a. (\(\frac{1}{2}\) x – 1) (2x – 3)

      = x2 - \(\frac{3}{2}\) x – 2x + 3

      = x2 - \(\frac{7}{2}\) x + 3

b. (x –7)(x –5)

      = x2 – 5x – 7x + 35

      = \(x^2\) – 12x + 35

c. (x - \(\frac{1}{2}\) )(x + \(\frac{1}{2}\) )(4x - 1)

      = (\(x^2\) + \(\frac{1}{2}\) x - \(\frac{1}{2}\) x - \(\frac{1}{4}\) )(4x - 1)

      = (\(x^2\) - \(\frac{1}{4}\) )(4x - 1)

      = 4\(x^3\) – \(x^2\) – x + \(\frac{1}{4}\)

a, \(\left(\frac{1}{2}x-1\right)\left(2x-3\right)=x^2-\frac{3}{2}x-2x+3=x^2-\frac{7}{2}x+3\)

b, \(\left(x-7\right)\left(x-5\right)=x^2-12x+35\)

c, \(\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(4x-1\right)=\left(x^2-\frac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\frac{1}{4}\)

\(\left(\frac{a}{b^2}+\frac{1}{a}-\frac{1}{b}\right):\left(\frac{b}{a}+\frac{a^2}{b^2}\right)=\left(\frac{a^2+b^2-ab}{ab^2}\right):\left(\frac{b^3+a^3}{ab^2}\right)=\frac{a^2+b^2-ab}{\left(a^3+b^3\right)}\)

\(=\frac{a^2+b^2-ab}{\left(a+b\right)\left(a^2+b^2-ab\right)}=\frac{1}{a+b}\)

\(\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4-x^3y+x^3y-x^2y^2+x^2y^2-xy^3+xy^3-y^4=x^4-y^4\)

3x³ - 5x² + 5x - 2

= 3x³ - 2x² - 3x² +2x + 3x-2

= x²(3x -2) - x(3x -2) + (3x-2)

= (3x-2)(x² -x +1) 

đề bài là j ạ?

HT K CHO MIK NHA

mình bảo

toán lớp 8 thì khó còn tuán lớp mấu giáo thì dễ

nhỉ

kham khảo nhé

k cho mik nha

ta có : \(\hept{\begin{cases}\left(a+b\right)\left(a^2-b^2\right)=a^3-b^3-ab^2+a^2b\\\left(b+c\right)\left(b^2-c^2\right)=b^3-c^3-bc^2+b^2c\\\left(a+c\right)\left(c^2-a^2\right)=c^3-a^3-ca^2+c^2a\end{cases}}\)

Vậy biểu thức bằng : \(a^2b-ab^2+b^2c-bc^2+c^2a-ca^2=ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b\right)-ca\left(b-c\right)=a\left(b-c\right)\left(a-b\right)+c\left(b-a\right)\left(b-c\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

đây nha bn