Question

Rút gọn bằng cách dùng hđt :

a) \(\left(x-1\right)^2-\left(x-2\right)\left(x+2\right)\)

b) \(\left(x^2+\frac{1}{x}+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3\)

Answer 1

a) \(\left(x-1\right)^2-\left(x-2\right)\left(x+2\right)=x^2-2x+1-x^2+4=5-2x\)

mình nghĩ là câu b bạn ghi đề sai vì như thế không có hằng đẳng thức nhé

b)\(\left(x^2+\frac{1}{3}x+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3=x^3-\frac{1}{27}-x^3+\frac{1}{27}+x^2-\frac{1}{3}x=x^2-\frac{1}{3}x\)

Answer 2

b,\(\left(x^2+\frac{1}{x}+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3\)

\(=\)\(\left(x-\frac{1}{3}\right)\left[\left(x^2+\frac{1}{x}+\frac{1}{9}\right)-\left(x-\frac{1}{3}\right)^2\right]\)

\(=\)\(\left(x-\frac{1}{3}\right)\left(x^2+\frac{1}{x}+\frac{1}{9}-x^2+\frac{2}{3}x-\frac{1}{9}\right)\)

\(=\left(x-\frac{1}{3}\right)\left(\frac{1}{x}+\frac{2}{3}x\right)\) \(=1+\frac{2}{3}x^2-\frac{1}{3x}-\frac{2}{9}x\)

Answer 3

a) (x-1)^2-(x-2)(x+2)=(x-1)^2-(x^2-4)=x^2-2x+1-x^2+4=-2x+5

b) (x^2+1/x+1/9)(x-1/3)-(x-1/3)^3=(x^3-1/27)-(x-1/3)^3=x^3-1/27-(x^3-3x^2*1/3+3x*1/9-1/27)

=x^3-1/27-x^3+x^2-1/2x+1/27=x^2-1/2x