(x+1)(x+2)(x+3)(x+4)-120 phân tích đa thức thành nhân tử
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HT
1
1 tháng 12 2021
=[(x+1)(x+4)][(x+2)(x+3)]-120
=(x2+5x+4)(x2+5x+6)-120 (1)
Đặt x2+5x+5=t ta có
(1)<=>(t-1)(t+1)-120
=t2-1-120
=t2-121
=(t-11)(t+11) (2)
Thay t=x2+5x+5 vào(2) ta có
(2)<=> (x2+5x+5 -11)(x2+5x+5 +11)=(x2+5x-6)(x2+5x+16)
thấy đúng thì tk hộ mình nha,chúc bạn học tốt
HT
1
LT
1
25 tháng 6 2016
(x-1) (x-2) (x+4) (x+5)-120
=x4+11-120
Nếu sai thì các bạn sửa cho mình,đừng ném đá
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
HN
0
LM
1
8 tháng 2 2019
\(x^4-14x^3+71x^2-154x+120\)
\(=x^4-2x^3-12x^3+24x^2+47x^2-94x-60x+120\)
\(=x^3\left(x-2\right)-12x^2\left(x-2\right)+47x\left(x-2\right)-60\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-12x^2+47x-60\right)\)
\(=\left(x-2\right)\left(x^3-3x^2-9x^2+27x+20x-60\right)\)
\(=\left(x-2\right)\left[x^2\left(x-3\right)-9x\left(x-3\right)+20\left(x-3\right)\right]\)
\(=\left(x-2\right)\left(x-3\right)\left(x^2-9x+20\right)\)
\(=\left(x-2\right)\left(x-3\right)\left(x^2-4x-5x+20\right)\)
\(=\left(x-2\right)\left(x-3\right)\left[x\left(x-4\right)-5\left(x-4\right)\right]\)
\(=\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)\)
6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
TN
2
1 tháng 7 2021
Ta có : \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(\left(x+1\right)\left(x+4\right)\right)\left(\left(x+2\right)\left(x+3\right)\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
- Đặt \(x^2+5x+5=a\)
\(=\left(a-1\right)\left(a+1\right)-24=a^2-1-24=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)-96\)
\(=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
\(=\left(x^2+5x+16\right)\left(x+6\right)\left(x-1\right)\)
bạn có thể giải thích kĩ hơn ko