Phân tích đa thức thành nhân tử bằng 2 cách x2+5x+4
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x2 + 5x + 4
= x2 + x + 4x + 4
(Tách 5x = x + 4x)
= x(x + 1) + 4(x + 1)
(có x + 1 là nhân tử chung)
= (x + 1)(x + 4)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+xy-5x-5y=\left(x^2+xy\right)-\left(5x+5y\right)\)
\(=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\)
x(x+y)-5(x+y) = (x-5)(x+y) bài này mik hok qua rùi mik đang ôn tập nên là mik tự làm chứ ko chép mấy câu trả lời khác
![](https://rs.olm.vn/images/avt/0.png?1311)
-Đặt \(t=\left(x^2-x+1\right)\)
\(\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2\)
\(=t^2-5xt+4x^2\)
\(=t^2-4xt-xt+4x^2\)
\(=t\left(t-4x\right)-x\left(t-4x\right)\)
\(=\left(t-4x\right)\left(t-x\right)\)
\(=\left(x^2-x+1-4x\right)\left(x^2-x+1-x\right)\)
\(=\left(x^2-5x+1\right)\left(x^2-2x +1\right)\)
\(=\left(x^2-5x+1\right)\left(x-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
8: \(=\left(x-2y\right)\cdot x\cdot\left(x+3\right)\)
9: \(=\left(5x+2\right)\left(x-3\right)-x\left(x-3\right)\)
\(=\left(x-3\right)\left(4x+2\right)\)
=2(2x+1)(x-3)
3: \(=2\left(x+2\right)\left(25x-15-x\right)\)
\(=2\left(x+2\right)\left(24x-15\right)\)
=6(x+2)(8x-5)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x2-xz-9y2+3yz
=(x2-9y2)-(xz-3yz)
=(x-3y)(x+3y)-z(x-3y)
=(x-3y)(x+3y-z)
b)x3-x2-5x+125
=x3-6x2+25x+5x2-30x+125
=x(x2-6x+25)+5(x2-6x+25)
=(x+5)(x2-6x+25)
c.x3+2x2-6x-27
=x3+5x2+9x-3x2-15x-27
=x(x2+5x+9)-3(x2+5x+9)
=(x-3)(x2+5x+9)
d. 12x3+4x2-27x-9
=12x3+4x2-27x-9
=4x2(3x+1)-9(3x+1)
=(4x2-9)(3x+1)
=(2x-3)(2x+3)(3x+1)
e.x4-25x2+20x-4
=x4+5x3-2x2-5x2-25x+10+2x2+10x-4
=x2(x2+5x-2)-5(x2+5x-2)+2(x2+5x-2)
=(x2-5x+2)(x2+5x-2)
f.x2(x2-6)-x2+9
=x4+x3-3x2-x3-x2+3x-3x2-3x+9
=x2(x2+x-3)-x(x2+x-3)-3(x2+x-3)
=(x2-x-3)(x2+x-3)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^3-5x^2+8x-3\)
\(\Leftrightarrow2x^3-x^2-4x^2+2x+6x+3\)
\(\Leftrightarrow x^2\cdot\left(2x-1\right)-2x\cdot\left(2x-1\right)+3\cdot\left(2x+1\right)\)
\(\Leftrightarrow\left(2x-1\right)\cdot\left(x^2-2x+3\right)\)
\(2x^3-5x^2+8x-3\)
\(=2x^3-x^2-4x^2+2x+6x-3\)
\(=x^2\left(2x-1\right)-2x\left(2x-1\right)+3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x^2-2x+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x2+5x+4=x2+4x+x+4=x(x+4)+(x+4)=(x+4)(x+1)
x2+5x+4=x2+2.x.5/2+25/4-9/4=(x+5/2)2-9/4=(x+5/2+3/2)(x+5/2-3/2)=(x+4)(x+1)