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\(8x^2+18x-5 \)
\(=8x^2-2x+20x-5\)
\(=2x\left(4x-1\right)+5\left(4x-1\right)\)
\(=\left(4x-1\right)\left(2x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(12x^3+8x^2-3x-2=4x^2\left(3x+2\right)-\left(3x+2\right)\)
\(=\left(3x+2\right)\left(4x^2-1\right)=\left(3x+2\right)\left(2x-1\right)\left(2x+1\right)\)
b) \(18x^3+27x^2-2x-3=9x^2\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(9x^2-1\right)=\left(2x+3\right)\left(3x-1\right)\left(3x+1\right)\)
c) \(8x^3+4x^2-34x+15=4x^2\left(2x-3\right)+8x\left(2x-3\right)-5\left(2x-3\right)\)
\(=\left(2x-3\right)\left(4x^2+8x-5\right)=\left(2x-3\right)\left(2x-1\right)\left(2x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(18x^2-36xy+18x^2-72z^2\)
\(=36x^2-36xy-72z^2\)
\(=36\left(x^2-xy-2z^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nếu thế how to phân tích cái 2x^2 =)))) đề sai
2x^2 phân tích như thế nào để đặt nhân tử chung hoặc nhóm hạng tử đây???? Xem lại đề bạn ơi
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\(4x^2-4x-35\) \(=\left(2x\right)^2-2.2x.1+1-36\)
\(=\left(2x-1\right)^2-6^2\)
\(=\left(2x-7\right)\left(2x+5\right)\)
\(18x^2-5x-2\) \(=\left(x-\frac{1}{2}\right)\left(x+\frac{2}{9}\right)\)
\(8x^3-26x^2+13x+5=\) \(8x^3-8x^2-18x^2+18x-5x+5\)
\(=8x^2\left(x-1\right)-18x\left(x-1\right)-5\left(x-1\right)\)
\(=\) \(\left(8x^2-18x-5\right)\left(x-1\right)\)
\(=\left(x-\frac{5}{2}\right)\left(x+\frac{1}{4}\right)\)\(\left(x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)
\(=9x^2.\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^2+3x-27=2x^2-6x+9x-27=2x\left(x-3\right)+9\left(x-3\right)=\left(2x+9\right)\left(x-3\right)\)
\(x^3-7x+6=x^3-x-6x+6=x\left(x^2-1\right)-6\left(x-1\right)=x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=\left(x-1\right)\left(x^2+x-6\right)\)
\(x^3+5x^2+8x+4=x^3+x^2+4x^2+8x+4=x^2\left(x+1\right)+4\left(x^2+2x+1\right)=x^2\left(x+1\right)+4\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^2+4x+4\right)=\left(x+1\right)\left(x+2\right)^2\)
\(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(27x^3-27x^2+18x-4\)
\(=27x^3-18x^2+12x-9x^2+6x-4\)
\(=3x\left(9x^2-6x+4\right)-\left(9x^2-6x+4\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
27x3 - 27x2 + 18x - 4
= 27x3 - 9x2 - 18x2 + 6x + 12x - 4
= 9x2 ( 3x - 1 ) - 6x ( 3x - 1 ) + 4 ( 3x - 1 )
= ( 9x2 - 6x + 4 ) ( 3x - 1 )
Trả lời:
8x3 - 24x2 + 18x
= 2x ( 4x2 - 12x + 9 )
= 2x [ (2x)2 - 2.2x.3 + 32 ]
= 2x ( 2x - 3 )2
mk cop mạng nha
2𝑥⋅(2𝑥−3)2
hok tốt