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a, \(P\left(x\right)=5x^3-3x+7-x=5x^3-4x+7\)
\(Q\left(x\right)=-5x^3+2x-3+2x-x^2-2=-5x^3-x^2+4x-5\)
b, \(M\left(x\right)=5x^3-4x+7-5x^3-x^2+4x-5=-x^2+2\)
c, Đặt \(M\left(x\right)+2=0\Rightarrow-x^2+4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
a: \(P\left(x\right)=5x^3-3x+7-x=5x^3-4x+7\)
\(Q\left(x\right)=-5x^3+2x-3+2x-x^2-2=-5x^3-x^2+4x-5\)
b: Ta có: \(M\left(x\right)=P\left(x\right)+Q\left(x\right)\)
\(=5x^3-4x+7-5x^3-x^2+4x-5\)
\(=-x^2+2\)
c: Đặt M(x)+2=0
\(\Leftrightarrow4-x^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`Q(x)=-5x^3+2x-3+2x-x^2-2`
`=-5x^3+4x-5`
`M(x)=P(x)+Q(x)`
`=5x^3-3x+7-5x^3+4x-5`
`=x+2`
`N(x)=P(x)-Q(x)`
`=5x^3-3x+7+5x^3-4x+5`
`=10x^3-7x+12`
b)Đặt `M(x)=0`
`<=>x+2=0`
`<=>x=-2`
Vậy M(x) có nghiệm `x=-2`
1k like đâu
a) \(P\left(x\right)=5x^3-3x+7-x\\ =5x^3+\left(-3x-x\right)+7\\ =5x^3-4x+7\\ Q\left(x\right)=-5x^3+2x-3+2x-x^2-2\\ =-5x^3+\left(2x+2x\right)+\left(-3-2\right)+x^2\\ =-5x^3+4x-5+x^2\)
\(M\left(x\right)=P\left(x\right)+Q\left(x\right)\\ =5x^3-4x+7+\left(-5x^3\right)+4x-5-x^2\\ =\left(5x^3-5x^3\right)+\left(-4x+4x\right)+\left(7-5\right)-x^2\\ =2-x^2\\ N\left(x\right)=P\left(x\right)-Q\left(x\right)\\ =5x^3-4x+7-\left(-5x^3+4x-5+x^2\right)\\ =5x^3-4x+7+5x^3-4x+5-x^2\\ =\left(5x^3+5x^3\right)+\left(-4x-4x\right)+\left(7+5\right)+x^{^2}\\ =10x^3-8x+12+x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(P\left(x\right)=5x^3-4x+7\)
\(Q\left(x\right)=-5x^3-x^2+4x-5\)
b: \(M\left(x\right)=-x^2+2\)
\(N\left(x\right)=10x^3+x^2-8x+12\)
c: Đặt M(x)=0
=>2-x2=0
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ab, \(M\left(x\right)=x+7-2x-5=-x+2\)
c, \(x+7=-\left(-2x-5\right)\Leftrightarrow x+7=2x+5\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
cái Q(x)=\(5x^2-4x^3-2x+7\)
mik ghi nhầm xin lổy đc chx
a) \(P\left(x\right)=6x^3-3x^2+5x-1\)
\(Q\left(x\right)=5x^2-4x^2-2x+7=\left(5x^2-4x^2\right)-2x+7=x^2-2x+7\) ( Kết quả này cũng giống như sắp xếp nhé)
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`@` `\text {Ans}`
`\downarrow`
`a)`
`P(x) =`\(3x^2+7+2x^4-3x^2-4-5x+2x^3\)
`= (3x^2 - 3x^2) + 2x^4 + 2x^3 - 5x + (7-4)`
`= 2x^4 + 2x^3 - 5x + 3`
`Q(x) =`\(3x^3+2x^2-x^4+x+x^3+4x-2+5x^4\)
