Bài 6: Phép trừ các phân thức đại số

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

\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)

\(=\dfrac{y^3+x^3}{x^2y^2}\)

Ngay ở dưới có ng` làm rồi đó bạn

Trước khi đăng sẽ thấy câu hỏi tương tự nên bn bấm vào để tham khảo nhé!

\(\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{x^2-6x+9}-\dfrac{x}{x^2-9}=\dfrac{18}{\left(x-3\right)^2\left(x+3\right)}+\dfrac{-3}{\left(x-3\right)^2}\)

\(\Rightarrow\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{x^2-2x.3+3^2}-\dfrac{x}{x^2-9}=\dfrac{18}{\left(x^2-9\right)\left(x-3\right)}-\dfrac{3}{\left(x-3\right)^2}\)

\(\Rightarrow\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{\left(x-3\right)^2}-\dfrac{x}{x^2-9}=\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{\left(x-3\right)^2}\)

\(\Rightarrow-\dfrac{x}{x^2-9}=0\)

Xảy ra khi \(x=0\)

Vậy \(x=0.\)

- giúp với

a)

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

b)

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

\(\dfrac{x^2+x+5}{x^3-27}-\dfrac{1-x}{3x-x^2}\)

\(=\dfrac{x^2+x+5}{\left(x-3\right)\left(x^2+x+9\right)}-\dfrac{1-x}{x\left(3-x\right)}\)

\(=\dfrac{\left(x^2+x+5\right)x}{x\left(x-3\right)\left(x^2+x+9\right)}+\dfrac{\left(1-x\right)\left(x^2+x+9\right)}{x\left(x-3\right)\left(x^2+x+9\right)}\)

Đến đây bạn nhân đa thức ra rồi tính nốt nhé!

\(=\dfrac{y}{x\left(y-5x\right)}+\dfrac{25x-15y}{\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{y\left(y+5x\right)+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{y^2+5xy+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{y^2+30xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)

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

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

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

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

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

Bài toán sai ngay với $x=y=0,5$

Đề đúng là thế này :

\(\dfrac{x}{y^3-1}+\dfrac{y}{x^3-1}-\dfrac{2\left(xy-2\right)}{x^2y^2+3}=0\)

\(\dfrac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}=\dfrac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}=\dfrac{y-5x}{x\left(y+5x\right)}\)

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

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

\(=\dfrac{x^3+x^2+6x-\left(x^3+3x^2+9x-x^2-3x-9\right)}{x\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\dfrac{x^3+x^2+6x-x^3-2x^2-6x+9}{x\left(x-3\right)\left(x^2+3x+9\right)}\)

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

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