phân tích đa thức thành nhân tử(x²+x+4)² +8x(x²+x+1)+15x²
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24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
KC
1
BQ
0
29 tháng 11 2017
x4+8x3+15x2-4x-2
= (x4+4x3+x2)+(4x3+16x2+4x)-(2x2+8x+2)
= x2.(x2+4x+1)+4x.(x2+4x+1) -2(x2+4x+1)
= (x2+4x+1).(x2+4x-2)
NB
2
BM
28 tháng 7 2017
a, \(\left(x^2-2x\right)\left(x^2-2x-1\right)-6\)
Đặt \(x^2-2x=a\)
Thay vào biểu thức ta đc:
\(a.\left(a-1\right)-6=a^2-a-6\) \(=a^2-3a+2a-6=a\left(a-3\right)+2\left(a-3\right)\)
\(=\left(a-3\right).\left(a+2\right)\)
\(\Rightarrow\left(x^2-2x\right)\left(x^2-2x-1\right)-6=\left(x^2-2x-3\right)\left(x^2-2x+2\right)\)
b, \(\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)
\(=\left[\left(x^2+x+4\right)^2+6x\left(x^2+x+4\right)+9x^2\right]+\left[2x\left(x^2+x+4\right)+6x^2\right]\)
\(=\left(x^2+x+4+3x\right)^2+2x\left(3x+x^2+x+4\right)\)
\(=\left(x^2+4x+4\right)\left(x^2+4x+4+2x\right)\) \(=\left(x+2\right)^2\left(x^2+6x+4\right)\)
BT
2
29 tháng 7 2016
1/ \(\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2=x^4+10x^3+32x^2+40x+16\)(làm tắt nhưng chắc bạn tự hiểu đc)
\(=\left(x^4+2x^3\right)+\left(4x^2+2x^3\right)+\left(12x^2+6x^3\right)+\left(4x^2+8x\right)+\left(12x^2+24x\right)+\left(8x+16\right)\)
\(=x^3\left(x+2\right)+2x^2\left(2+x\right)+6x^2\left(2+x\right)+4x\left(x+2\right)+12x\left(x+2\right)+8\left(x+2\right)\)
\(=\left(x+2\right)\left(x^3+2x^2+6x^2+4x+12x+8\right)=\left(x+2\right)\left(x^3+8x^2+16x+8\right)\)
\(=\left(x+2\right)\left[\left(x^3+2x^2\right)+\left(6x^2+12x\right)+\left(4x+8\right)\right]=\left(x+2\right)\left[x^2\left(x+2\right)+6x\left(x+2\right)+4\left(x+2\right)\right]\)
\(=\left(x+2\right)\left(x+2\right)\left(x^2+6x+4\right)\)
29 tháng 7 2016
2/ \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=x^4+20x^3+140x^2+400x+400\)
\(=\left(x^4+10x^3+20x^2\right)+\left(10x^3+100x^2+200x\right)+\left(20x^2+200x+400\right)\)
\(=x^2\left(x^2+10x+20\right)+10x\left(x^2+10x+20\right)+20\left(x^2+10x+20\right)\)
\(=\left(x^2+10x+20\right)\left(x^2+10x+20\right)=\left(x^2+10x+20\right)^2\)
\(=\left(x^2+x+4\right)^2+3x\left(x^2+x+4\right)+5x\left(x^2+x+4\right)+15x^2\\ =\left(x^2+x+4\right)\left(x^2+x+4+3x\right)+5x\left(x^2+x+4+3x\right)\\ =\left(x^2+x+4+3x\right)\left(x^2+x+4+5x\right)\\ =\left(x^2+4x+4\right)\left(x^2+6x+4\right)\\ =\left(x+2\right)^2\left(x^2+6x+4\right)\)
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