Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(...=P\left(x\right)=2x^4-x^4+3x^3+4x^2-3x^2+3x-x+3\)
\(P\left(x\right)=x^4+3x^3+x^2+2x+3\)
\(...=Q\left(x\right)=x^4+x^3+3x^2-x^2+4x+4-2\)
\(Q\left(x\right)=x^4+x^3+2x^2+4x+2\)
b) \(P\left(x\right)+Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)+\left(x^4+x^3+2x^2+4x+2\right)\)
\(\Rightarrow P\left(x\right)+Q\left(x\right)=2x^4+4x^3+3x^2+6x+5\)
\(P\left(x\right)-Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)-\left(x^4+x^3+2x^2+4x+2\right)\)
\(\)\(\Rightarrow P\left(x\right)-Q\left(x\right)=x^4+3x^3+x^2+2x+3-x^4-x^3-2x^2-4x-2\)
\(\Rightarrow P\left(x\right)-Q\left(x\right)=2x^3-x^2-2x+1\)
Bài 1:
a) Ta có: \(P\left(x\right)=3x^4+2x^2-3x^4-2x^2+2x-5\)
\(=\left(3x^4-3x^4\right)+\left(2x^2-2x^2\right)+2x-5\)
\(=2x-5\)
Bài 1:
b)
\(P\left(-1\right)=2\cdot\left(-1\right)-5=-2-5=-7\)
\(P\left(3\right)=2\cdot3-5=6-5=1\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(P(x) = 5x^3 + 3 - 3x^2 + x^4 - 2x - 2 + 2x^2 + x\)
`= x^4 + 5x^3 + (-3x^2 + 2x^2) + (-2x+x) + (3-2)`
`= x^4 + 5x^3 - x^2 - x + 1`
\(Q(x) = 2x^4 + x^2 + 2x + 2 - 3x^2 - 5x + 2x^3 - x^4\)
`= (2x^4 - x^4) + 2x^3 + (x^2 - 3x^2) + (2x-5x) + 2`
`= x^4 + 2x^3 - 2x^2 - 3x +2`
`b)`
`P(x)+Q(x) = (x^4 + 5x^3 - x^2 - x + 1) + (x^4 + 2x^3 - 2x^2 - 3x +2)`
`= x^4 + 5x^3 - x^2 - x + 1 + x^4 + 2x^3 - 2x^2 - 3x +2`
`= (x^4+x^4)+(5x^3 + 2x^3) + (-x^2 - 2x^2) + (-x-3x) + (1+2)`
`= 2x^4 + 7x^3 - 3x^2 - 4x + 3`
`P(x)-Q(x)=(x^4 + 5x^3 - x^2 - x + 1) - (x^4 + 2x^3 - 2x^2 - 3x +2)`
`= x^4 + 5x^3 - x^2 - x + 1 - x^4 - 2x^3 + 2x^2 + 3x -2`
`= (x^4 - x^4) + (5x^3 - 2x^3) + (-x^2+2x^2)+(-x+3x)+(1-2)`
`= 3x^3 + x^2 + 2x - 1`
`Q(x)-P(x) = (x^4 + 2x^3 - 2x^2 - 3x +2)-(x^4 + 5x^3 - x^2 - x + 1)`
`= x^4 + 2x^3 - 2x^2 - 3x +2-x^4 - 5x^3 + x^2 + x - 1`
`= (x^4-x^4)+(2x^3 - 5x^3)+(-2x^2+x^2)+(-3x+x)+(2-1)`
`= -3x^3 - x^2 - 2x + 1`
`@` `\text {Kaizuu lv u.}`
`a,`
`F(x)=4x^4-2+2x^3+2x^4-5x+4x^3-9`
`F(x)=(2x^4+4x^4)+(2x^3+4x^3)-5x+(-2-9)`
`F(x)=6x^4+6x^3-5x-11`
`b,`
`K(x)=F(x)+G(x)`
`K(x)=(6x^4+6x^3-5x-11)+(6x^4+6x^3-x^2-5x-27)`
`K(x)=6x^4+6x^3-5x-11+6x^4+6x^3-x^2-5x-27`
`K(x)=(6x^4+6x^4)+(6x^3+6x^3)-x^2+(-5x-5x)+(-11-27)`
`K(x)=12x^4+12x^3-x^2-10x-38`
`c,`
`H(x)=F(x)-G(x)`
`H(x)=(6x^4+6x^3-5x-11)-(6x^4+6x^3-x^2-5x-27)`
`H(x)=6x^4+6x^3-5x-11-6x^4-6x^3+x^2+5x+27`
`H(x)=(6x^4-6x^4)+(6x^3-6x^3)+x^2+(-5x+5x)+(-11+27)`
`H(x)=x^2+16`
Đặt `x^2+16=0`
Ta có: \(x^2\ge0\text{ }\forall\text{ }x\)
`->`\(x^2+16\ge16>0\text{ }\forall\text{ }x\)
`->` Đa thức `H(x)` vô nghiệm.
a: \(F\left(x\right)=x^5-3x^2+x^3-x^2-2x+5\)
\(=x^5+x^3-4x^2-2x+5\)
\(G\left(x\right)=x^5-x^4+x^2-3x+x^2+1\)
\(=x^5-x^4+2x^2-3x+1\)
b: Ta có: \(H\left(x\right)=F\left(x\right)+G\left(x\right)\)
\(=x^5+x^3-4x^2-2x+5+x^5-x^4+2x^2-3x+1\)
\(=2x^5-x^4+x^3-2x^2-5x+6\)
a: f(x)=3x^4+2x^3+6x^2-x+2
g(x)=-3x^4-2x^3-5x^2+x-6
b: H(x)=f(x)+g(x)
=3x^4+2x^3+6x^2-x+2-3x^4-2x^3-5x^2+x-6
=x^2-4
f(x)-g(x)
=3x^4+2x^3+6x^2-x+2+3x^4+2x^3+5x^2-x+6
=6x^4+4x^3+11x^2-2x+8
c: H(x)=0
=>x^2-4=0
=>x=2 hoặc x=-2
\(a) f ( x ) = 2 x ^4 + 3 x ^2 − x + 1 − x ^2 − x ^4 − 6 x ^3\)
\(= ( 2 x ^4 − x ^4 ) − 6 x ^3 + ( 3 x ^2 − x ^2 ) − x + 1\)
\(= x ^4 − 6 x ^3 + 2 x ^2 − x + 1\)
\(g ( x ) = 10 x ^3 + 3 − x ^4 − 4 x ^3 + 4 x − 2 x ^2\)
\(= − x ^4 + ( 10 x ^3 − 4 x ^3 ) − 2 x ^2 + 4 x + 3\)
\(= − x ^4 + 6 x ^3 − 2 x ^2 + 4 x + 3\)
\(b) f ( x ) + g ( x ) = x ^4 − 6 x ^3 + 2 x ^2 − x + 1 − x ^4 + 6 x ^3 − 2 x ^2 + 4 x + 3\)
\(= ( x ^4 − x ^4 ) + ( − 6 x ^3 + 6 x ^3 ) + ( 2 x ^2 − 2 x ^2 ) + ( − x + 4 x ) + ( 1 + 3 )\)
\(= 3 x + 4\)
c)Có \(h ( x ) = f ( x ) + g ( x ) = 3 x + 4\)
\(Cho h ( x ) = 0 ⇒ 3 x + 4 = 0\)
\(⇒ 3 x = − 4\)
\(⇒ x = − \frac{4 }{3} \)
Vậy \(x=-\frac{4}{3}\) là nghiệm của \(h ( x ) \)