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`@` `\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) P(x) = 5x5 - 4x2 + 7x + 15
Q(x) = 5x5 - 4x2 + 3x + 8
b) Có: P(x) - Q(x) = 4x + 7
P(x) - Q(x) = 0 <=> x = \(-\dfrac{-7}{4}\)
`a,```P(x) = 8x^5 +7x -6x^2 -3x^5 +2x^2+15`
`= (8x^5 -3x^5 ) +(-6x^2+2x^2) +7x+15`
`=5x^5 -4x^2 +7x+15`
`Q(x) =4x^5 +3x-2x^2 +x^5 -2x^2+8`
`=(4x^5+x^5) +(-2x^2 -2x^2)+3x+8`
`= 5x^5 - 4x^2 +3x+8`
`b, P(x) -Q(x)=(5x^5 -4x^2 +7x+15)-(5x^5 - 4x^2 +3x+8)`
`= 5x^5 -4x^2 +7x+15-5x^5 +4x^2 -3x-8`
`= (5x^5-5x^5)+(-4x^2+4x^2) +(7x-3x)+(15-8)`
`= 0 + 0 +4x + 7`
`=4x+7`
Thu gọn
A(x) = 5 +3x2 – x - 2x2
A(x) = 5+x2-x
A(x) = x2-x+5
B(x) = 3x + 3 – x – x2
B(x) = ( 3x-x) + 3 - x2
B(x) = 2x+3-x2
B(x)= -x2 + 2x+3
\(a)A\left(x\right)=5+3x^2-x-2x^2\)
\(A\left(x\right)=\left(3x^2-2x^2\right)-x+5\)
\(A\left(x\right)=x^2-x+5\)
\(B\left(x\right)=3x+3-x-x^2\)
\(B\left(x\right)=-x^2+\left(3x-x\right)+3\)
\(B\left(x\right)=-x^2+2x+3\)
\(b)C\left(x\right)=A\left(x\right)+B\left(x\right)\)
\(C\left(x\right)=\left(x^2-x+5\right)+\left(-x^2+2x+3\right)\)
\(C\left(x\right)=x^2-x+5+-x^2+2x+3\)
\(C\left(x\right)=\left(x^2-x^2\right)+\left(-x+2x\right)+\left(5+3\right)\)
\(C\left(x\right)=-x+8\)
\(c)D\left(x\right)=A\left(x\right)-B\left(x\right)\)
\(D\left(x\right)=\left(x^2-x+5\right)-\left(-x^2+2x+3\right)\)
\(D\left(x\right)=x^2-x+5+x^2-2x-3\)
\(D\left(x\right)=\left(x^2+x^2\right)+\left(-x-2x\right)+\left(5-3\right)\)
\(D\left(x\right)=2x^2-3x+2\)
a) \(A\left(x\right)=5+3x^2-x-2x^2\)
\(A\left(x\right)=5+\left(3x^2-2x^2\right)-x\)
\(A\left(x\right)=5+x^2-x\)
\(A\left(x\right)=x^2-x+5\)
\(B\left(x\right)=3x+3-x-x^2\)
\(B\left(x\right)=\left(3x-x\right)+3-x^2\)
\(B\left(x\right)=2x+3-x^2\)
\(B\left(x\right)=-x^2+2x+3\)
b) Ta có \(C\left(x\right)=A\left(x\right)+B\left(x\right)\)
\(\begin{matrix}\Rightarrow A\left(x\right)=x^2-x+5\\^+B\left(x\right)=-x^2+2x+3\\\overline{A\left(x\right)+B\left(x\right)=0+x+8}\end{matrix}\)
Vậy \(C\left(x\right)=x+8\)
c) Ta có \(D\left(x\right)=A\left(x\right)-B\left(x\right)\)
\(\begin{matrix}\Rightarrow A\left(x\right)=x^2-x+5\\^-B\left(x\right)=-x^2+2x+3\\\overline{A\left(x\right)-B\left(x\right)=2x^2-3x+2}\end{matrix}\)
Vậy \(D\left(x\right)=2x^2-3x+2\)
Ở câu b, \(A\left(x\right)+B\left(x\right)=0+x+8\) số 0 bạn bỏ rồi để khoảng trống \(A\left(x\right)+B\left(x\right)=\) \(x+8\) như vậy nha, với các dấu \(=\) ở câu b và c với cái số bạn đặt thẳng hàng nha (các từ in đậm bạn không cần ghi)
a. Ta có: A(x) = x5 + x2 + 5x + 6 - x5 - 3x - 5
= x2 + 2x + 1 (0.5 điểm)
B(x) = x4 + 2x2 - 3x - 3 - x4 - x2 + 3x + 4 = x2 + 1 (0.5 điểm)
Câu 1:
a.
$M(x)=2x^3-(x^2+3x^2)+(8-1)=2x^3-4x^2+7$
$N(x)=2x^3-4x^2+3x+(5+7)=2x^3-4x^2+3x+12$
b.
$M(x)+N(x)=(2x^3-4x^2+7)+(2x^3-4x^2+3x+12)$
$=2x^3-4x^2+7+2x^3-4x^2+3x+12=4x^3-8x^2+3x+19$
c.
$3x^2(5x^2-x+2)=3x^2.5x^2-3x^2.x+3x^2.2$
$=15x^4-3x^3+6x^2$
Câu 2:
a.
$A(x)=4x^5+5x^4-(2x^2+x^2)+(7x+x)=4x^5+5x^4-3x^2+8x$
$B(x)=-3x^5-x^4-2x^3+x-2$
b.
$M(x)=A(x)+B(x)=(4x^5+5x^4-3x^2+8x)+(-3x^5-x^4-2x^3+x-2)$
$=4x^5+5x^4-3x^2+8x-3x^5-x^4-2x^3+x-2$
$=(4x^5-3x^5)+(5x^4-x^4)-2x^3-3x^2+(8x+x)-2$
$=x^5+4x^4-2x^3-3x^2+9x-2$
c.
$5x^3(2x^2-3x+10)=5x^3.2x^2-5x^3.3x+5x^3.10$
$=10x^5-15x^4+50x^3$