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a) \(A\left(x\right)=2x^3+2-3x^2+1=2x^3-3x^2+3\)
Có bậc là 3
\(B\left(x\right)=2x^2+3x^3-x-6=3x^3+2x^2-x-6\)
Có bậc 3
b) Thay \(x=2\) vào A(x) ta được:
\(2\cdot2^3-3\cdot2^2+3=2\cdot8-3\cdot4+3=16-12+3=7\)
Vậy giá trị của A(x) tại x=2 là 7
c) \(A\left(x\right)+B\left(x\right)\)
\(=2x^3-3x^2+3+3x^3+2x^2-x-6\)
\(=5x^3-x^2-x-3\)
\(A\left(x\right)-B\left(x\right)\)
\(=\left(2x^3-3x^2+3\right)-\left(2x^2+3x^3-x-6\right)\)
\(=2x^3-3x^2+3-2x^2-3x^3+x+6\)
\(=-x^3-5x^2+x+9\)
a: A(x)=2x^3-3x^2+3
Bậc là 3
B(x)=3x^3+2x^2-x-6
Bậc là 3
b: A(2)=2*2^3-3*2^2+3=7
c; A(x)+B(x)
=2x^3-3x^2+3+3x^3+2x^2-x-6
=5x^3-x^2-x-3
A(x)-B(x)
=2x^3-3x^2+3-3x^3-2x^2+x+6
=-x^3-5x^2+x+9
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\(Q=(x^3-3x^2+2x+1)(-x^2)-x(2x^2-3x+1)\)
\(=-x^5+3x^4-2x^3-x^2-2x^3+3x^2-x\)
\(=-x^5+3x^4-4x^3+2x^2-x\)
Hệ số của \(x^3:-4\)
Hệ số của \(x^2:2\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a: x^3-7x-6
=x^3-x-6x-6
=x(x-1)(x+1)-6(x+1)
=(x+1)(x^2-x-6)
=(x-3)(x+2)(x+1)
b: =2x^3+x^2-2x^2-x+6x+3
=x^2(2x+1)-x(2x+1)+3(2x+1)
=(2x+1)(x^2-x+3)
c: =2x^3-3x^2-2x^2+3x+2x-3
=x^2(2x-3)-x(2x-3)+(2x-3)
=(2x-3)(x^2-x+1)
d: =2x^3+x^2+2x^2+x+2x+1
=(2x+1)(x^2+x+1)
e: =3x^3+x^2-3x^2-x+6x+2
=(3x+1)(x^2-x+2)
f: =27x^3-9x^2-18x^2+6x+12x-4
=(3x-1)(9x^2-6x+4)
a) \(x^3-7x-6\)
\(=x^3-x-6x-6\)
\(=\left(x^3-x\right)-\left(6x+6\right)\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
b) \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(x^2-x+3\right)\left(2x+1\right)\)
c) \(2x^3-5x^2+5x+1\)
\(=2x^3-3x^2-2x^2+3x+2x-3\)
\(=\left(2x^3-3x^2\right)-\left(2x^2-3x\right)+\left(2x-3\right)\)
\(=x^2\left(2x-3\right)-x\left(2x-3\right)+\left(2x-3\right)\)
\(=\left(x^2-x+1\right)\left(2x-3\right)\)
d) \(2x^3+3x^2+3x+1\)
\(=2x^3+x^2+2x^2+x+2x+1\)
\(=\left(2x^3+x^2\right)+\left(2x^2+x\right)+\left(2x+1\right)\)
\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(3x^3-2x^2+5x+2\)
\(=3x^3+x^2-3x^2-x+6x+2\)
\(=\left(3x^3+x^2\right)-\left(3x^2+x\right)+\left(6x+2\right)\)
\(=x^2\left(3x+1\right)-x\left(3x+1\right)+2\left(3x+1\right)\)
\(=\left(3x-1\right)\left(x^2-x+2\right)\)
f) \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
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a: \(=\dfrac{1}{4}\cdot\dfrac{1}{2}\cdot x^4y=\dfrac{1}{8}x^4y\)
Bậc là 5
b: \(=\dfrac{-1}{3}\cdot\dfrac{3}{2}\cdot x^2y\cdot xy^3=\dfrac{-1}{2}x^3y^4\)
Bậc là 7
c: \(=\dfrac{3}{4}\cdot x^6y^4\)
Bậc là 10
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Bài `1:`
`a)3x^3+6x^2=3x^2(x+2)`
`b)x^2-y^2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)`
Bài `2:`
`a)(2x-1)^2-25=0`
`<=>(2x-1-5)(2x-1+5)=0`
`<=>(2x-6)(2x+4)=0`
`<=>[(x=3),(x=-2):}`
`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`
`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`
`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`
`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`
Answer:
\(3x^2-\frac{1}{2}x+1+2x-x^2\)
\(=\left(3x^2-x^2\right)+\left(2x-\frac{1}{2}x\right)+1\)
\(=x^2.\left(3-1\right)+\left(2-\frac{1}{2}\right)x+1\)
\(=2x^2+\frac{3}{2}x+1\)
Vậy \(2x^2\) có bậc là 2; \(\frac{3}{2}x\) có bậc là 1; 1 có bậc là 0.