![](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)
\(5-7x^2=\left(\sqrt{5}\right)^2-\left(x\sqrt{7}\right)^2\)
\(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)
\(3+4x=\left(\sqrt{3}\right)^2-\left(2\sqrt{x}\right)^2\) ( do x<0 )
\(=\left(\sqrt{3}-2\sqrt{x}\right)\left(3+2\sqrt{x}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x3 + 7x - 6
= x3 - x - 6x - 6
= x3 - x - 6 (x+1)
= x (x2 - 1) - 6 (x+1)
= (x + 1) ( x (x - 1) - 6 )
= ( x + 1) ((x2 - x - 6))
= (x + 1) ((x2 + 2 - 3 - 6))
= (x + 1) (x(x +2) - 3 ( x + 2))
= (x + 1)(x + 2)(x + 3)
![](https://rs.olm.vn/images/avt/0.png?1311)
xét \(x\ne0\)ta có :
\(M=\)\(^{x^2\cdot\left(x^2+6x+7-\frac{6}{x}+\frac{1}{x^2}\right)}\)
Đặt \(x-\frac{1}{x}=t\Rightarrow t^2=x^2-2+\frac{1}{x^2}\Leftrightarrow t^2+2=x^2+\frac{1}{x^2}\)
Do đó \(M=x^2\cdot\left(t^2+2+6t+7\right)\Leftrightarrow x^2\cdot\left(t^2+6t+9\right)\)
\(\Leftrightarrow M=x^2\cdot\left(t+3\right)^2\)
M=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=x^2(x^2+3x-1)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x^2+3x-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4+6x^3+7x^2-6x+1\)
\(=x^4+6x^3+9x^2-2x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=x^4+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2\right)^2+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(P\left(x\right)=x^4+6x^3+7x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=x^4+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
4x^2 - 7x -2 = 4x^2 - 8x + x - 2 = 4x(x - 2) + (x - 2) = (x -2)(4x + 1)
Bài làm:
\(5-7x^2=\left(\sqrt{5}-\sqrt{7}x\right)\left(\sqrt{5}+\sqrt{7}x\right)\)
Bài làm :
Ta có hằng đẳng thức sau :
\(a^2-b^2=\left(a+b\right)\left(a-b\right)\)
Vậy theo hằng đẳng thức trên thì :
\(5-7x^2=\left(\sqrt{5}+\sqrt{7}x\right)\left(\sqrt{5}+\sqrt{7}x\right)\)
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!