Bài 8: Phép chia các phân thức đại số

\(18,\\ \dfrac{1-9x^2}{x^2+4x}:\dfrac{2-6x}{3x}\left(x\ne0;x\ne-4;x\ne\dfrac{1}{3}\right)\\ =\dfrac{\left(1-3x\right)\left(1+3x\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-3x\right)}=\dfrac{3\left(1+3x\right)}{2\left(x+4\right)}\\ \dfrac{27-x^3}{5x+10}:\dfrac{x-3}{3x+6}\left(x\ne-2;x\ne3\right)=\dfrac{\left(3-x\right)\left(x^2+3x+9\right)}{5\left(x+2\right)}:\dfrac{3\left(x+2\right)}{-\left(3-x\right)}\\ =\dfrac{-3\left(x^2+3x+9\right)}{5}\)

\(19,\\ \dfrac{4x^2}{25y^2}:\dfrac{6x}{5y}:\dfrac{2x}{9y}\left(x,y\ne0\right)=\dfrac{4x^2\cdot5y\cdot9y}{25y^2\cdot6x\cdot2x}=\dfrac{3}{5}\)

a) \(\left(6.8^4-5.8^3+8^2\right):8^2\)

\(=\left(6.8^4:8^2\right)-\left(5.8^3:8^2\right)+\left(8^2:8^2\right)\)

\(=6.8^2-5.8+1\)

\(=6.64-40+1\)

\(=384-40+1\)

\(=344+1=345\)

\(a,=6\cdot8^2-5\cdot8+1=6\cdot64-40+1=345\\ b,=7\cdot5^6-8\cdot5^2+5=...\\ c,=\left(5\cdot3^4+3^5-2\cdot3^3\right):3^2=5\cdot3^2+3^3-2\cdot3=45+27-6=66\\ d,=\left(3\cdot2^4+2^6+3\cdot2^8\right):2^3=3\cdot2+2^3+3\cdot2^5=6+8+96=110\)

\(a,=x^2-x^3-11x\\ b,=xy-2y^2+3x\\ c,=2xy-\dfrac{9}{2}x+\dfrac{25}{2}y^2\\ d,=x^2y-4y^2+6x\)

a: \(\dfrac{x^3-x^4-11x^2}{x}=x^2-x^3-11x\)

b: \(\dfrac{2x^2y^2-4xy^3+6x^2y}{2xy}=xy-2y^2+3x\)

a: \(\dfrac{x^2+2x+1}{x+1}=x+1\)

b: \(\dfrac{8x^3-27y^3}{2x-3y}=4x^2+6xy+9y^2\)

c: \(\dfrac{8x^3+y^3}{2x+y}=4x^2-2xy+y^2\)

\(C=\dfrac{-2x^2y^2+4xy-6xy^3}{xy}=-2xy+4-6y^2\)

\(=-2\cdot1\cdot4+4-6\cdot4^2\)

\(=-8+4-96\)

\(=-100\)

Bài 1:
a. Để $A\vdots B$ thì: \(\left\{\begin{matrix} 4\geq n\\ 2n\geq 7\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} n\leq 4\\ n\geq 3,5\end{matrix}\right.\)

Vì $n$ là số tự nhiên $n=4$

b.

Để $A\vdots B$ thì: \(\left\{\begin{matrix} 8\geq 3n\\ 2n\geq 4\\ \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} n\leq \frac{8}{3}\\ n\geq 2\end{matrix}\right.\). Mà $n$ là số tự nhiên nên $n=2$