Giải bài 4 trang 8

Đề bài

Thu gọn mỗi đa thức sau:

a) $x^2{y}^5+2x{y}^2-x^2{y}^5+\frac{24}{35}x{y}^2$

b) $-11y^2{z}^3-22x{y}^3{z}^3+2y^2{z}^3-33x{y}^3{z}^3-72$

c) $\frac{\sqrt{4}}{41}{x}^2{y}^4{z}^3+x^2{y}^{4}z+\frac{39}{41}{x}^2{y}^4{z}^3-x^2{y}^{4}z+z^{18}$

Lời giải chi tiết

a) Ta có:

$x^2{y}^5+2x{y}^2-x^2{y}^5+\frac{24}{35}x{y}^2=\left(x^2{y}^5-x^2{y}^5\right)+\left(2x{y}^2+\frac{24}{35}x{y}^2\right)=\frac{94}{35}x{y}^2$

b) Ta có:

$\begin{array}{c}-11y^2{z}^3-22x{y}^3{z}^3+2y^2{z}^3-33x{y}^3{z}^3-72\\ =\left(-11y^2{z}^3+2y^2{z}^3\right)+\left(-22x{y}^3{z}^3-33x{y}^3{z}^3\right)-72\\ =-9y^2{z}^3-55x{y}^3{z}^3-72\end{array}$

c) Ta có:

$\begin{array}{c}\frac{\sqrt{4}}{41}{x}^2{y}^4{z}^3+x^2{y}^{4}z+\frac{39}{41}{x}^2{y}^4{z}^3-x^2{y}^{4}z+z^{18}\\ =\left(\frac{\sqrt{4}}{41}{x}^2{y}^4{z}^3+\frac{39}{41}{x}^2{y}^4{z}^3\right)+\left(x^2{y}^{4}z-x^2{y}^{4}z\right)+z^{18}\\ =x^2{y}^4{z}^3+z^{18}\end{array}$