`= (5x^4 - x^4) + (3x^3 + x^3) + 2x^2 + (x + 4x)- 2`
`= 4x^4 + 4x^3 + 2x^2 + 5x - 2`
`b)`
`P(-1) = 2*(-1)^4 + 2*(-1)^3 - 5*(-1) + 3`
`= 2*1 + 2*(-1) + 5 + 3`
`= 2 - 2 + 5 + 3`
`= 8`
___
`Q(0) = 4*0^4 + 4*0^3 + 2*0^2 + 5*0 - 2`
`= 4*0 + 4*0 + 2*0 + 5*0 - 2`
`= -2`
`c)`
`G(x) = P(x) + Q(x)`
`=> G(x) = 2x^4 + 2x^3 - 5x + 3 + 4x^4 + 4x^3 + 2x^2 + 5x - 2`
`= (2x^4 + 4x^4) + (2x^3 + 4x^3) + 2x^2 + (-5x + 5x) + (3 - 2)`
`= 6x^4 + 6x^3 + 2x^2 + 1`
`d)`
`G(x) = 6x^4 + 6x^3 + 2x^2 + 1`
Vì `x^4 \ge 0 AA x`
`x^2 \ge 0 AA x`
`=> 6x^4 + 2x^2 \ge 0 AA x`
`=> 6x^4 + 6x^3 + 2x^2 + 1 \ge 0`
`=> G(x)` luôn dương `AA` `x`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) P(x)=5x3 - 3x - x + 7
Q(x)=-5x3- x2 + 2x + 2x -3 - 2
b) P(x) + Q(x) = ( 5x3- 3x - x + 7)+ ( -5x3- x2 + 2x + 2x - 3 - 2 )
=5x3 - 3x - x + 7 - 5x3 - x2 + 2x + 2x - 3 - 2
=(5x3-5x3)+(-x2)+(-3x-x+2x+2x)+(7-3-2)
=> M = -x2+2
P(x)-Q(x)= (5x3-3x-x+7)-(-5x3-x2+2x+2x-3-2)
= 5x3-3x-x+7+5x3-x2+2x+2x-3-2
=(5x3+5x3)+(-x2)+(-3x-x+2x+2x)+(7-3-2)
=> N =10x3 -x2 +2
c)-x2+2=0
-x2=0+2
-x2=2
=>-x2=\(-\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P(x) = 5x3 - 3x + 7 - x = 5x3 + ( -3x - x ) + 7 = 5x3 - 4x + 7
Q(x) = -5x3 + 2x - 3 + 2x - x2 - 2 = -5x3 + ( 2x + 2x ) - x2 + ( -3 - 2 ) = -5x3 + 4x - x2 - 5
M(x) = P(x) + Q(x)
= 5x3 - 4x + 7 + ( -5x3 + 4x - x2 - 5 )
= ( 5x3 - 5x3 ) + ( 4x - 4x ) - x2 + ( 7 - 5 )
= -x2 + 2
N(x) = P(x) - Q(x)
= ( 5x3 - 4x + 7 ) - ( -5x3 + 4x - x2 - 5 )
= 5x3 - 4x + 7 + 5x3 - 4x + x2 + 5
= ( 5x3 + 5x3 ) + ( -4x - 4x ) + x2 + ( 7 + 5 )
= 10x3 - 8x + x2 + 12
M(x) = 0 <=> -x2 + 2 = 0
<=> -x2 = -2
<=> x2 = 2
<=> x = \(\pm\sqrt{2}\)
Vậy nghiệm của M(x) là \(\pm\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-5.\left(\dfrac{1}{2}\right)^3+3\left(\dfrac{1}{2}\right)^2+\dfrac{2}{2}+5-5\left(\dfrac{1}{2}\right)^3+6\left(\dfrac{1}{2}\right)^2+\dfrac{2}{2}+5\)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-\dfrac{5.1}{8}+\dfrac{3.1}{4}+6-\dfrac{5.1}{8}+\dfrac{6.1}{4}+6\)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-\dfrac{5}{8}+\dfrac{3}{4}+6-\dfrac{5}{8}+\dfrac{3}{2}+6\)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: P(x)=x^4-2x^4-5x^3-7x^2+2x-1
=-x^4-5x^3-7x^2+2x-1
Q(x)=3x^4-2x^4+5x^3+6x^2-2x+5
=x^4+5x^3+6x^2-2x+5
Rồi em cần làm gì với hai đa thức đó?