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=(a+b+c)(a2+b2+c2−ab−bc−ca)
=(a+b+c)(a2+2ab+b2−ab−ac+c2)−3ab(a+b+c)
=(a+b)3+c3−3ab(a+b+c)
=a3+3ab(a+b)+b3+c3−3abc−3ab(a+b
a3+b3+c3−3abc
a: \(=20x^2-10x^3+25x^2\)
b: \(=12x^3+18x^2-10x-15\)
f: \(=\dfrac{2x^3-10x^2-11x^2+55x+12x-60}{x-5}=2x^2-11x+12\)
a: \(=\dfrac{x^2-5x+x+4}{x\left(x-2\right)}=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{x-2}{x}\)
b: \(=\dfrac{x^2-6x+9+4x^2+8x-4x^2-8x}{\left(x-3\right)\left(x+2\right)}\)
\(=\dfrac{x-3}{x+2}\)
a) \(=\dfrac{x\left(x-5\right)+x+4}{x\left(x-2\right)}=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)
b) \(=\dfrac{\left(x-3\right)^2+4x\left(x+2\right)-8x-4x^2}{\left(x+2\right)\left(x-3\right)}=\dfrac{x^2-6x+9+4x^2+8x-8x-4x^2}{\left(x+2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-6x+9}{\left(x+2\right)\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+2\right)\left(x-3\right)}=\dfrac{x-3}{x+2}\)
a,3xn(6xn−3+1)−2xn(9xn−3−1)a,3xn(6xn−3+1)−2xn(9xn−3−1)
=xn[3(6xn−3+1)−2(9xn−3−1)]=xn[3(6xn−3+1)−2(9xn−3−1)]
=xn(18xn−3+3−18xn−3+2)=xn(18xn−3+3−18xn−3+2)
=5xn=5xn
b,5n+1−4.5nb,5n+1−4.5n=5n.5+5n.4=5n(5+4)=45n=5n.5+5n.4=5n(5+4)=45n
c,62.64−43(36−1)c,62.64−43(36−1)
=66−43.36+43=66−43.36+43
=26.36−43.36+43=26.36−43.36+43
=36(26−43)+43=36(26−43)+43
=36[(22)3−43]+43=36.0+43=43=64
~Hok tốt~
TL:
a)
=\(18x^{2n-3}+3x^n-18x^{2n-3}+2x^n\)
=\(6x^n\)
b)
=\(5^n.5-4.5^n\)
=\(5^n\left(5-4\right)\)
=\(5^n\)
vậy.......
hc tốt