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\(\dfrac{y}{xy-5x^2}-\dfrac{15y-25x}{y^2-25x^2}\)
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BN
1
4 tháng 12 2018
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 ~
4 tháng 12 2018
a) \(\dfrac{y}{xy-5x^2}-\dfrac{15y-25x}{y^2-25x^2}=\dfrac{y}{x\left(y-5x\right)}-\dfrac{15y-25x}{\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\dfrac{x\left(15y-25x\right)}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y-5x}{x\left(y+5x\right)}\)
29 tháng 11 2022
b: \(=\dfrac{2}{x+2y}-\dfrac{1}{2y-x}+\dfrac{4y}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2x-4y+x+2y+4y}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{3x+2y}{\left(x-2y\right)\left(x+2y\right)}\)
NN
28 tháng 11 2017
3) \(\dfrac{y}{xy-5y^2}-\dfrac{15y-25x}{y^2-25x^2}\) MTC: \(\left(xy-5y^2\right)\left(y^2-25x^2\right)\)
\(=\dfrac{y\left(y^2-25x^2\right)}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}-\dfrac{\left(xy-5y^2\right)\left(15y-25x\right)}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}\)
\(=\dfrac{y\left(y^2-25x^2\right)-\left(xy-5y^2\right)\left(15y-25x\right)}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}\)
\(=\dfrac{\left(y^3-25x^2y\right)-\left(15xy^2-25x^2y-75y^3+125xy^2\right)}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}\)
\(=\dfrac{y^3-25x^2y-15xy^2+25x^2y+75y^3-125xy^2}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}\)
\(=\dfrac{76y^3-140xy^2}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}\)
NN
28 tháng 11 2017
4) \(\dfrac{4-2x+x^2}{2+x}-2-x\)
\(=\dfrac{4-2x+x^2}{2+x}-\dfrac{2+x}{1}\)
\(=\dfrac{4-2x+x^2}{2+x}-\dfrac{\left(2+x\right)\left(2+x\right)}{2+x}\)
\(=\dfrac{4-2x+x^2-\left(2+x\right)\left(2+x\right)}{2+x}\)
\(=\dfrac{4-2x+x^2-\left(2+x\right)^2}{2+x}\)
\(=\dfrac{4-2x+x^2-\left(4+4x+x^2\right)}{2+x}\)
\(=\dfrac{4-2x+x^2-4-4x-x^2}{2+x}\)
\(=\dfrac{-6x}{2+x}\)
NL
1
27 tháng 11 2018
\(\frac{y}{x\left(y-5x\right)}-\frac{15y-25x}{\left(y+5x\right)\cdot\left(y-5x\right)}\)
=\(\frac{y^2+5xy-15xy+25x^2}{x\left(y+5x\right)\left(y-5x\right)}\)
=\(\frac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}=\frac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y-5x}{xy+5x^2}\)
NN
27 tháng 11 2017
1) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{x-1}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x-1}+1\) MTC: \(\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2\left(x-1\right)+2x+\left(x+1\right)+\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^3-x^2+2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\)
b) \(\dfrac{1}{x^3-x}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{x^2-1}\)
\(=\dfrac{1}{x\left(x^2-1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\) MTC: \(x\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1-\left(x+1\right)+2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1-x-1+2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
DN
2
9 tháng 12 2018
Xét:
\(\left(3x-2y\right)\left(25x^2-9y^2\right)\)
\(=\left(3x-2y\right)\left(5x-3y\right)\left(5x+3y\right)\)
\(=\left(5x-3y\right)\left(15x^2+9xy-10xy-6y^2\right)\)
\(=\left(5x-3y\right)\left(15x^2-xy-6y^2\right)\)
Từ đó dễ dàng suy ra tích chéo = nhau => đpcm
9 tháng 12 2018
ta có : \(VP=\dfrac{15x^2-xy-6y^2}{25x^2-9y^2}=\dfrac{\left(3x-2y\right)\left(5x+3y\right)}{\left(5x-3y\right)\left(5x+3y\right)}=\dfrac{3x-2y}{5x-3y}=VT\)
N
3
3 tháng 9 2020
\(\frac{y}{xy-5x^2}-\frac{15x-25x}{y^2-25x^2}\)
ĐKXĐ : \(\hept{\begin{cases}x,y\ne0\\y\ne\pm5x\end{cases}}\)
\(=\frac{y}{x\left(y-5x\right)}-\frac{-10x}{\left(y-5x\right)\left(y+5x\right)}\)
\(=\frac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\frac{-10xx}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\frac{y^2+5xy+10x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
A
3 tháng 9 2020
\(\frac{y}{xy-5x^2}-\frac{-10x}{y^2-25x^2}=\frac{y^3-25x^2y}{\left(xy-5x^2\right)\left(y^2-25x^2\right)}-\frac{-10x^2y+50x^3}{\left(y^2-25x^2\right)\left(xy-5x^2\right)}\)
\(=\frac{y^3-25x^2y+10x^2y-50x^3}{\left(xy-5x^2\right)\left(y^2-25x^2\right)}=\frac{y^3-15x^2y-50x^3}{\left(xy-5x^2\right)\left(y^2-25x^2\right)}=\frac{y^3-50x^3}{x\left(y-5x\right)^2\left(y+5x\right)}\)
\(=\dfrac{y}{x\left(y-5x\right)}+\dfrac{25x-15y}{\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y\left(y+5x\right)+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2+5xy+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2+30xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)