Giải bài 2 trang 16

Đề bài

Phân tích đa các đa thức sau thành nhân tử:

a) $100-x^2$;                     

b) $4x^2-y^2$;                         

c) ${\left(x+y\right)}^2-\frac{1}{4}{y}^2$;

d) ${\left(x-y\right)}^2-{\left(y-z\right)}^2$;  

e) $x^2-{\left(1+2x\right)}^2$;                

g) $x^4-16$.

Lời giải chi tiết

a) $100-x^2$

$={10}^2-x^2$

$=\left(10-x\right)\left(10+x\right)$

b) $4x^2-y^2$

$={\left(2x\right)}^2-y^2$

$=\left(2x-y\right)\left(2x+y\right)$

c) ${\left(x+y\right)}^2-\frac{1}{4}{y}^2$

$={\left(x+y\right)}^2-{\left(\frac{1}{2}y\right)}^2$

$=\left(x+y-\frac{1}{2}y\right)\left(x+y+\frac{1}{2}y\right)$

$=\left(x+\frac{y}{2}\right)\left(x+\frac{3y}{2}\right)$

d) ${\left(x-y\right)}^2-{\left(y-z\right)}^2$

$=\left(x-y+y-z\right)\left(x-y-y+z\right)$

$=\left(x-z\right)\left(x-2y+z\right)$

e) $x^2-{\left(1+2x\right)}^2$

$=\left(x+1+2x\right)\left(x-1-2x\right)$

$=\left(3x+1\right)\left(-x-1\right)$

$=-\left(3x+1\right)\left(x+1\right)$

g) $x^4-16$

$={\left(x^2\right)}^2-4^2$

$=\left(x^2-4\right)\left(x^2+4\right)$

$=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)$