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a)lát đề
b)x3+x+2
=x3-x2+2x+x2-x+2
=x(x2-x+2)+(x2-x+2)
=(x+1)(x2-x+2)
c)x4+64
=(x2)2+82+2x2*8-2x2*8
=(x2+8)2-(4x)2
=(x2-4x+8)(x2+4x+8) .
![](https://rs.olm.vn/images/avt/0.png?1311)
1) x2 - 4x + 3
= x2 - x - 3x + 3
= (x2 - x) - (3x - 3)
= x.(x - 1) - 3.(x - 1)
= (x - 1).(x - 3)
2) x2 - x - 6
= x2 + 2x - 3x - 6
= (x2 + 2x) - (3x + 6)
= x.(x + 2) - 3.(x + 2)
= (x + 2).(x - 3)
3) x2 + 5x + 4
= x2 + x + 4x + x
= (x2 + x) + (4x + x)
= x.(x + 1) + 4.(x + 1)
= (x + 1).(x + 4)
4) x2 + 5x + 6
= x2 + 2x + 3x + 6
= (x2 + 2x) + (3x + 6)
= x.(x + 2) + 3.(x + 2)
= (x + 2).(x + 3)
a,=x^2+x+3x+3
=x(x+1)+3(x+1)
=(x+3)(x+1)
b,x^2-3x+2x-6
=x(x-3)+2(x-3)
=(x+2)(x-3)
2 câu còn lại từ lm nha.........
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt H \(=x^4-5x^3+7x^2-6\)
Gỉa sử : \(H=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)
\(=x^4+cx^3+dx^2+ax^{3\:}+acx^2+adx+bx^2+bcx+bd\)
\(=x^4+\left(a+c\right)x^3+\left(ac+b+d\right)x^2+\left(ad+bc\right)x+bd\)
\(\Leftrightarrow\hept{\begin{cases}a+c=-5\\ac+b+d=7\\ad+bc=0\end{cases}}\)
\(\left\{bd=6\right\}\)
\(\Leftrightarrow\hept{\begin{cases}a=-3\\b=3\\c=-2\end{cases}}\)
\(\left\{d=-2\right\}\)
\(\Rightarrow H=\left(x^2-3x+3\right)\left(x^2-2x-2\right)\)
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2+x\right)^2-2x^2-2x-15\)
\(=\left(x^2+x\right)^2-\left(2x^2+2x+15\right)\)
\(=\left(x^2+x\right)^2-\left[\left(2x^2+2x\right)+15\right]\)
\(=\left(x^2+x\right)^2-\left[2.\left(x^2+x\right)+15\right]\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\) \(\left(1\right)\)
đặt \(x^2+x=t\)
\(\left(1\right)\)\(=\) \(t^2-2t-15\)
\(=\left(t-1\right)^2-16\)
\(=\left(t-1-4\right)\left(t-1+4\right)\)
\(=\left(t-5\right)\left(t+3\right)\)
thay \(t=x^2+x\) ta có
\(\left(1\right)=\left(x^2+x-5\right)\left(x^2+x+3\right)\)
các câu còn lại tương tự nha
học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(Q\left(x\right)=x^4-x^3-10x^2+2x+4\)
Giả sử nhân tử khi phân tích P(x) là \(P\left(x\right)=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)
Khai triển : \(P\left(x\right)=x^4+cx^3+dx^2+ax^3+acx^2+adx+bx^2+bcx+bd\)
\(=x^4+x^3\left(c+a\right)+x^2\left(d+ac+b\right)+x\left(ad+bc\right)+bd\)
Áp dụng hệ số bất định : \(\begin{cases}c+a=-1\\d+ac+b=-10\\ad+bc=2\\bd=4\end{cases}\) . Giải ra được \(\begin{cases}a=-3\\b=-2\\c=2\\d=-2\end{cases}\)
Vậy \(P\left(x\right)=\left(x^2-3x-2\right)\left(x^2+2x-2\right)\)
Giả sử:
\(P\left(x\right)=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)
\(=x^4+cx^3+dx^2+ax^3+acx^2+adx+bx^2+bcx+bd\)
\(=x^4+\left(a+c\right)x^3+\left(d+ac+b\right)x^2+\left(ad+bc\right)x+bd\)
Ta có:
\(\begin{cases}a+c=-1\\d+ac+b=-10\\ad+bc=2\\bd=4\end{cases}\) \(\Rightarrow\begin{cases}a=1\\b=1\\d=4\\c=-15\end{cases}\)
\(\Rightarrow P\left(x\right)=\left(x^2+x+1\right)\left(x^2-15x+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-4x+3=x^2-3x-x+3=x\left(x-3\right)-\left(x-3\right)=\left(x-1\right)\left(x-3\right)\)
\(x^2+5x+4=x^2+4x+x+4=x\left(x+4\right)+\left(x+4\right)=\left(x+1\right)\left(x+4\right)\)
\(x^2-x-6=x^2-3x+2x-6=x\left(x-3\right)+2\left(x-3\right)=\left(x+2\right)\left(x-3\right)\)
\(x^4+4=\left(x^2\right)^2+2.x^2.2+2^2-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2\right)^2-\left(2x^2\right)=\left(x^2+2+2x\right)\left(x^2-2-2x\right)\)
d= x(x3 +x2 +x +5 ) -2