Phân tích thành nhân tử (x+1).(x+2).(x+3).(x+4)-4
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12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
T
16 tháng 9 2023
Mình bổ sung nhé:
\(=\left(x+1\right)\left(x^4+x^3+x^2-x^3+1\right)\)
\(=\left(x+1\right)\left[x^2\left(x^2+x+1\right)-\left(x^3-1\right)\right]\)
\(=\left(x+1\right)\left[x^2\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
16 tháng 9 2023
=x^3(x^2+x+1)+(x^2+x+1)
=(x^2+x+1)(x^3+1)
=(x^2+x+1)(x+1)(x^2-x+1)
26 tháng 8 2023
a: =4(x-2)(x+1)+4(x-2)^2+(x+1)^2
=(2x-4)^2+2*(2x-4)(x+1)+(x+1)^2
=(2x-4+x+1)^2=(3x-3)^2=9(x-1)^2
b: =x^7(x^2-1)-x^5(x+1)+x^3(x+1)+(x^2-1)
=(x+1)[x^7(x-1)-x^5+x^3+x-1]
=(x+1)[x^7(x-1)-x^3(x-1)(x+1)+(x-1)]
=(x+1)(x-1)(x^7-x^4-x^3+1)
=(x+1)(x-1)(x^3-1)(x^4-1)
=(x+1)(x-1)^2*(x^2+x+1)(x^2+1)(x-1)(x+1)
=(x+1)^2*(x-1)^3*(x^2+1)(x^2+x+1)
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 \(x^2+5x+5=a\)
\(=\left(a-1\right)\left(a+1\right)-24=a^2-1-24=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
HT
1
30 tháng 11 2021
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)-96\)
\(=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
\(=\left(x^2+5x+16\right)\left(x+6\right)\left(x-1\right)\)
T
20 tháng 8 2023
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\) (sửa đề)
\(=\left[\left(x+1\right)\left(x+4\right)\right].\left[\left(x+2\right).\left(x+3\right)\right]-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt \(y=x^2+5x+4\), thay vào đa thức, ta được:
\(y\left(y+2\right)-24\)
\(=y^2+2y-24\)
\(=\left(y^2+2y+1\right)-25\)
\(=\left(y+1\right)^2-5^2\)
\(=\left(y+1-5\right)\left(y+1+5\right)\)
\(=\left(y-4\right)\left(y+6\right)\)
\(=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
(x+1).(x+2).(x+3).(x+4)-4
=(x+1)(x+4)(x+2)(x+3)-4
=(x2+5x+4)(x2+5x+6)-4
Đặt t=x2+5x+4 ta được:
t.(t+2)-4
=t2+2t-4
Vẫn sai đề