phân tích đa thức thành nhân tử
x4+4
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`x^3 - 5x^2 + 8x + 4`
`= x^3 + x^2 + 4x^2 + 4x + 4x + 4`
`= x^2(x + 1) + 4x(x + 1) + 4(x + 1)`
`= (x + 1)(x^2 + 4x 4)`
`= (x + 1)(x + 2)^2`
đây là kq phân tích đa thức thành nhân tử
(x-1)*(2*x+1)*(x^2-x+2)
=x4+2008x2+2008x-x+2008
=(x4-x)+(2008x2+2008x+2008)
=x(x3-1)+2008(x2+x+1)
=x(x-1)(x2+x+1)+2008(x2+x+1)
=(x2+x++1)(x2-x+2008)
\(\left(x^2-1\right)^2-4\left(x^2-1\right)+3=0\)
\(\left(x^2-1\right)\left(x^2-1-4\right)=-3\)
\(\orbr{\begin{cases}x^2-1=-3\\x^2-5=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=-2\\x^2=2\end{cases}}\Rightarrow\orbr{\begin{cases}\left(kotontai\right)\\x=\sqrt{2}\end{cases}}\)
vay \(x=\sqrt{2}\)
(X2-1)2-4(X2-1)+3=0
(X²-1)(X²-1-4)=-3
1,
X²-1=--3
X²=2
X=√2=2
2,
X²-5=-3
X²=-3-5
X²=--2
X=√-2=2
(x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5)(x+3)(x+4)-24
=(x2+7x+10)(x2+7x+12)-24
Đặt t=x2+7x+10 ta được:
t.(t+2)-24
=t2+2t-24
=t2-4t+6t-24
=t.(t-4)+6.(t-4)
=(t-4)(t+6)
thay t= x2+7x+10 ta được:
(x2+7x+6)(x2+7x+16)
=(x2+x+6x+6)(x2+7x+16)
=[x.(x+1)+6.(x+1)](x2+7x+16)
=(x+1)(x+6)(x2+7x+16)
Vậy (x+2)(x+3)(x+4)(x+5)-24=(x+1)(x+6)(x2+7x+16)
= [(x+2).(x+5)]. [(x+3).(x+4)] - 24 = (x2 + 5x + 2x+ 10). (x2 + 4x+3x+12) - 24
= (x2 + 7x + 10).(x2 + 7x + 12) - 24
= (x2 + 7x + 10). [(x2 + 7x + 10)+ 2] - 24 = (x2 + 7x + 10)2 + 2. (x2 + 7x + 10) - 24
= (x2 + 7x + 10)2 + 6 (x2 + 7x + 10) - 4(x2 + 7x + 10) - 24
= [ (x2 + 7x + 10)2 + 6 (x2 + 7x + 10)] - [4(x2 + 7x + 10) + 24]
= (x2 + 7x + 10) . [(x2 + 7x + 10) + 6] - 4. [(x2 + 7x + 10) + 6]
= (x2 + 7x + 10 - 4). [(x2 + 7x + 10) + 6] = (x2 + 7x + 6). (x2 + 7x + 16)
= (x2 + x+ 6x + 6). (x2 + 7x + 16) = [x(x+1) + 6.(x+1)]. (x2 + 7x + 16) = (x+6).(x+1).(x2 + 7x + 16)
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
x4+4 = (x2)2+22 = x4 + 2.x2.2 + 4 – 4x2
= (x2 + 2)2 – (2x)2 = (x2-2x+2)(x2+2x+2)
Ta có: \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)