p=(2+x/2-x+4x^2-2-x/2+x):x^2-3x/2x^2-x^ rút gọn p
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(P_{\left(x\right)}=2x^3-2x+x^2+3x+2\)
\(P_{\left(x\right)}=2x^3+x^2+x+2\)
\(Q_{\left(x\right)}=4x^3-3x^2-3x+4x-3x^3+4x^2+1\)
\(Q_{\left(x\right)}=x^3+x^2+x+1\)
b) \(P_{\left(x\right)}+Q_{\left(x\right)}=\left(2x^3+x^2+x+2\right)+\left(x^3+x^2++x+1\right)\)
\(=3x^3+2x^2+2x+3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(P\left(x\right)=2x^3+x^2+x+2\)
\(Q\left(x\right)=x^3+x^2+x+1\)
b: \(P\left(-1\right)=2\cdot\left(-1\right)+1-1+2=0\)
\(Q\left(-1\right)=-1+1-1+1=0\)
Do đó: x=-1 là nghiệm chung của P(x), Q(x)
\(P\left(x\right)=2x^3-2x+x^2+3x+2\)
\(P\left(x\right)=2x^3+x^2+x+2\)
\(Q\left(x\right)=4x^3-3x^2-3x+4x-3x^3+4x^2+1\)
\(Q\left(x\right)=x^3+x^2+x+1\)
__________________________________________________
\(P\left(-1\right)=2.\left(-1\right)^3+\left(-1\right)^2+\left(-1\right)+2\)
\(P\left(-1\right)=0\)
\(Q\left(-1\right)=\left(-1\right)^3+\left(-1\right)^2+\left(-1\right)+1\)
\(Q\left(-1\right)=0\)
Vậy x = -1 là nghiệm của P(x),Q(x)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x(3x3-x)-4x2(x-x2+1)+(x-3x2)x
=6x4-2x2+4x4-4x3-4x2+x2-3x3
=(6x4+4x4)+(-4x3-3x3)+(-4x2-x2)
=10x4-7x3-5x2
\(2x\left(3x^3-x\right)-4x^2\left(x-x^2+1\right)+\left(x-3x^2\right)x\\ =6x^4-2x^2+4x^4-4x^3-4x^2+x^2-3x^3\\ =\left(6x^4+4x^4\right)+\left(-4x^3\right)-3x^3+\left(-4x^2-x^2\right)\)
\(=10x^4-7x^3-5x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)
\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)
\(=6x^2-3x+\dfrac{5}{2}\)
b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)
\(=3x-y-y-x+2x^2-2x\)
\(=2x^2-2y\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28
b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10
c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x
d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( x + 2 )( x2 - 2x + 4 ) - ( 18 + x3 )
= x3 + 8 - 18 - x3 = -10
b) ( 2x - y )( 4x2 + 2xy + y2 ) - ( 2x + y )( 4x2 - 2xy + y2 )
= 8x3 - y3 - ( 8x3 + y3 )
= 8x3 - y3 - 8x3 - y3 = -2y3
c) ( x - 3 )( x + 3 ) - ( x + 5 )( x - 1 )
= x2 - 9 - ( x2 + 4x - 5 )
= x2 - 9 - x2 - 4x + 5 = -4x - 4
d) ( 3x - 2 )2 + ( x + 1 )2 + 2( 3x - 2 )( x + 1 )
= ( 3x - 2 + x + 1 )2
= ( 4x - 1 )2