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![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(=\left(\dfrac{1}{2}-2x\right)\left(\dfrac{1}{4}+x+4x^2\right)\)
c: \(=\left(x-\dfrac{1}{2}\right)^2\)
d: \(=\left(x+\dfrac{1}{2}\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(1.24^2-0.24^2\\ =\left(\dfrac{31}{25}\right)^2-\left(\dfrac{6}{25}\right)^2\\ =\left(\dfrac{31}{25}-\dfrac{6}{25}\right)\left(\dfrac{31}{25}+\dfrac{6}{25}\right)=\dfrac{37}{25}\)
b)
\(\dfrac{1}{8}-8x^3\\ =\left(\dfrac{1}{2}\right)^3-\left(2x\right)^3\\ =\left(\dfrac{1}{2}-2x\right)\left(\dfrac{1}{4}+2x+4x^2\right)\)
c)
\(x^2-x+\dfrac{1}{4}\\ =x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\\ =\left(x-\dfrac{1}{2}\right)^2\)
d)
\(x^2+x+\dfrac{1}{4}\\ =x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\\ =\left(x+\dfrac{1}{2}\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
viết biểu thức sau dưới dạng tích
m^2-n^2
=(m-n)(m+n)
Hang dang thuc so 3
![](https://rs.olm.vn/images/avt/0.png?1311)
`9x^2+4y^2-12xy+6x-4y+1`
`=(3x)^2-2.3x.2y+(2y)^2+2(3x-2y)+1`
`=(3x-2y)^2+2(3x-2y)+1`
`=(3x-2y+1)^2`
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(81+\left(9-x^2\right)^2\)
Tương tự: \(81+-x^4+18x^2+-81\)
\(=\left(-x^4\right)+\left(18x^2\right)+\left(81+-81\right)\)
\(=-x^4+18x^2\)
\(81-\left(9-x^2\right)^2\)
\(=\)\(9^2-\left(3^2-x^2\right)^2\)
\(=\)\(9^2-\left(9-x^2\right)^2\)
\(=\)\(\left(9-9+x\right)\left(9+9-x\right)\)
\(=\)\(x\left(18-x\right)\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
1.24 x 1.24 - 0.24 x 0.24
Đúng nhé , tk nhé , mơn nhìu !!!
~ HOK TỐT ~
Ta có :
\(1,24^2-0,24^2=\left(1,24+0,24\right).\left(1,24-0,24\right)\)
~