Bài 10: Chia đơn thức cho đơn thức

$x^{2}yz:xyz=\left(x^2:x\right)\left(y:y\right)\left(z:z\right)=x$

$x^3{y}^4:x^{3}y=\left(x^3:x^3\right)\left(y^4:y\right)=y^3$

${\left(x+y\right)}^2:\left(x+y\right)=x+y$

Ta có: y – x= -1.(x – y) nên ${\left(y-x\right)}^4=\left[-1.{\left(x-y\right)}^4\right]={\left(-1\right)}^4.{\left(x-y\right)}^4={\left(x-y\right)}^4$

${\left(x-y\right)}^5:{\left(y-x\right)}^4$ = ${\left(x-y\right)}^5:{\left(x-y\right)}^4$ = x – y

${\left(x-y+z\right)}^4:{\left(x-y+z\right)}^3=(x-y+z)$

$18x^2{y}^{2}z : 6xyz=\left(18:6\right)\left(x^2:x\right)\left(y^2:y\right)\left(z:z\right)=3xy$

$5a^{3}b : \left(-2a^{2}b\right)=\left[5:\left(-2\right)\right].\left(a^3:a^2\right)\left(b:b\right)=-5/2.a$

$27x^4{y}^{2}z : 9x^{4}y=\left(27:9\right)\left(x^4:x^4\right)\left(y^2:y\right).z=3yz$

$x^4:x^n$ = $x^{4-n}$ là phép chia hết nên 4 – n $\ge$ 0 ⇒ 0 $\le$ n ≤ 4

suy ra: n ∈ {0; 1; 2; 3; 4}

$x^n:x^3$ = $x^{n-3}$ là phép chia hết nên n – 3 $\ge$ 0 ⇒ n $\ge$ 3

$5x^n{y}^3 : 4x^2{y}^2$ = 5/4 $\left(x^n : x^2\right)\left(y^3 : y^2\right)$ = 5/4 $x^{n-2}.y$ là phép chia hết

Suy ra: n – 2 $\ge$ 0 ⇒ n $\ge$ 2

 $x^n{y}^{n+1} : x^2{y}^5$ = $\left(x^n:x^2\right)\left(y^{n+1}:y^5\right)=x^{n-2}.y^{n-4}$ là phép chia hết

Ta có: ${\left(-x^2{y}^5\right)}^2\hspace{0.17em}: \left(-x^2{y}^5\right)$ = $-x^2{y}^5$

Thay x = 1/2 và y = - 1 vào biểu thức ta được:

$-{\left(1/2\right)}^5.{\left(-1\right)}^5=-1/4.\left(-1\right)=1/4$

$$\begin{gathered} {\left(-x^3 y^2 z\right)}^4:{\left(-x{y}^2 z\right)}^3 \\ =\left[{\left(-1\right)}^4.x^{3.4} y^{2.4} z^4\right]:\left(-x^{1.3} y^{2.3} z^{1.3}\right) \\ =x^{12} y^8 z^4 : \left(-x^3 y^6 z^3\right)=-x^9 y^2 z \end{gathered}$$