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?1 . Có . Mẫu thức chung : 12x2y3z đơn giản hơn
?2 . \(\dfrac{3}{x^2-5x}=\dfrac{3}{x\left(x-5\right)}=\dfrac{6}{2x\left(x-5\right)}\)
\(\dfrac{5}{2x-10}=\dfrac{5}{2\left(x-5\right)}=\dfrac{5x}{2x\left(x-5\right)}\)
?3 . \(\dfrac{3}{x^2-5x}=\dfrac{3}{x\left(x-5\right)}=\dfrac{6}{2x\left(x-5\right)}\)
\(\dfrac{-5}{10-2x}=\dfrac{5}{2x-10}=\dfrac{5}{2\left(x-5\right)}=\dfrac{5x}{2x\left(x-5\right)}\)
MTC : \(y^3-z^2y\)
\(\frac{x}{y^2-yz}=\frac{x}{y\left(y-z\right)}=\frac{x\left(y+z\right)}{y\left(y-z\right)\left(y+z\right)}=\frac{xy+xz}{y^3-z^2y}\)
\(\frac{z}{y^2+yz}=\frac{z}{y\left(y+z\right)}=\frac{z\left(y-z\right)}{y\left(y+z\right)\left(y-z\right)}=\frac{yz-z^2}{y^3-z^2y}\)
\(\frac{y}{y^2-z^2}=\frac{y}{\left(y-z\right)\left(y+z\right)}=\frac{y^2}{y^3-z^2y}\)
a) MTC : \(\left(x+1\right)\left(x^2-x+1\right)\)
Quy đồng :
\(\frac{x-1}{x^3+1}=\frac{x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2x}{x^2-x+1}=\frac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2}{x+1}=\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
b ) MTC : \(10x\left(2y-x\right)\left(2y+x\right)\)
\(\frac{7}{5x}=\frac{7.2.\left(2y-x\right)\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=\frac{-4.10x.\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}=\frac{-40x\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
c ) MTC : \(\left(x+2\right)^3\)
\(\frac{6x^2}{x^3+6x^2+12x+8}=\frac{6x^2}{\left(x+2\right)^3}\)
\(\frac{3x}{x^2+4x+4}=\frac{3x}{\left(x+2\right)^2}=\frac{3x\left(x+2\right)}{\left(x+2\right)^3}\)
\(\frac{2}{2x+4}=\frac{1}{x+2}=\frac{\left(x+2\right)^2}{\left(x+2\right)^3}\)
Có thể chọn mẫu thức chung là 12x2y3 z hoặc 24x3y4z
Chọn mẫu thức chung là 12x2y3z đơn giản hơn