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NH
2
3 tháng 8 2015
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
TL
1
23 tháng 9 2019
a) x^7+x^2 +1 =x^7 - x^4+x^4 +x^2+1
= (x^7 - x^4) +[ (x^2)^2 +x^2 +1]
= x^4(x^3 -1)+(x^2 - 1)
= x^4 ( x-1)(x^2 +x +1)+ (x-1)(x+1)
= (x-1)[ x^4( x^2+x+1)+(x+1)]
= (x-1)(x^6 +x^5+x^4+x+1)
b) x^8 +x+1 = x^8 -x^2+x^2 +x+1
= (x^8-x^2) +(x^2 +x+1)
=x^2(x^6 -1) +(x^2+x+1)
=x^2[ (x^3)^2 -1)+(x^2+x+1)
= x^2 (x^3-1)(x^3+1) +(x^2 +x+1)
= x^2(x-1)(x^2+x+1)(x^3+1) +(x^2 +x+1)
= (x^2+x+1)[ x^2(x-1)(x^3+1) +1]
ND
1
28 tháng 1 2019
\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)
\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)
\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)
Đặt \(q=x^2+6x-7\)ta có :
\(A=q\left(q-9\right)+8\)
\(A=q^2-9q+8\)
\(A=q^2-q-8q+8\)
\(A=q\left(q-1\right)-8\left(q-1\right)\)
\(A=\left(q-1\right)\left(q-8\right)\)
Thay \(q=x^2+6x-7\)vào A ta được :
\(A=\left(x^2+6x-7-1\right)\left(x^2+6x-7-8\right)\)
\(A=\left(x^2+6x-8\right)\left(x^2+6x-15\right)\)
SH
0
TM
4
NC
17 tháng 8 2020
a) \(x^5-x^4-1\)
\(=\left(x^5+x^2\right)-\left(x^4+x\right)-\left(x^2-x+1\right)\)
\(=x^2\left(x^3+1\right)-x\left(x^3+1\right)-\left(x^2-x+1\right)\)
\(=x^2\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3+x^2-x^2-x-1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)
NC
17 tháng 8 2020
b) \(x^8+x^7+1\)
\(=\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^6-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^3-1\right)\left(x^3+1\right)+x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[\left(x^3-x^2\right)\left(x^3+1\right)+\left(x^2-x\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left[\left(x^3-x\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
24 tháng 12 2016
đề sai nha bạn
mình sửa đề cho:
\(A=\left(x+1\right)\left(x+2\right)\left(x+7\right)\left(x+8\right)+8\)
\(A=\left(x+1\right)\left(x+8\right)\left(x+2\right)\left(x+7\right)+8\)
\(A=\left(x^2+9x+8\right)\left(x^2+9x+14\right)+8\)
Đặt \(x^2+9x+8=a\)
\(\Rightarrow A=a\left(a+6\right)+8=a^2+6a+8=\left(a+2\right)\left(a+4\right)\)
\(\Rightarrow A=\left(x^2+9x+8+2\right)\left(x^2+9x+8+4\right)=\left(x^2+9x+10\right)\left(x^2+9x+12\right)\)
a.
\(x^7+x^2+1=\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x\left(x^6-1\right)+\left(x^2+x+1\right)=x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^3+1\right)\right]\)
\(=\left(x^2+x+1\right)\left[\left(x^2-x\right)\left(x^3+1\right)\right]\)
\(=\left(x^2+x+1\right)\left(x^5+x^2-x^4-x\right)\)
b.
\(x^8+x+1=\left(x^8-x^5\right)+\left(x^5-x^2\right)+\left(x^2+x+1\right)\)
\(=x^5\left(x^3-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^5\left(x-1\right)\left(x^2+x+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x^5\left(x-1\right)+x^2\left(x-1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left[\left(x-1\right)x^2\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left[\left(x^3-x^2\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
a) \(x^7+x^2+1=x^7+x^6+x^5-x^5+x^4-x^4+x^3-x^3+2x^2\)\(-x^2+x-x+1\)
\(=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)\)\(-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)\)\(-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^5-x^4+x^2-x+1\right)\left(x^2+x+1\right)\)
b) \(x^8+x+1=x^8-x^2+\left(x^2+x+1\right)=x^2\left(x^6-1\right)\)\(+\left(x^2+x+1\right)\)
\(=x^2\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x^2\left(x^3+1\right)\left(x-1\right)+1\right]\)