Phân tích thành nhân tử \(1+a\sqrt{a}\)
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)
\(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a\sqrt{b}+\sqrt{ab}+\sqrt{a}+1\)
\(=\sqrt{ab}\cdot\sqrt{a}+\sqrt{ab}+\sqrt{a}+1\)
\(=\left(\sqrt{ab}\cdot\sqrt{a}+\sqrt{ab}\right)+\left(\sqrt{a}+1\right)\)
\(=\sqrt{ab}\cdot\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{ab}+1\right)\left(\sqrt{a}+1\right)\)
a√b + √(ab) + √a + 1
= [a√b + √(ab)] + (√a + 1)
= √(ab)(√a + 1) + (√a + 1)
= (√a + 1)[√(ab) + 1]
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a\sqrt{a}+2a+\sqrt{a}+2=\left(a\sqrt{a}+2a\right)+\left(\sqrt{a}+2\right)\)
\(=a\left(\sqrt{a}+2\right)+\left(\sqrt{a}+2\right)=\left(\sqrt{a}+2\right)\left(a+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\sqrt{\left(a+b\right)\left(a-b\right)}+\sqrt{a-b
}\)
\(=\sqrt{a-b}\cdot\sqrt{a+b}+\sqrt{a-b}\)
\(=\sqrt{a-b}\cdot\left(\sqrt{a+b}+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{a}b\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)1)
\(\left(\sqrt{a}b+1\right)\left(\sqrt{a}+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)
\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)
\(=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}\)
\(=\dfrac{-7xy\cdot\sqrt{3xy}}{xy}\)
\(=-7\sqrt{3}\cdot\sqrt{xy}\)
b) Ta có: \(ab+b\sqrt{a}+\sqrt{a}+1\)
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
$a)-7xy.\sqrt{\dfrac{3}{xy}}$
$=-7.\sqrt{x^2y^2.\dfrac{3}{xy}}(do \,x,y>0a\to xy>0)$
$=-7.\sqrt{\dfrac{xy}{3}}$
$b)ab+b\sqrt{a}+\sqrt{a}+1(a \ge 0)$
$=b\sqrt{a}(\sqrt{a}+1)+\sqrt{a}+1$
$=(\sqrt{a}+1)(b\sqrt{a}+1)$
\(\left(\sqrt{a+1}\right).\left(a-\sqrt{a+1}\right)\)