Tính:
\(a,\dfrac{-5}{4+2y}+\dfrac{y-2}{2y+y^2}\)
\(b,\dfrac{x-1}{x^2-2xy}+\dfrac{3}{2xy-x^2}\)
K
Khách
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Những câu hỏi liên quan
21 tháng 3 2023
a: \(A=31x^2y^3-2xy^3+\dfrac{1}{4}x^2y^2+2\)
\(B=2xy^3+\dfrac{3}{4}x^2y^2-31x^2y^3-x^2-5\)
P=\(A+B=x^2y^2-x^2-3\)
\(A-B=62x^2y^3-4xy^3-\dfrac{1}{2}x^2y^2+x^2+7\)
b: Khi x=6 và y=-1/3 thì \(P=\left(6\cdot\dfrac{-1}{3}\right)^2-6^2-3=4-36-3=1-36=-35\)
AH
Akai Haruma
Giáo viên
11 tháng 12 2018
Lời giải:
a) \(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}=\frac{-5}{2(y+2)}+\frac{y-2}{y(y+2)}=\frac{-5y}{2y(y+2)}+\frac{2(y-2)}{2y(y+2)}\)
\(=\frac{-5y+2(y-2)}{2y(y+2)}=\frac{-(3y+4)}{2y(y+2)}\)
b)
\(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}=\frac{x-1}{x^2-2xy}-\frac{3}{x^2-2xy}=\frac{x-1-3}{x^2-2xy}=\frac{x-4}{x(x-2y)}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
a.
Với \(y=0\) không phải nghiệm
Với \(y\ne0\Rightarrow\left\{{}\begin{matrix}3x+2=\dfrac{5}{y}\\2x\left(x+y\right)+y=\dfrac{5}{y}\end{matrix}\right.\)
\(\Rightarrow3x+2=2x\left(x+y\right)+y\)
\(\Leftrightarrow2x^2+\left(2y-3\right)x+y-2=0\)
\(\Delta=\left(2y-3\right)^2-8\left(y-2\right)=\left(2y-5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2y+3+2y-5}{4}=-\dfrac{1}{2}\\x=\dfrac{-2y+3-2y+5}{4}=-y+2\end{matrix}\right.\)
Thế vào pt đầu ...
Câu b chắc chắn đề sai
19 tháng 11 2021
\(a,VT=\dfrac{x^2+2xy+4-3x^2-3xy}{\left(x+y\right)\left(x+2y\right)}=\dfrac{-2x^2-xy+4}{\left(x+y\right)\left(x-2y\right)}=VP\\ b,VP=\dfrac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}=VT\)
NN
3 tháng 12 2017
a) \(\dfrac{3x-2}{2xy}+\dfrac{7x+2}{2xy}\)
\(=\dfrac{\left(3x-2\right)+\left(7x+2\right)}{2xy}\)
\(=\dfrac{3x-2+7x+2}{2xy}\)
\(=\dfrac{10x}{2xy}\)
\(=\dfrac{5}{y}\)
b) \(\dfrac{5x+y^2}{x^2y}+\dfrac{x^2-5y}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}+\dfrac{x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)+x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3+x^3-5xy}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
c) \(\dfrac{3x-2}{2xy}-\dfrac{7x-y}{2xy}\)
\(=\dfrac{\left(3x-2\right)-\left(7x-y\right)}{2xy}\)
\(=\dfrac{3x-2-7x+y}{2xy}\)
\(=\dfrac{-2-4x+y}{2xy}\)
d) \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
e) \(\dfrac{16xy}{3x-1}.\dfrac{3-9x}{12xy^3}\)
\(=\dfrac{16xy\left(3-9x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{4\left(3-9x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4\left(9x-3\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4.3\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-12}{3y^2}\)
\(=\dfrac{-4}{y^2}\)
f) \(\dfrac{8xy}{3x-1}:\dfrac{12xy^3}{5-15x}\)
\(=\dfrac{8xy}{3x-1}.\dfrac{5-15x}{12xy^3}\)
\(=\dfrac{8xy\left(5-15x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{2\left(5-15x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2\left(15x-5\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2.5\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-10}{3y^2}\)
a)\(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}\)
\(\frac{-5}{2\left(2+y\right)}+\frac{y-2}{y\left(2+y\right)}\)
\(\frac{-5y}{2y\left(2+y\right)}+\frac{2y-4}{2y\left(2+y\right)}\)
\(\frac{-5y+2y-4}{2y\left(2+y\right)}\)
\(\frac{-3y-4}{2y\left(2+y\right)}\)
b)\(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}\)
\(\frac{x-1}{x\left(x-2y\right)}+\frac{3}{x\left(2y-x\right)}\)
\(\frac{x-1}{x\left(x-2y\right)}+\frac{-3}{x\left(x-2y\right)}\)
\(\frac{x-1-3}{x\left(x-2y\right)}\)
\(\frac{x-4}{x\left(x-2\right)}\)