Đáp án:
BÀI 2
a) 3x²y+6xy
=3xy(x+2)
b)3y²-3z²+3x²+6xy
=3(x+y+z)(x+y-z)
BÀI 3
a) (-2x)(2x²-2x+1)
=-4x³+4x²-2x
b) $\frac{2}{x+1}$ + $\frac{2x}{x+1}$
=2
c) ($\frac{1}{x-2}$ - $\frac{1}{x+2}$ ) . $\frac{x^2+4x+4}{4}$
=$\frac{x(x+2)}{2(x-2) }$
Giải thích các bước giải:
BÀI 2
a) 3x²y+6xy
=3xy(x+2)
b)3y²-3z²+3x²+6xy
=3(y²-z²+x²+2xy)
=3(x²+2xy+y²-z²)
=3(x+y)²-z²
=3(x+y+z)(x+y-z)
BÀI 3
a) (-2x)(2x²-2x+1)
=-4x³+4x²-2x
b) $\frac{2}{x+1}$ + $\frac{2x}{x+1}$
= $\frac{2+2x}{x+1}$
=$\frac{2(x+1)}{x+1}$
=2
c) ($\frac{1}{x-2}$ - $\frac{1}{x+2}$ ) . $\frac{x^2+4x+4}{4}$
= ($\frac{x+2}{(x-2)(x+2) }$ - $\frac{x-2}{(x+2)(x-2)}$ ) . $\frac{x^2+4x+4}{4}$
=$\frac{(x+2)-(2-x)}{(x-2)(x+2) }$ . $\frac{x^2+4x+4}{4}$
=$\frac{x+2-2+x}{(x-2)(x+2) }$ . $\frac{x^2+4x+4}{4}$
= $\frac{2x}{(x-2)(x+2) }$ . $\frac{x^2+4x+4}{4}$
= $\frac{2x(x^2+4x+4)}{4(x-2)(x+2) }$
=$\frac{2x(x+2)²}{4(x-2)(x+2) }$
=$\frac{x(x+2)}{2(x-2) }$
