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a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}\left(ĐK:x\ne0;x\ne-4\right)\)
\(=\frac{6}{x\left(x+4\right)}+\frac{3}{2\left(x+4\right)}=\frac{12+3x}{2x\left(x+4\right)}=\frac{3\left(4+x\right)}{2x\left(x+4\right)}=\frac{3}{2x}\)
b) \(\frac{4xy-5}{140x^3y}-\frac{6y^2-5}{10x^3y}\left(ĐK:x,y\ne0\right)\)
\(=\frac{4xy-5-6y^2+5}{10x^3y}=\frac{2y\left(2x-3y\right)}{10x^3y}=\frac{2x-3y}{5x^3}\)
\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)
\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)
ĐK: ...
c) \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow5x+25=0\)
\(\Leftrightarrow x=-5\)( ko t/m )
d) tương tự, ngại tính lắm
e) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1}{x^3-1}-\frac{3x^2}{x^3-1}=\frac{2x\left(x-1\right)}{x^3-1}\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=\frac{-1}{4}\left(c\right)\end{matrix}\right.\)
\(\frac{4xy-5}{10x^3y}-\frac{6y^2-5}{10x^3y}=\frac{\left(4xy-5\right)-\left(6y^2-5\right)}{10x^3y}=\frac{4xy-6y^2}{10x^3y}=\frac{2y\left(2x-3y\right)}{2y.5x^3}=\frac{2x-3y}{5x^3}\)
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\)
\(=\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x+1\right)\left(2x-1\right)}:\frac{4x}{10x-5}\)
\(=\frac{\left(2x+1+2x-1\right)\left(2x+1-2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\times\frac{10x-5}{4x}\)
\(=\frac{4x.2}{\left(2x+1\right)\left(2x-1\right)}\times\frac{5\left(2x-1\right)}{4x}\)
\(=\frac{10}{2x+1}\)
\(a,\frac{4xy-5}{10x^3y}-\frac{6y^2-5}{10x^3y}=\frac{\left(4xy-5\right)-\left(6y^2-5\right)}{10x^3y}=\frac{4xy-5-6y^2+5}{10x^3y}=\frac{4xy-6y^2}{10x^3y}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\)
\(=\left(\frac{2x+1}{2x-1}+\frac{2x-1}{2x-1}\right):\frac{4x}{10x-5}\)
\(=\frac{2x+1+2x-1}{2x-1}:\frac{4x}{10x-5}\)
\(=\frac{4x}{2x-1}.\frac{10x-5}{4x}\)
\(=\frac{10x-5}{2x-1}\)
\(=\frac{5\left(2x-1\right)}{2x-1}\)
\(=\frac{5}{1}=5\)