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\(\dfrac{2}{x^3-y^3}=\dfrac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)};\dfrac{1}{x^2-y^2}=\dfrac{1}{\left(x-y\right)\left(x+y\right)}\)MTC: (x-y)((x+y)(x2+xy+y2)
\(\dfrac{2}{x^3-y^3}=\dfrac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)\(\dfrac{1}{x+y}=\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\dfrac{2x+1}{x^2-y^2}=\dfrac{2x+1}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(2x+1\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
a: \(\dfrac{1}{6x^2y^3}=\dfrac{7x^2}{42x^4y^3}\)
\(\dfrac{-5}{21xy^2}=\dfrac{-10x^3y}{42x^4y^3}\)
\(\dfrac{3}{14x^4y}=\dfrac{3\cdot3y}{42x^4y^3}=\dfrac{9y}{42x^4y^3}\)
b: \(\dfrac{2}{x^3-y^3}=\dfrac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\dfrac{2x+1}{x^2-y^2}=\dfrac{\left(2x+1\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(2x+1\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\dfrac{2}{x^3-y^3}=\dfrac{2x+2y}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\dfrac{2x+1}{x^2-y^2}=\dfrac{\left(2x+1\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
Ta có \(\frac{2}{x^3-y^3}=\frac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{2x-1}{x^2-y^2}=\frac{2x+1}{\left(x+y\right)\left(x-y\right)}\)
\(\frac{1}{x+y}\) giữ nguyên
MTC: \(\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
Các nhân tử phụ tương ứng là : \(\left(x+y\right);\left(x-y\right)\left(x^2+xy+y^2\right);\left(x^2+xy+y^2\right)\)
Ta có:
\(\frac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\frac{2.\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{1}{x+y}=\frac{1.\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{2x+1}{\left(x+y\right)\left(x-y\right)}=\frac{\left(2x+1\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)