phân tích đa thức a^4+a^3+a^2+a thành nhân tử
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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(a^4-a^3-a^2+a\)
\(=a^3\left(a-1\right)-a\left(a-1\right)\)
\(=\left(a-1\right)\left(a^3-a\right)\)
Ta có: \(\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)-3\)
\(=\left(a+1\right)\left(a+4\right)\left(a+2\right)\left(a+3\right)-3\)
\(=\left(a^2+5a+4\right)\left(a^2+5a+6\right)-3\)
\(=\left(a^2+5a+5\right)^2-1-3\)
\(=\left(a^2+5a+5\right)^2-4\)
\(=\left(a^2+5a+7\right)\left(a^2+5a+3\right)\)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
ta có a^4 + a^2 + 1 = a ^ 4 + 2a^2 + 1 - a^2 = (a^2+1)^2 - a^2 = (a^2- a + 1)(a^2 + a + 1)
\(a^4+a^2+1\)
\(=a^4+2a^2+1-a^2\)
\(=\left(a^2+1\right)^2-a^2\)
\(=\left(a^2+1-a\right)\left(a^2+a+1\right)\)
\(x^3+x^2+4\)
\(=x^3+2x^2-x^2+4\)
\(=x^2\left(x+2\right)-\left(x^2-4\right)\)
\(=x^2\left(x+2\right)-\left(x+2\right)\left(x-2\right)\)
\(=\left(x+2\right)\left(x^2-x+2\right)\)
x3-2x-4
=x3-4x+2x-4
=x(x2-4)+2(x-2)
=x(x-2)(x+2)+2(x-2)
=(x^2+2x+2)(x-2)
\(a)\left(x^2+2x\right)\left(x^2+2x+4\right)+3\)
Để đơn giản hơn cũng như là dễ nhìn hơn thì ta :
Đặt : \(x^2+2x=a\)
Do đó ta có đa thức :
\(a.\left(a+4\right)+3=a^2+4a+3\)
\(=a^2+a+3a+3\)
\(=a\left(a+1\right)+3\left(a+1\right)\)
\(=\left(a+1\right)\left(a+3\right)\)
\(=\left(x^2+2x+1\right)\left(x^2+2x+3\right)\)
\(=\left(x+1\right)^2.\left(x^2+2x+3\right)\)
Hoặc bạn có thể đặt \(x^2+2x+2=t\)
Thì \(P=\left(x^2+2x\right)\left(x^2+2x+4\right)+3\)
\(P=\left(t-2\right)\left(t+2\right)+3\)
\(P=t^2-4+3\)
\(P=t^2-1\)
\(P=\left(t-1\right)\left(t+1\right)\)
\(P=\left(x^2+2x+1\right)\left(x^2+2x+3\right)\)
\(P=\left(x+1\right)^2\left(x^2+2x+3\right)\)
\(a^4+a^3+a^2+a\)
\(=a^3\left(a+1\right)+a\left(a+1\right)\)
\(=\left(a+1\right)\left(a^3+a\right)\)
nha !!!
Trả lời:
\(a^4+a^3+a^2+a\)
\(=\left(a^4+a^3\right)+\left(a^2+a\right)\)
\(=a^3\left(a+1\right)+a\left(a+1\right)\)
\(=a\left(a+1\right)\left(a^2+1\right)\)