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Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
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=[(x+1)(x+4)][(x+2)(x+3)]+8=(x2+5x+4)(x2+5x+6)+8
Đặt x2+5x+4=t
Ta có : t(t+2)+8=t2+2t-8=(t-2)(t+4)
k mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-8\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-8\)
Đặt \(x^2+5x+5=t\)
Khi đó: \(A=\left(t-1\right)\left(t+1\right)-8\)
\(=t^2-9=\left(t-3\right)\left(t+3\right)\)
\(=\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)
Chúc bạn học tốt.
A=\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-8\)
A=\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-8\)
A=\(\left(x^2+5x +4\right)\left(x^2+5x+6\right)-8\)
Đặt \(x^2+5x+4=x\)ta có:
x(x+2)-8=\(x^2+2x-8\)=\(\left(x+1\right)^2-9\)=(x+1-3)(x+1+3)=(x-2)(x+4)=\(\left(x^2+5x+4-2\right)\left(x^2+5x+4+4\right)\)=\(\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(27x^3-\frac{1}{8}=\left(3x-\frac{1}{2}\right)\left(9x^2+\frac{3}{2}x+\frac{1}{4}\right)\)
(Nếu đúng thì click cho tớ với nhá!)
\(27x^3-\frac{1}{8}\)
\(=\left(3x\right)^3-\left(\frac{1}{2}\right)^3\)
\(=\left(3x-\frac{1}{2}\right)\left(9x^2+3x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right)\)
\(=\left(3x-\frac{1}{2}\right)\left(9x^2+\frac{3}{2}x+\frac{1}{4}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^8+x+1\)
\(=x^8-x^7+x^5-x^4+x^2+x^7-x^6+x^4-x^3+x+x^6-x^5+x^3-x^2+1\)
\(=\left(x^8-x^7+x^5-x^4+x^2\right)+\left(x^7-x^6+x^4-x^3+x\right)+\left(x^6-x^5+x^3-x^2+1\right)\)
\(=x^2\left(x^6-x^5+x^3-x^2+1\right)+x\left(x^6-x^5+x^3-x^2+1\right)+\left(x^6-x^5+x^3-x^2+1\right)\)
\(=\left(x^6-x^5+x^3-x^2+1\right)\left(x^2+x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^9 + x^8 + x^7 - x^3 + 1
= x^7 ( x^2 + x + 1 ) - ( x^3 - 1 )
= x^7 ( x^2 + x + 1 ) - ( x - 1 )(x^2 + x + 1 )
= ( x^7 - x + 1 )(x^2 + x + 1 )
![](https://rs.olm.vn/images/avt/0.png?1311)
(x+1)(x+2)(x+3)(x+4)-8
=[(x+1).(x+4)].[(x+2).(x+3)]-8
=(x2+5x+4).(x2+5x+6)-8
Đặt (x2+5x+4)=t =>(x2+5x+6)=t+2
Thay vào biểu thức ta có:
(x2+5x+4).(x2+5x+6)-8
t.(t+2)-8
=t2+2t+1-9
=(t+1)2-32
=(x2+5x+4+1)-32
=(x2+5x+5+3).(x2+5x+5-3)
=(x2+5x+8).(x2+5x+2)
=
ta làm như sau :
\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-8.\)
\(\Rightarrow\left(x^2+5X+4\right)\left(x^2+5x+6\right)-8\)
Đặt \(x^2+5x+4=t\)
\(\Leftrightarrow t\left(t+2\right)-8\)
\(\Leftrightarrow t^2+2t-8\Leftrightarrow t^2+2t+1-9\)
\(\Leftrightarrow\left(t+1\right)^2-3^2\)
\(\Leftrightarrow\left(t-2\right)\left(t+4\right)\)
\(\Leftrightarrow\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)
\(x^3-\left(\frac{1}{2}\right)^3\)
\(=\left(x-\frac{1}{2}\right)\left(x^2+x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right)\)
\(=\left(x-\frac{1}{2}\right)\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)\)