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Thu gọn Q(x) = x4 + 7x2 + 1
Khi đó R(x) = Q(x) - P(x) = 4x2 + 3x + 2. Chọn A
`P(x)=\(4x^2+x^3-2x+3-x-x^3+3x-2x^2\)
`= (x^3-x^3)+(4x^2-2x^2)+(-2x-x+3x)+3`
`= 2x^2+3`
`Q(x)=`\(3x^2-3x+2-x^3+2x-x^2\)
`= -x^3+(3x^2-x^2)+(-3x+2x)+2`
`= -x^3+2x^2-x+2`
`P(x)-Q(x)-R(x)=0`
`-> P(X)-Q(x)=R(x)`
`-> R(x)=P(x)-Q(x)`
`-> R(x)=(2x^2+3)-(-x^3+2x^2-x+2)`
`-> R(x)=2x^2+3+x^3-2x^2+x-2`
`= x^3+(2x^2-2x^2)+x+(3-2)`
`= x^3+x+1`
`@`\(\text{dn inactive.}\)
a: P(x)-Q(x)-R(x)=0
=>R(x)=P(x)-Q(x)
=2x^2+3+x^3-2x^2+x-2
=x^3+x+1
`M(x)=P(x)+Q(x)`
`=x^4-5x+2x^2+1+5x+x^2+5-3x^2+x^4`
`=2x^4+6`
Đặt `M(x)=0`
`<=>2x^4+6=0`
`<=>x^4=-3`(vô lý vì `x^4>=0`)
a) Ta có M(x)=P(x)+Q(x)
=(\(x^4-5x+2x^2+1\))+(\(5x+x^2+5-3x^2+x^4\))
=\(x^4-5x+2x^2+1\)+\(5x+x^2+5-3x^2+x^4\)
=(\(x^4+x^4\))+(-5x+5x)+(\(2x^2\)+\(x^2\)-\(3x^2\))+(1+5)
=\(2x^4\)+6
Vậy M(x)=\(2x^4+6\)
b)Vì 2x\(^4\)\(\ge\) 0 với \(\forall\) x
nên \(2x^4+6\) \(\ge\)0 với \(\forall\)x
\(\Rightarrow\)M(x) \(\ge\) 0 với \(\forall\) x
Vậy M(x) vô nghiệm
`@` `\text {Ans}`
`\downarrow`
`a)`
Thu gọn:
`P(x)=`\(5x^4 + 3x^2 - 3x^5 + 2x - x^2 - 4 +2x^5\)
`= (-3x^5 + 2x^5) + 5x^4 + (3x^2 - x^2) + 2x - 4`
`= -x^5 + 5x^4 + 2x^2 + 2x - 4`
`Q(x) =`\(x^5 - 4x^4 + 7x - 2 + x^2 - x^3 + 3x^4 - 2x^2\)
`= x^5 + (-4x^4 + 3x^4) - x^3 + (x^2 - 2x^2) + 7x - 2`
`= x^5 - x^4 - x^3 - x^2 + 7x - 2`
`@` Tổng:
`P(x)+Q(x)=`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) + (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)
`= -x^5 + 5x^4 + 2x^2 + 2x - 4 + x^5 - x^4 - x^3 - x^2 + 7x - 2`
`= (-x^5 + x^5) - x^3 + (5x^4 - x^4) + (2x^2 - x^2) + (2x + 7x) + (-4-2)`
`= 4x^4 - x^3 + x^2 + 9x - 6`
`@` Hiệu:
`P(x) - Q(x) =`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) - (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)
`= -x^5 + 5x^4 + 2x^2 + 2x - 4 - x^5 + x^4 + x^3 + x^2 - 7x + 2`
`= (-x^5 - x^5) + (5x^4 + x^4) + x^3 + (2x^2 + x^2) + (2x - 7x) + (-4+2)`
`= -2x^5 + 6x^4 + x^3 + 3x^2 - 5x - 2`
`b)`
`@` Thu gọn:
\(H (x) = ( 3x^5 - 2x^3 + 8x + 9) - ( 3x^5 - x^4 + 1 - x^2 + 7x)\)
`= 3x^5 - 2x^3 + 8x + 9 - 3x^5 + x^4 - 1 + x^2 - 7x`
`= (3x^5 - 3x^5) + x^4 - 2x^3 - x^2 + (8x + 7x) + (9+1)`
`= x^4 - 2x^3 - x^2 + 15x + 10`
\(R( x) = x^4 + 7x^3 - 4 - 4x ( x^2 + 1) + 6x\)
`= x^4 + 7x^3 - 4 - 4x^3 - 4x + 6x`
`= x^4 + (7x^3 - 4x^3) + (-4x + 6x) - 4`
`= x^4 + 3x^3 + 2x - 4`
`@` Tổng:
`H(x)+R(x)=` \((x^4 - 2x^3 - x^2 + 15x + 10)+(x^4 + 3x^3 + 2x - 4)\)
`= x^4 - 2x^3 - x^2 + 15x + 10+x^4 + 3x^3 + 2x - 4`
`= (x^4 + x^4) + (-2x^3 + 3x^3) - x^2 + (15x + 2x) + (10-4)`
`= 2x^4 + x^3 - x^2 + 17x + 6`
`@` Hiệu:
`H(x) - R(x) =`\((x^4 - 2x^3 - x^2 + 15x + 10)-(x^4 + 3x^3 + 2x - 4)\)
`=x^4 - 2x^3 - x^2 + 15x + 10-x^4 - 3x^3 - 2x + 4`
`= (x^4 - x^4) + (-2x^3 - 3x^3) - x^2 + (15x - 2x) + (10+4)`
`= -5x^3 - x^2 + 13x + 14`
`@` `\text {# Kaizuu lv u.}`
Ta có
P ( x ) = 2 x 3 − 3 x + x 5 − 4 x 3 + 4 x − x 5 + x 2 − 2 = x 5 − x 5 + 2 x 3 − 4 x 3 + x 2 + ( 4 x − 3 x ) − 2 = − 2 x 3 + x 2 + x − 2 Và Q ( x ) = x 3 − 2 x 2 + 3 x + 1 + 2 x 2 = x 3 + − 2 x 2 + 2 x 2 + 3 x + 1 = x 3 + 3 x + 1
Khi đó
M ( x ) = P ( x ) + Q ( x ) = − 2 x 3 + x 2 + x − 2 + x 3 + 3 x + 1 = − 2 x 3 + x 2 + x − 2 + x 3 + 3 x + 1 = − 2 x 3 + x 3 + x 2 + ( x + 3 x ) − 2 + 1 = − x 3 + x 2 + 4 x − 1
Bậc của M ( x ) = - x 3 + x 2 + 4 x - 1 l à 3
Chọn đáp án C
`@` `\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.}`
Ta có:
Vì : P(x) + Q(x) = x5 – 2x2 + 1
Suy ra Q(x) = x5 – 2x2 + 1– P(x).