![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Ta có: \(Q=2x\left(\dfrac{1}{2}x^2+y\right)-x\left(x^2+y\right)+xy\left(x^3-1\right)\)
\(=x^3+2xy-x^3-xy+x^3y-xy\)
\(=x^3y\)
\(=10^3\cdot\dfrac{-1}{10}=1000\cdot\dfrac{-1}{10}=-100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{8}\) + \(\dfrac{5}{7}\) = \(\dfrac{7}{56}\) + \(\dfrac{40}{56}\) = \(\dfrac{47}{56}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 3:
a: Thay x=-3 vào A, ta được:
\(A=\dfrac{-3-4}{-3+5}=\dfrac{-7}{2}\)
b: \(B=\dfrac{2x-8+x+20}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x+12}{\left(x+4\right)\left(x-4\right)}=\dfrac{3}{x-4}\)
c: \(M=A\cdot B=\dfrac{x-4}{x+5}\cdot\dfrac{3}{x-4}=\dfrac{3}{x+5}\)
Để M nguyên thì \(x+5\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{-6;-2;-8\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: \(x\notin\left\{10;-10;\sqrt{10};-\sqrt{10}\right\}\)
b: \(A=\dfrac{5x^3+50x+2x^2+20+5x^3-50x-2x^2+20}{\left(x^2-10\right)\left(x^2+10\right)}\cdot\dfrac{x^2-100}{x^2+4}\)
\(=\dfrac{10x^3+40}{\left(x^2-10\right)\left(x^2+10\right)}\cdot\dfrac{x^2-100}{x^2+4}\)