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![](https://rs.olm.vn/images/avt/0.png?1311)
a: P(x)=5x^3+3x^2-2x-5
\(Q\left(x\right)=5x^3+2x^2-2x+4\)
b: P(x)-Q(x)=x^2-9
P(x)+Q(x)=10x^3+5x^2-4x-1
c: P(x)-Q(x)=0
=>x^2-9=0
=>x=3; x=-3
d: C=A*B=-7/2x^6y^4
![](https://rs.olm.vn/images/avt/0.png?1311)
a: P(1)=2-3-4=-5
b: \(P\left(x\right)+Q\left(x\right)=3x^2-6x+1\)
\(P\left(x\right)-Q\left(x\right)=x^2-9\)
c: Đặt H(x)=0
=>(x-3)(x+3)=0
=>x=3 hoặc x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(P\left(x\right)=4x^3+2x-3+2x-2x^2-1\\ =4x^3-2x^2+\left(2x+2x\right)+\left(-3-1\right)\\ =4x^3-2x^2+4x-4\)
Bậc của P(x) là 3
\(Q\left(x\right)=6x^3-3x+5-2x+3x^2\\ =6x^3+3x^2+\left(-3x-2x\right)+5\\ =6x^3+3x^2-5x+5\)
Bậc của Q(x) là 3
b, \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=4x^3-2x^2+4x-4+6x^3+3x^2-5x+5\\ =\left(4x^3+6x^3\right)+\left(-2x^2+3x^2\right)+\left(4x-5x\right)+\left(-4+5\right)\\ =10x^3+x^2-x+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(P\left(1\right)=2-3-4=-5\)
b, \(H\left(x\right)=P\left(x\right)-Q\left(x\right)=x^2-9\)
c, Ta có \(H\left(x\right)=\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=3;x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(P\left(x\right)=A\left(x\right)+B\left(x\right)=2x^2-x^3+x^3-x^2-3x+4=x^2-3x+4\)
b: Theo đề, ta có: Q(-1)=0
\(\Leftrightarrow5-5+a^2-a=0\)
=>a(a-1)=0
=>a=0 hoặc a=1
a, \(P\left(x\right)=2x^2-x^3+x^3-x^2+4-3x=x^2-3x+4\)
b, Ta có \(Q\left(-1\right)=5-5+a^2+a=a^2+a=0\)
\(\Leftrightarrow a\left(a+1\right)=0\Leftrightarrow a=0;a=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: P(x)=5x^2-4x+7
Sửa đề: Q(x)=-5x^3-x^2+4x-5
Q(x)+P(x)+5x^2-2=0
=>5x^2-4x+7-5x^3-x^2+4x-5+5x^2-2=0
=>5x^3=0
=>x=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-5.\left(\dfrac{1}{2}\right)^3+3\left(\dfrac{1}{2}\right)^2+\dfrac{2}{2}+5-5\left(\dfrac{1}{2}\right)^3+6\left(\dfrac{1}{2}\right)^2+\dfrac{2}{2}+5\)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-\dfrac{5.1}{8}+\dfrac{3.1}{4}+6-\dfrac{5.1}{8}+\dfrac{6.1}{4}+6\)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-\dfrac{5}{8}+\dfrac{3}{4}+6-\dfrac{5}{8}+\dfrac{3}{2}+6\)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=13\)
\(Ta\)\(có\)
\(P\left(x\right)=-F\left(x\right)\)
\(\Leftrightarrow-x^3+2x^2+x-1=\left(x^3-2x^2+3x+5\right)\)
\(\Leftrightarrow-x^3+2x^2+x-1=-x^3+2x^2-3x-5\)
\(\Leftrightarrow x-1=-3x-5\)
\(\Leftrightarrow4x=-4\)
\(\Leftrightarrow x=-1\)
\(Vậy\)\(......\)
sorry \(-x^3+2x^2+x-1=-\left(x^3-2x+3x+5\right)\)
mik thiếu dấu trừ nha