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Đặt \(x^2+3x+1=t\)
\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)
\(=t^2+t-6=\left(t^2-2t\right)+\left(3t-6\right)\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1=a\)ta có :
\(a\left(a+1\right)-6\)
\(=a^2+a-6\)
\(=a^2+6a-a-6\)
\(=\left(a^2+6a\right)-\left(a+6\right)\)
\(=a\left(a+6\right)-\left(a+6\right)\)
\(=\left(a+6\right)\left(a-1\right)\)
Thay \(a=x^2+3x+1\)vào A ta có :
\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)
\(=\left(x^2+3x+7\right)\left(x^2+3x\right)\)
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\(x^3-3x^2+6x-4\)
\(=x^3-2x^2+4x-x^2+2x-4\)
\(=\left(x^3-2x^2+4x\right)-\left(x^2-2x+4\right)\)
\(=x\left(x^2-2x+4\right)-\left(x^2-2x+4\right)\)
\(=\left(x-1\right)\left(x^2-2x+4\right)\)
x^3 - 3x^2 + 6x - 4
<=> x^3-3x^2+3x-1+3x-3
<=>(x-1)^3+3(x-1)
<=>(x-1)+((x-1)^2+3)
<=>(x-1)+(x^2-2x+4)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-3x^2+3x-1-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
\(x^3-3x^2+3x-1-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
Rất vui vì giúp đc bạn <3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-3x^2+3x-1-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
\(x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3\\ =\left(x-1-y\right)\text{[ (x-1)^2+y(x-1)+y^2}\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^2 - 7xy + 10y^2
= (x^2 - 2xy) - (5xy - 10y^2)
= x(x - 2y) - 5y( x - 2y)
= (x - 5y)(x - 2y)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3+3x^2-3x-1=\left(x^3-1\right)+\left(3x^2-3x\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left[\left(x^2+x+1\right)+3x\right]=\left(x-1\right)\left(x^2+4x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^3-3x^2+3x-1 = (x^3+1)-(3x^2-3x)
(Mình sẽ có hằng đẳng thức x^3+1 cũng giống như x^3+1^3 vì 1^3=1 nhé )
= ( x^3+1^3)- (3x^2-3x )
=(x-1)* (x^2+ x*1 + 1^2) -( 3x^2-3x)( Chuyển sang hằng đăng thức )
=(x-1 ) *(x^2+ x + 1 ) - 3x(x+1)
=(x-1)*(x^2+x+1-3x )
CÓ MỘT BƯỚC LÀ VÌ DẤU TRỪ Ở TRƯỚC NÊN ĐỔI X+1 THÀNH X-1 NHÉ
Nếu đúng k dùm minha j , cảm ơn
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\(x^3+y^3-3x^2+3x-1\\=(x^3-3x^2+3x-1)+y^3\\=(x-1)^3+y^3\\=(x-1+y)[(x-1)^2-(x-1)y+y^2]\\=(x+y-1)(x^2-2x+1-xy+y+y^2)\)
\(x^3-3x^2+1-3x\)
\(=\left(x^3+1\right)-\left(3x^2+3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
Chúc bạn học tốt.
\(x^3-3x^2+1-3x\)
\(=\left(x^3+1\right)-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3\left(x^2+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3\left(x+1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left[x^2-x+1-3\left(x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2-x+1-3x-3\right)\)
\(=\left(x+1\right)\left(x^2-4x-2\right)\)