Question

Answer 1

Bài 1:

a: \(\dfrac{x^2+y^2-1-2xy}{x^2-y^2+1-2x}\)

\(=\dfrac{\left(x^2-2xy+y^2\right)-1}{\left(x^2-2x+1\right)-y^2}\)

\(=\dfrac{\left(x-y\right)^2-1}{\left(x-1\right)^2-y^2}\)

\(=\dfrac{\left(x-y-1\right)\left(x-y+1\right)}{\left(x-1-y\right)\left(x-1+y\right)}\)

\(=\dfrac{x-y+1}{x+y-1}\)

b: \(\left(a-3\right)\left(a^2+3a+9\right)\)

\(=\left(a-3\right)\left(a^2+a\cdot3+3^2\right)\)

\(=a^3-3^3=a^3-27\)

c: \(5x^3+4x^2-3x\left(2x^2+7x-1\right)\)

\(=5x^3+4x^2-6x^3-21x^2+3x\)

\(=-x^2-17x^2+3x\)

d: \(\dfrac{14xy^5\left(3y-2x\right)}{21x^2y\left(2x-3y\right)^2}\)

\(=\dfrac{7xy\cdot\left(3y-2x\right)\cdot2y^4}{7xy\cdot\left(3y-2x\right)^2\cdot3x}\)

\(=\dfrac{2y^4}{3x\left(3y-2x\right)}\)

e: Sửa đề:\(\dfrac{x^3-x^2-x+1}{x^2-2x+1}\)

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

\(=\dfrac{x^2\left(x-1\right)-\left(x-1\right)}{\left(x^2-2x+1\right)}\)

\(=\dfrac{\left(x-1\right)\left(x^2-1\right)}{\left(x^2-2x+1\right)}\)

\(=\dfrac{\left(x-1\right)\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)

\(=x+1\)

Bài 2:

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

\(A=\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\right)\cdot\dfrac{4x^2-4}{5}\)

\(=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x^2-1\right)}{5}\)

\(=\dfrac{\left(x+1\right)^2+3\cdot2-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{x^2+2x+1+6-\left(x^2+2x-3\right)}{1}\cdot\dfrac{2}{5}\)

\(=\dfrac{x^2+2x+7-x^2-2x+3}{1}\cdot\dfrac{2}{5}\)

\(=10\cdot\dfrac{2}{5}=4\)

b: ĐKXĐ: \(y\notin\left\{1;-1\right\}\)

\(B=\dfrac{3}{3y+3}+\dfrac{10}{5-5y}+\dfrac{5y-1}{y^2-1}\)

\(=\dfrac{3}{3\left(y+1\right)}-\dfrac{10}{5\left(y-1\right)}+\dfrac{5y-1}{\left(y-1\right)\left(y+1\right)}\)

\(=\dfrac{1}{y+1}-\dfrac{2}{y-1}+\dfrac{5y-1}{\left(y-1\right)\left(y+1\right)}\)

\(=\dfrac{y-1-2\left(y+1\right)+5y-1}{\left(y+1\right)\left(y-1\right)}\)

\(=\dfrac{6y-2-2y-2}{\left(y+1\right)\left(y-1\right)}=\dfrac{4y-4}{\left(y+1\right)\left(y-1\right)}=\dfrac{4}{y+1}\)

c: ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

\(\dfrac{1}{3x-2}-\dfrac{4}{3x+2}-\dfrac{3x-6}{4-9x^2}\)

\(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{9x^2-4}\)

\(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{3x+2-4\left(3x-2\right)+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{6x-4-12x+8}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=-\dfrac{2}{3x+2}\)