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1.
\(f\left(x\right)=2x^4+6x^3+8x^2+12x+1\)
2.
\(h\left(x\right)=\left(2x^4+6x^3+8x^2+12x+1\right)-\left(2x^4+6x^3+17x^2+12x-26\right)\)
\(=-9x^2+27\)
3.
\(h\left(x\right)=0\Leftrightarrow-9x^2+27=0\)
\(\Leftrightarrow x^2=3\Rightarrow x=\pm\sqrt{3}\)
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\(A\left(x\right)=x^5+3x^3-x^5+x-1=3x^3+x-1\)
Bậc : 4
\(B\left(x\right)=3x^3-2x^2-1\)
Bậc : 5
\(A\left(x\right)+B\left(x\right)=3x^3+x-1+3x^3-2x^2-1\)
\(=6x^3-2x^2+x-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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a: \(A\left(2\right)=2^5-2\cdot2^4+5\cdot2-3=32-32+10-3=7\)
\(B\left(-1\right)=-\left(-1\right)^5+3\cdot\left(-1\right)^3+5\cdot\left(-1\right)+11=1-3-5+11=4\)
b: Ta có: A(x)+B(x)
\(=x^5-2x^4+5x-3-x^5+3x^3+5x+11\)
\(=-2x^4+3x^3+10x+8\)
Ta có: A(x)-B(x)
\(=x^5-2x^4+5x-3+x^5-3x^3-5x-11\)
\(=2x^5-2x^4-3x^3-14\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
Vì : P(x) + Q(x) = x5 – 2x2 + 1
Suy ra Q(x) = x5 – 2x2 + 1– P(x).
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Ta có: f(x) + g(x) – h(x)
= (x5 – 4x3 + x2 – 2x + 1) + (x5 – 2x4 + x2 – 5x + 3) – (x4 – 3x2 + 2x – 5)
= x5 – 4x3 + x2 – 2x + 1 + x5 – 2x4 + x2 – 5x + 3 – x4 + 3x2 - 2x + 5
= (x5 +x5) – (2x4 + x4) – 4x3 + (x2 + x2 + 3x2)- (2x + 5x + 2x) + (1 + 3 + 5)
= (1 + 1)x5 – (2 + 1)x4 – 4x3 + (1 + 1 + 3)x2 - (2 + 5 + 2)x + (1 + 3 + 5)
= 2x5 – 3x4 – 4x3 + 5x2 – 9x + 9
Vì 5^(x-2).(x+3)=1
=> 5^(x-2).(x+3)=5^0
=> (x-2).(x+3)=0
=> x = 2 hoặc -3