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=> 2 f(x) = 6x^4 - 3x^2 - 5 + 4x^4 - 6x^3 + 7x^2 + 8x - 9
= 10x^4 - 6x^3 + 4x^2 + 8x - 14
=> 2.f ( x ) = 2 ( 5x^4 - 3x^3 + 2x^2 + 4x - 7 )
=> ( fx) = 5x^4 - 3x^3 + 2x^2 + 4x - 7
g(x) tự tìm
ta có:
f(x) + g(x) = 6x^4 - 3x^2 - 5
f(x) - g(x) = 4x^4 - 6x^3 + 7x^2 + 8x - 9
công hai vế lại với nhau ta được:
f(x)+g(x)+f(x)-g(x)=6x^4 - 3x^2 - 5 + 4x^4 - 6x^3 + 7x^2 + 8x - 9
=>2f(x)=6x4+4x4-6x3-3x2+7x2+8x-5-9
2f(x)=10x4-6x3+4x2+8x-14
2f(x)=2.(5x4-3x3+2x2+4x-7)
=>f(x)=5x4-3x3+2x2+4x-7
=>g(x)=6x^4 - 3x^2 - 5 -(5x4-3x3+2x2+4x-7)
=6x4-3x2-5-5x4+3x3-2x2-4x+7
=6x4-5x4+3x3-3x2-2x2-4x-5+7
=x4+3x3-5x2-4x+2
A(x)=4x4−6x2−7x3−5x−6
B(x)=−5x2+7x3+5x+4−4x4
a/ - Tính:
M(x)=A(x)+B(x)
M(x)=4x4+6x2−7x3−5x−6−5x2+7x3+5x+4−4x4
M(x)=x2−2
- Tìm nghiệm:
M(x)=x2−2=0⇔x2=2⇔x=−√2;x=√2
b/ C(x)+B(x)=A(x)⇒C(x)=A(x)−B(x)
C(x)=4x4−6x2−7x3−5x−6−(−5x2+7x3+5x+4−4x4)
C(x)=4x4−6x2−7x3−5x−6+5x2−7x3−5x−4+4x4
C(x)=8x4−14x3−x2−10x−10
cho đa thức : A(x)=4x^4+6x^2-7x^3-5x-6 và B(x)=-5x^2+x^3+5x+4-4x^4
a)Tính M(x)=A(x)+B(x) rồi tính nghiệm của đa thức M(x)
b)tìm đa thức C(x)sao cho C(x)|+B(x)=A(x)
a) dễ tự làm
b) A(x) có bậc 6
hệ số: -1 ; 5 ; 6 ; 9 ; 4 ; 3
B(x) có bậc 6
hệ số: 2 ; -5 ; 3 ; 4 ; 7
c) bó tay
d) cx bó tay
a) G(x) = 2x5-4x4-10x3+3x2-4x-8
H(x) = x5-2x4-5x3+x2+7x-4
b) G(x)+H(x)=3x5-6x4-15x3+4x2+3x-12
G(x)-H(x) =x5-2x4-5x3+2x2-11x-4
c) G(x) = 2H(x)
2x5-4x4-10x3+3x2-4x-8=2( x5-2x4-5x3+x2+7x-4)
2x5-4x4-10x3+3x2-4x-8-2( x5-2x4-5x3+x2+7x-4)=0
2x5-4x4-10x3+3x2-4x-8-2x5+4x4+10x3-2x2-14x+8=0
x2-18x=0
x(x-18)=0
x=0 hoặc x-18=0
x=18
a) A(x) = 5x4 - 5 + 6x3 + x4 - 5x - 12
= (5x4 + x4) + (- 5 - 12) + 6x3 - 5x
= 6x4 - 17 + 6x3 - 5x
= 6x4 + 6x3 - 5x - 17
B(x) = 8x4 + 2x3 - 2x4 + 4x3 - 5x - 15 - 2x2
= (8x4 - 2x4) + (2x3 + 4x3) - 5x - 15 - 2x2
= 4x4 + 6x3 - 5x - 15 - 2x2
= 4x4 + 6x3 - 2x2 - 5x - 15
b) C(x) = A(x) - B(x)
= 6x4 + 6x3 - 5x - 17 - (4x4 + 6x3 - 2x2 - 5x - 15)
= 6x4 + 6x3 - 5x - 17 - 4x4 - 6x3 + 2x2 + 5x + 15
= ( 6x4 - 4x4) + ( 6x3 - 6x3) + (- 5x + 5x) + (-17 + 15) + 2x2
= 2x4 - 2 + 2x2
= 2x4 + 2x2 - 2
Bài 1 ( a )
\(A_x=-4x^5-x^3+4x^2+5x+9+4x^5-6x^2-2\)
\(=-x^3-2x^2+5x-7\)
\(B_x=-3x^4-2x^3+10x^2-8x+5x^3-7-2x^3+8x\)
\(=-3x^4+x^3+10x^2-7\)
Bài 1 ( b )
\(P_x=\left(-x^3-2x^2+5x-7\right)+\left(3x^4+x^3+10x-7\right)\)
\(=-x^3-2x^2+5x-7+3x^4+x^3+10x-7\)
\(=3x^4-2x^2+15x-14\)
\(Q_x=\left(-x^3-2x^2+5x-7\right)-\left(3x^4+x^3+10x-7\right)\)
\(=-x^3-2x^2+5x-7-3x^4-x^3-10x+7\)
\(=-3x^4-2x^3-5x\)
`@` `\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.}`
`A(x) + B(x) = 6x^4 - 3x^2 - 5`
`A(x) - B(x) = 4x^4 - 6x^3 + 7x^2 + 8x - 9`
Áp dụng bài toán tổng hiệu ta có:
`A(x) = [(6x^4 - 3x^2 - 5) + (4x^4 - 6x^3 + 7x^2 + 8x - 9)] : 2`
`= (6x^4 - 3x^2 - 5 + 4x^4 - 6x^3 + 7x^2 + 8x - 9) : 2`
`= (10x^4 - 6x^3 + 4x^2 + 8x - 14) : 2`
`= 5x^4 - 3x^3 + 2x^2 + 4x - 7`
`B(x) = (6x^4 - 3x^2 - 5) - (5x^4 - 3x^3 + 2x^2 + 4x - 7)`
`= 6x^4 - 3x^2 - 5 - 5x^4 + 3x^3 - 2x^2 - 4x + 7`
`= x^4 + 3x^3 - 5x^2 - 4x + 2`
Vậy ....
\(2A\left(x\right)=\left(6x^4-3x^2-5\right)+\left(4x^4-6x^3+7x^2+8x-9\right)\\ =\left(6x^4+4x^4\right)-6x^3+\left(-3x^2+7x^2\right)+8x+\left(-5-9\right)\\ =10x^4-6x^3+4x^2+8x-14\\ =>A\left(x\right)=5x^4-3x^3+2x^2+4x-7\)
\(=>B\left(x\right)=\left(6x^4-3x^2-5\right)-A\left(x\right)\\ =\left(6x^4-3x^2-5\right)-\left(5x^4-3x^3+2x^2+4x-7\right)\\ =\left(6x^4-5x^4\right)+3x^3+\left(-3x^2-2x^2\right)-4x+\left(-5+7\right)\\ =x^4+3x^3-5x^2-4x+2\)