c1:
a,\(=6x^3y^4\)
b,\(=x^3-2x^2+5x\)
c, \(=x-2\)
d, \(=x-1\)
c2:
a, \(=5xy\left(x-2y\right)\)
b,\(=3\left(x+3\right)-\left(x^2-9\right)=3\left(x+3\right)-\left(x-3\right)\left(x+3\right)=\left(x+3\right)\left(3-x+3\right)=\left(x+3\right)\left(6-x\right)\)
c,\(=xy\left(x-y\right)+z\left(x-y\right)=\left(x-y\right)\left(xy+z\right)\)
c3:
a,bt A được xác định khi x\(_{^{ }\ne}\)+-2
b,rút gọn:
\(=\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)
\(=\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-x^2-2x+2x-4}{\left(x-2\right)\left(x+2\right)}=-\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)
c, thay x=1 vào bt A , ta được:
\(\dfrac{-4}{\left(1-2\right)\left(1+2\right)}\)\(=\dfrac{4}{3}\)
c5:
M=\(a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)
thay a+b=1 vào bt M , ta được:
\(M=1-3ab+3ab\left(1-2ab\right)+6a^2b^2\)
\(=1\)
