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a: \(\frac{x+1}{x-y}+\frac{x-1}{x-y}+\frac{x+3}{x-y}\)
\(=\frac{x+1+x-1+x+3}{x-y}=\frac{3x+3}{x-y}\)
b: \(\frac{5xy^2-3z}{3xy}+\frac{4x^2y+3z}{3xy}=\frac{5xy^2-3z+4x^2y+3z}{3xy}\)
\(=\frac{5xy^2+4x^2y}{3xy}=\frac{xy\left(5y+4x\right)}{3xy}=\frac{4x+5y}{3}\)
c: \(\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}\)
\(=\frac{x+9}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x\left(x+3\right)}\)
\(=\frac{x\left(x+9\right)-3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2+9x-3x+9}{x\left(x+3\right)\left(x-3\right)}\)
\(=\frac{x^2+6x+9}{x\left(x+3\right)\left(x-3\right)}=\frac{\left(x+3\right)^2}{x\left(x+3\right)\left(x-3\right)}=\frac{x+3}{x\left(x-3\right)}\)
d: \(\frac{x}{5x+5}-\frac{x-10}{10x-10}\)
\(=\frac{x}{5\left(x+1\right)}-\frac{x-10}{10\left(x-1\right)}\)
\(=\frac{2x\left(x-1\right)-\left(x-10\right)\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}=\frac{2x^2-2x-\left(x^2+x-10x-10\right)}{10\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x^2-2x-x^2+9x+10}{10\left(x-1\right)\left(x+1\right)}=\frac{x^2+7x+10}{10\left(x-1\right)\left(x+1\right)}\)
(x-6)(x+6)/2x+10 * -3(x-6)= 3x+18/2x+10
(x-3y)(x+3y)/x^2y^2* 3xy/2(x-3y)=3x+9y/2xy
3(x-y)(x+y)/5xy * -15x^2y/2(X-y)=-9x/2
a) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)
b) \(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3=\frac{1}{27}x^3+8y^3\)
c) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-\left(3y\right)^3=x^3-27y^3\)
d) \(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{2}x^2+\frac{1}{9}\right)=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3=x^6-\frac{1}{9}\)
( x + 4 )( x2 - 4x + 16 ) = x3 + 43 = x3 + 64
( 1/3x + 2y )( 1/9x2 - 2/3xy + 4y2 ) = ( 1/3x )3 - ( 2y )3 = 1/27x3 - 8y3
( x - 3y )( x2 + 3xy + 9y2 ) = x3 - ( 3y )3 = x3 - 27y3
( x2 - 1/3 )( x4 + 1/3x2 + 1/9 ) = ( x2 )3 - ( 1/3 )3 = x6 - 1/27
HĐT số 6 + 7 bạn nhé ^^
1, \(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)\(=\frac{4y.y}{11x^2.x^2}.\frac{-3x^2}{2.4y}\)\(=\frac{y}{11x^2}.\frac{-3}{2}=\frac{-3y}{22x^2}\)
2, \(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)\(=\frac{4x^2}{5y^2}.\frac{5y}{6x}.\frac{3y}{2x}\)\(=\frac{2x.2x}{5y.y}.\frac{5y}{3.2x}.\frac{3y}{2x}\)\(=\frac{2x}{y}.\frac{1}{3}.\frac{3y}{2x}\)
\(\frac{2x}{3y}.\frac{3y}{2x}=1\)
3, \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)\(=\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}\)\(=\frac{\left(x+2\right)}{3}.\frac{1}{2}=\frac{x+2}{6}\)
4, \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\left(-\frac{2\left(x-2\right)}{x+2}\right)=\frac{5}{4}.\frac{-2}{1}=-\frac{5}{2}\)
5, \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{3}{-\left(x-6\right)}=\frac{x+6}{2\left(x+5\right)}.\frac{-3}{1}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
6, \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}=\frac{\left(x-3y\right)\left(x+3y\right)}{\left(xy\right)^2}.\frac{3xy}{2\left(x-3y\right)}=\frac{x+3y}{xy}.\frac{3}{2}=\frac{3\left(x+3y\right)}{2xy}\)
7, \(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}=\frac{3\left(x-y\right)\left(x+y\right)}{5xy}.\frac{5xy.3x}{-2\left(x-y\right)}=\frac{3\left(x+y\right)}{1}.\frac{3x}{-2}=\frac{-9x\left(x+y\right)}{2}\)