Question

Bài 1:

a) Rút gọn phân thức: \(\frac{2x^4+6x^3+18x^2}{x^4-27x}\)

b) Cộng, trừ phân thức sau: \(\frac{x+\frac{1}{3}}{1-x^2}\) +\(\frac{5}{3x-3}\) +\(\frac{1}{3x+3}\)

Bài 2: cho biết a+b =7 và a.b = 3 . Tính (a-b)\(^2\)

Bài 3: Thu gọn

\(\frac{x^2+1}{x^2-3x}\) + \(\frac{3}{x}\) - \(\frac{x}{x-3}\)

Mong m.n giúp ạ!!

Answer 1

Bài 1:

a)

\(\frac{2x^4+6x^3+18x^2}{x^4-27x}=\frac{2x^2(x^2+3x+9)}{x(x^3-27)}=\frac{2x(x^2+3x+9)}{(x-3)(x^2+3x+9)}=\frac{2x}{x-3}\)

b)

\(\frac{x+\frac{1}{3}}{1-x^2}+\frac{5}{3x-3}+\frac{1}{3x+3}=\frac{3x+1}{3(1-x)(1+x)}-\frac{5(x+1)}{3(1-x)(1+x)}+\frac{1-x}{3(x+1)(1-x)}\)

\(=\frac{3x+1-5(x+1)+1-x}{3(1-x)(1+x)}=\frac{-3x-3}{3(1-x)(1+x)}=\frac{-3(x+1)}{3(1-x)(1+x)}=\frac{1}{x-1}\)

Answer 2

Bài 2:

$(a-b)^2=a^2+b^2-2ab=(a^2+b^2+2ab)-4ab=(a+b)^2-4ab$

$=7^2-4.3=37$

Answer 3

Bài 3:

ĐKXĐ: $x\neq 0; x\neq 3$

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

\(=\frac{x^2+1+3(x-3)-x^2}{x(x-3)}=\frac{3x-8}{x(x-3)}\)