Câu hỏi của Mộc Gia Linh

Question

Rút gọn: 
$\left(-\dfrac{1}{53}x^4y^5\right)^7.\left(-53xy^2\right)8$

Việt Anh · Accepted Answer

$\left(-\dfrac{1}{53}x^4y^5\right)^7.\left(-53xy^2\right).8$

$=\left[\left(-\dfrac{1}{53}\right)^7x^{4+7}y^{5+7}\right].\left(-53xy^2\right).8$

$=\left[\left(-\dfrac{1}{53^7}\right)x^{11}y^{12}\right].\left(-53xy^2\right).8$

$\Rightarrow\left[\left(-\dfrac{1}{53^7}\right).\left(-53\right)\right].\left(x^{11}.x\right).\left(y^{12}.y^2\right).8$

$\Rightarrow\left(\dfrac{53}{53^7}\right)x^{12}y^{14}.8$

$\Rightarrow\dfrac{1}{53^6}x^{12}y^{14}.8$Câu trả lời: $\left(-\dfrac{1}{53}x^4y^5\right)^7.\left(-53xy^2\right).8$

$=\left[\left(-\dfrac{1}{53}\right)^7x^{4+7}y^{5+7}\right].\left(-53xy^2\right).8$

$=\left[\left(-\dfrac{1}{53^7}\right)x^{11}y^{12}\right].\left(-53xy^2\right).8$

$\Rightarrow\left[\left(-\dfrac{1}{53^7}\right).\left(-53\right)\right].\left(x^{11}.x\right).\left(y^{12}.y^2\right).8$

$\Rightarrow\left(\dfrac{53}{53^7}\right)x^{12}y^{14}.8$

$\Rightarrow\dfrac{1}{53^6}x^{12}y^{14}.8$

Bài 3: Đơn thức