\(\dfrac{4x^2-3x+5}{x^3-1}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)

\(\Leftrightarrow\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)

\(ĐKXĐ:x\ne1\)

\(\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{(1+2x)\left(x-1\right)}{(x^2+x+1)\left(x-1\right)}-\dfrac{6\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\)

\(\Rightarrow4x^2-3x+5-\left(1+2x\right)\left(x-1\right)-6\left(x^2+x+1\right)\)

\(\Rightarrow4x^2-3x+5-\left(x-1+2x^2-2x\right)-6x^2-6x-6\)

\(\Rightarrow4x^2-3x+5-x+1-2x^2+2x-6x^2-6x-6\)

\(\Rightarrow-4x^2-8x\)

⇒-4x(x-4)

bn đăng lại ở toán nha vì ở đây sẽ khó có ng lm cho bn.

(hihahu :V lm 1 câu thoy, lười)

b) \(\left(y-2\right)\left(y+2\right)\left(y^2+4\right)\)

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

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

\(=\left(y^2\right)^2-4^2=y^4-16\)

Bạn thu gọn các đa thức rồi thay thế vào sẽ tính ra ngay nha!

\(\dfrac{5x-3}{x^2-9}-\dfrac{x}{x-3}=\dfrac{2x-1}{x+3}\\ĐKXĐ:x\ne3;-3\\ \Leftrightarrow \dfrac{5x-3}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(2x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\\ \Rightarrow5x-3-x^2+3x=2x^2-6x-x+3\\ \Leftrightarrow8x-3-x^2=2x^2-7x+3\\ \Leftrightarrow8x+7x-x^2-2x^2=3+3\\ \Leftrightarrow15x-3x^2=6\\ \Leftrightarrow3x\left(5-x\right)=6\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)