Phân tich đa thức thành nhân tử
\(x^5+x+1\)
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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`
\(\left(x^2-1\right)^2-4\left(x^2-1\right)+3=0\)
\(\left(x^2-1\right)\left(x^2-1-4\right)=-3\)
\(\orbr{\begin{cases}x^2-1=-3\\x^2-5=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=-2\\x^2=2\end{cases}}\Rightarrow\orbr{\begin{cases}\left(kotontai\right)\\x=\sqrt{2}\end{cases}}\)
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
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
đây là kq phân tích đa thức thành nhân tử
(x-1)*(2*x+1)*(x^2-x+2)
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
x5-x4-1=x5-x3-x2-x4+x2+x+x3-x-1
=x2.(x3-x-1)-x.(x3-x-1)+(x3-x-1)
=(x3-x-1)(x2-x+1)
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
(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)
Lời giải:
$x^5+x-1=(x^5+x^2)-(x^2-x+1)$
$=x^2(x^3+1)-(x^2-x+1)=x^2(x+1)(x^2-x+1)-(x^2-x+1)$
$=(x^2-x+1)[x^2(x+1)-1]=(x^2-x+1)(x^3+x^2-1)$
Sau khi phân tích thì đa thức có dạng ( x2 + ax + 1 )( x3 + bx2 + cx + 1 )
=> x5 + x + 1 = ( x2 + ax + 1 )( x3 + bx2 + cx + 1 )
=> x5 + x + 1 = x5 + bx4 + cx3 + x2 + ax4 + abx3 + acx2 + ax + x3 + bx2 + cx + 1
=> x5 + x + 1 = x5 + ( a + b )x4 + ( ab + c + 1 )x3 + ( ac + b + 1 )x2 + ( c + a )x + 1
Đồng nhất hệ số ta có :
a + b = 0 ; ab + c + 1 = 0 ; ac + b + 1 = 0 ; c + a = 1
Giải hệ này ta được : a = 1 ; b = -1 ; c = 0
=> x5 + x + 1 = ( x2 + x + 1 )( x3 - x2 + 1 )
\(x^5+x+1=\left(x^5-x^2\right)+\left(x+x^2+1\right)=x^2\left(x^3-1\right)+\left(x^2+x+1\right)=x^2\left(x-1\right)\left(x^2+x+1\right)\)
\(+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)