x\left(5x-1\right)}\)\(-\)\(\frac{25x-15}{\left(5x-1\right)\left(5x+1\right)}\)
\(=\)\(\frac{-1\left(5x+1\right)}{x\left(5x-1\right)\left(5x+1\right)}\)\(-\)\(\frac{\left(25x^{ }-15\right)x}{\left(5x-1\right)\left(5x+1\right)x}\)
\(=\) \(\frac{-5x-1-25x^2+15x}{x\left(5x-1\right)\left(5x+1\right)}\)\(=\)\(\frac{-25x^2+10x-1}{x\left(5x-1\right)\left(5x+1\right)}\)
\(=\)\(\frac{-\left(5x-1\right)^2}{x\left(5x-1\right)\left(5x+1\right)}\)\(=\)\(\frac{-5x+1}{5x^2+x}\)
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\(\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}\)
\(=\frac{1}{x\left(1-5x\right)}-\frac{25x-15}{\left(5x-1\right)\left(5x+1\right)}\)
\(=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{\left(1-5x\right)\left(5x+1\right)}\)
\(=\frac{5x+1}{x\left(1-5x\right)\left(5x+1\right)}+\frac{\left(25x-15\right)x}{x\left(1-5x\right)\left(5x+1\right)}\)
\(=\frac{5x+1+25x^2+15x}{x\left(1-5x\right)\left(5x-1\right)}\)
\(=\frac{25x^2-10x+1}{x\left(1-5x\right)\left(5x+1\right)}\)
\(=\frac{\left(5x-1\right)^2}{x\left(1-5x\right)\left(5x+1\right)}\)
\(=\frac{\left(1-5x\right)^2}{x\left(1-5x\right)\left(5x+1\right)}\)
\(=\frac{1-5x}{x\left(5x+1\right)}\)
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Thực hiện phép tính
a) \(\frac{x+2y}{8x^2y^5}\)- \(\frac{3x^2+2}{12x^4y^4}\)
b) \(\frac{y}{xy-5x^2}\)- \(\frac{15y-25x}{y^2-25x^2}\)
c)\(\frac{4-x}{x^3+2x}\)- \(\frac{x+5}{x^3-x^2+2x-2}\)
a, \(\frac{x+2y}{8x^2y^5}-\frac{3x^2+2}{12x^4y^4}\)
=\(\frac{\left(x+2y\right)3x^2}{24x^4y^5}-\frac{\left(3x^2+2\right)2y}{24x^4y^5}\)
=\(\frac{3x^3+6x^2y}{24x^4y^5}-\frac{6x^2y+4y}{24x^4y^5}\)
=\(\frac{3x^3+6x^2y-6x^2y-4y}{24x^4y^5}\)
=\(\frac{3x^3-4y}{24x^4y^5}\)
b,\(\frac{y}{xy-5x^2}-\frac{15y-25x}{y^2-25x^2}\)
=\(\frac{y}{x\left(y-5x\right)}-\frac{15y-25x}{\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\frac{\left(15y-25x\right)x}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2+5xy}{x\left(y-5x\right)\left(y+5x\right)}-\frac{15xy-25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y-5x}{x\left(y+5x\right)}\)
c,\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x^3-x^2\right)+\left(2x-2\right)}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\frac{\left(x+5\right)x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{4x-4-x^2+x}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x^2+5x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2\left(x^2+2\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2}{x\left(x-1\right)}\)
Thực hiện phép tính
\(\frac{y}{xy-5x^2}\) - \(\frac{15x-25x}{y^2-25x^2}\)
\(\frac{y}{xy-5x^2}-\frac{15x-25x}{y^2-25x^2}\)
ĐKXĐ : \(\hept{\begin{cases}x,y\ne0\\y\ne\pm5x\end{cases}}\)
\(=\frac{y}{x\left(y-5x\right)}-\frac{-10x}{\left(y-5x\right)\left(y+5x\right)}\)
\(=\frac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\frac{-10xx}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\frac{y^2+5xy+10x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(\frac{y}{xy-5x^2}-\frac{-10x}{y^2-25x^2}=\frac{y^3-25x^2y}{\left(xy-5x^2\right)\left(y^2-25x^2\right)}-\frac{-10x^2y+50x^3}{\left(y^2-25x^2\right)\left(xy-5x^2\right)}\)
\(=\frac{y^3-25x^2y+10x^2y-50x^3}{\left(xy-5x^2\right)\left(y^2-25x^2\right)}=\frac{y^3-15x^2y-50x^3}{\left(xy-5x^2\right)\left(y^2-25x^2\right)}=\frac{y^3-50x^3}{x\left(y-5x\right)^2\left(y+5x\right)}\)
\(\frac{y}{xy-5x^2}-\frac{15x-25x}{y^2-25x^2}\)
\(=\frac{y}{x.\left(y-5x\right)}-\frac{15x-25x}{y^2-\left(5x\right)^2}\)
\(=\frac{y}{x.\left(y-5x\right)}-\frac{15x-25x}{\left(y-5x\right).\left(y+5x\right)}\)
\(=\frac{y.\left(y+5x\right)}{x.\left(y-5x\right).\left(y+5x\right)}-\frac{x.\left(15x-25x\right)}{x.\left(y-5x\right).\left(y+5x\right)}\)
\(=\frac{y^2+5xy}{x.\left(y-5x\right).\left(y+5x\right)}-\frac{15x^2-25x^2}{x.\left(y-5x\right).\left(y+5x\right)}\)
\(=\frac{y^2+5xy}{x.\left(y-5x\right).\left(y+5x\right)}+\frac{-\left(15x^2-25x^2\right)}{x.\left(y-5x\right).\left(y+5x\right)}\)
\(=\frac{y^2+5xy-15x^2+25x^2}{x.\left(y-5x\right).\left(y+5x\right)}\)
Đến đoạn này thì chịu.
