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\(=x\left(3x-7y\right)-\left(3x-7y\right)=\left(x-1\right)\left(3x-7y\right)\)
3x2 - 7xy - 3x + 7y
= (3x2-3x) - (7xy-7y)
= 3x(x-1) - 7y(x-1)
= (3x-7y)(x-1)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) 3x4-3x3+9x3-9x2-24x2+24x-48x+48
=3x3(x-1)+9x2(x-1)-24x(x-1)-48(x-1)
=(x-1)(3x3+9x2-24x-48)
=3(x-1)(x3+3x2-8x-16)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=3\left(x^2-8x+16\right)=3\left(x-4\right)^2\\ b,=5\left(x^2-1\right)=5\left(x-1\right)\left(x+1\right)\\ c,=\left(x+y\right)^2-9=\left(x+y+3\right)\left(x+y-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-24x-8\)
\(x^2-2.12x+144-152\)
\(\Leftrightarrow\left(x-12\right)^2-152\)
\(\Leftrightarrow\left(x-12-\sqrt{152}\right)\left(x-12+\sqrt{152}\right)\)
P/s tham khảo nha
tên của mày nói lên tất cả , ko cần bít đề đúng hay sai , tao cứ đăng thôi :))
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2.
a) 4x(x-1)-6x+6
= 4x(x-1)-6(x-1)
= (4x-6)(x-1)
3.
a) 6x2-24x=0
6x(x-4)=0
TH1: 6x=0 TH2: x-4=0
x=0 x=4
Vậy x\(\in\){0;4}
2. a. \(4x\left(x-1\right)-6x+6\)
\(=4x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(4x-6\right)\left(x-1\right)\)
3. a. \(6x^2-24x=0\)
\(\Leftrightarrow6x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-3\right)\left(x-10\right)\left(x-5\right)\left(x-6\right)-24x^2\)
\(=\left(x^2+30-13x\right)\left(x^2+30-11x\right)-24x^2\)
\(=\left(x^2+30x-12x-x\right)\left(x^2+30x-12x+x\right)-24x^2\)
\(=\left(x^2+30-12x\right)^2-x^2-24x^2\)
\(=\left(x^2-12x+30\right)^2-\left(5x\right)^2\)
\(=\left(x^2-12x+30+5x\right)\left(x^2-12x+30-5x\right)\)
\(=\left(x^2-7x+30\right)\left(x^2-17x+30\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Dễ thì tự mà làm đi ??? ( Đừng có ném gạch )
==
#Thiên_Hy
===
\(3x^2+48+24x-12y^2\)
\(=-3\left(-x^2-16-8x+4y^2\right)\)
\(=-3\left[\left(2y\right)^2-\left(x^2+8x+16\right)\right]\)
\(=-3\left[\left(2y\right)^2-\left(x+4\right)^2\right]\)
\(=-3\left(2y-x-4\right)\left(2y+x+4\right)\)