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a) \(\frac{x-1}{2x}+\frac{2x+1}{2x}+\frac{1-5x}{6x}\)
\(=\frac{3x-3}{6x}+\frac{6x+3}{6x}+\frac{1-5x}{6x}\)
\(=\frac{3x-3+6x+3+1-5x}{6x}\)
\(=\frac{4x+1}{6x}\)
a.\(\frac{1-3x}{2}-\frac{x+3}{2}=\frac{1-3x-x-3}{2}=\frac{1-4x-3}{2}=\frac{-4x-2}{2}=\frac{-2\left(2x+1\right)}{2}=-2x-1\)
b. \(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}=\frac{2\left(x^2-y^2\right)+2y^2}{x}=\frac{2x^2-2y^2+2y^2}{x}=2x\)
c. \(\frac{3x+1}{x+y}-\frac{2x-3}{x+y}=\frac{3x+1-2x+3}{x+y}=\frac{x+4}{x+y}\)
d. \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}=\frac{xy}{2x-y}-\frac{1-x^2}{2x-y}=\frac{xy-1+x^2}{2x-y}\)
e. \(\frac{4x-1}{3x^2y}-\frac{7x-1}{3x^2y}=\frac{4x-1-7x+1}{3x^2y}=\frac{-3x}{3x^2y}=\frac{-1}{xy}\)
a) \(\frac{x+7}{2x+3}-\frac{5}{2x+3}=\frac{x+7-5}{2x+3}=\frac{x+2}{2x+3}\)
b) \(\frac{m^2}{3\left(m+3\right)}+\frac{2m+3}{m+3}=\frac{m^2}{3\left(m+3\right)}+\frac{\left(2m+3\right).3}{3.\left(m+3\right)}\)
\(=\frac{m^2+6m+9}{3\left(m+3\right)}=\frac{\left(m+3\right)^2}{3\left(m+3\right)}=\frac{m+3}{3}\)
c) \(\frac{x^2-4}{x+5}.\frac{2x+10}{x+2}=\frac{\left(x-2\right)\left(x+2\right).2.\left(x+5\right)}{\left(x+5\right).\left(x+2\right)}=\left(x-2\right).2=2x-4\)
d) \(\frac{3+6y}{y^2-2y+1}:\frac{2y+1}{y-1}=\frac{3\left(2y+1\right)}{\left(y-1\right)^2}.\frac{y-1}{2y+1}=\frac{3}{y-1}\)
\(a,\frac{x+7}{2x+3}-\frac{5}{2x+3}\)
\(=\frac{x+7-5}{2x+3}\)
\(=\frac{x+2}{2x+3}\)
\(b,\frac{m^2}{3\left(m+3\right)}+\frac{2m+3}{m+3}\)
\(=\frac{m^2}{3\left(m+3\right)}+\frac{3\left(2m+3\right)}{3\left(m+3\right)}\)
\(=\frac{m^2+6m+9}{3\left(m+3\right)}\)
\(=\frac{\left(m+3\right)^2}{3\left(m+3\right)}\)
\(=\frac{m+3}{3}\)
\(c,\frac{x^2-4}{x+5}.\frac{2x+10}{x+2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{x+5}.\frac{2\left(x+5\right)}{x+2}\)
\(=2\left(x-2\right)\)
d, nghịch đảo lên rồi làm tương tự nha
a, \(\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}\)
b, \(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}=\frac{x+3}{\left(x-1\right)\left(x+2\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)}{x\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{x\left(x-1\right)}\)
\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)
tương tự đến hết nha a hay cj gì đps !
ai giups tui vs