Thực hiện phép tính:
a, \(\dfrac{y}{xy-5x^2}-\dfrac{15y-25x}{y^2-25x^2}\)
\(b,\dfrac{2x}{x^2+2xy}-\dfrac{y}{2y^2-xy}+\dfrac{4y}{x^2-4y^2}\)
kha sdaif dòng mik xin phép trình bày bằng lời ạ :
a) tìm MTC rồi quy đồng lên làm bình thường ại , tử cộng tử mấu giữ nguyên
b) cx vậy ạ tách mẫu tìm MTC rồi ....
~ hok tốt ~
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Thực hiện phép tính:
a, \(\dfrac{y}{xy-5x^2}-\dfrac{15y-25x}{y^2-25x^2}\)
\(b,\dfrac{2x}{x^2+2xy}-\dfrac{y}{2y^2-xy}+\dfrac{4y}{x^2-4y^2}\)
a) \(\dfrac{y}{xy-5x^2}-\dfrac{15y-25x}{y^2-25x^2}=\dfrac{y}{x\left(y-5x\right)}-\dfrac{15y-25x}{\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\dfrac{x\left(15y-25x\right)}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y-5x}{x\left(y+5x\right)}\)
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b: \(=\dfrac{2}{x+2y}-\dfrac{1}{2y-x}+\dfrac{4y}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2x-4y+x+2y+4y}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{3x+2y}{\left(x-2y\right)\left(x+2y\right)}\)
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công trừ phân thức
\(\frac{5x^2+y^2}{xy}-\frac{3x-2y}{xy}\)
\(\frac{5x^2+y^2}{xy}-\frac{3x-2y}{xy}\)
\(=\frac{5x^2+y^2-3x-2y}{xy}\)
Tham khảo nhé~
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Giải hệ phương trình sau:
\(\hept{\begin{cases}\frac{25x^2-y^2}{20x-4y-3\left(5x+y\right)}=3\\\frac{25x^2-y^2}{2\left(5x-y\right)+10x+2y}=1\end{cases}}\)
Làm được rút gọn cái mẫu xong tự nhiên bí :>
\(\hept{\begin{cases}\frac{25x^2-y^2}{20x-4y-3\left(5x+y\right)}=3\\\frac{25x^2-y^2}{2\left(5x-y\right)+10x+2y}=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{\left(5x-y\right)\left(5x+y\right)}{4\left(5x-y\right)-3\left(5x+y\right)}=3\\\frac{\left(5x-y\right)\left(5x+y\right)}{2\left(5x-y\right)+2\left(5x+y\right)}=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{4\left(5x-y\right)-3\left(5x+y\right)}{\left(5x-y\right)\left(5x+y\right)}=\frac{1}{3}\\\frac{2\left(5x-y\right)+2\left(5x+y\right)}{\left(5x-y\right)\left(5x+y\right)}=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{4}{5x+y}-\frac{3}{5x-y}=\frac{1}{3}\\\frac{2}{5x+y}+\frac{2}{5x-y}=1\end{cases}}\)
Đặt: \(\hept{\begin{cases}\frac{1}{5x+y}=a\\\frac{1}{5x-y}=b\end{cases}}\)thì hệ thành
\(\hept{\begin{cases}4a-3b=\frac{1}{3}\\2a+2b=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=\frac{11}{42}\\b=\frac{5}{21}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{5x+y}=\frac{11}{42}\\\frac{1}{5x-y}=\frac{5}{21}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{441}{550}\\y=-\frac{21}{110}\end{cases}}\)
PS: Bí thì bỏ chứ đăng lên làm gì :3
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