Bài 2.
\(a)(x+1)^2+1+x\\=(x+1)^2+ (x+1)\\=(x+1)(x+1+1)\\=(x+1)(x+2)\\---\\b)(x+y)^3-(x-y)^3\\=[(x+y)-(x-y)][(x+y)^2+(x+y)(x-y)+(x-y)^2]\\=(x+y-x+y)(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2)\\=2y(3x^2+y^2)\)
\(---\\c)2x^2-4x+2-2y^2\\=2[(x^2-2x+1)-y^2]\\=2[(x-1)^2-y^2)\\=2(x-1-y)(x-1+y)\\---\\d)49y^2-x^2+6x-9\\=49y^2-(x^2-6x+9)\\=(7y)^2-(x-3)^2\\=(7y-x+3)(7y+x-3)\)
\(Toru\)
Bài 2:
a) \(\left(x+1\right)^2+1+x\)
\(=\left(x+1\right)^2+\left(x+1\right)\)
\(=\left(x+1\right)\left(x+1+1\right)\)
\(=\left(x+1\right)\left(x+2\right)\)
b) \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
c) \(2x^2-4x+2-2y^2\)
\(=2\left(x^2-2x+1-y^2\right)\)
\(=2\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=2\left[\left(x-1\right)^2-y^2\right]\)
\(=2\left(x-y-1\right)\left(x+y-1\right)\)
d) \(49y^2-x^2+6x-9\)
\(=49y^2-\left(x^2-6x+9\right)\)
\(=\left(7y\right)^2-\left(x-3\right)^2\)
\(=\left(7y-x+3\right)\left(7y+x-3\right)\)
Bài 3.
\(P=\dfrac{2}{x^2-x}+\dfrac{2}{x^2+x+1}+\dfrac{4x}{1-x^3}\left(ĐK:x\ne0;x\ne1\right)\)
\(=\dfrac{2}{x\left(x-1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{4x}{x^3-1}\)
\(=\dfrac{2\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2x\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{4x^2}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2x^2+2x+2+2x^2-2x-4x^2}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2}{x\left(x^3-1\right)}\)
\(=\dfrac{2}{x^4-x}\)
Vậy \(P=\dfrac{2}{x^4-x}\).
\(b,\) Thay \(x=2\) vào \(P\), ta được:
\(P=\dfrac{2}{2^4-2}\)
\(=\dfrac{2}{16-2}\)
\(=\dfrac{2}{14}\)
\(=\dfrac{1}{7}\)
#\(Toru\)
