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R(x)=x4+2x3-x2+x-3
Với x=1 ta có
R(x)=1+2-1+1-3=0
Với x=2 ta có
R(x)=16+16-4+2-3=27
Với x=-1 ta có
R(x)=1+(-2)-1+1-3=-4
Với x=0 ta có
R(x)=0+0-0+0-3=-3
Vậy chỉ có 1 là nghiệm cua R(x)
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\(\left(\dfrac{1}{4}\right)^{2n}=\left(\dfrac{1}{8}\right)^2\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2.2n}=\left(\dfrac{1}{2}\right)^{3.2}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{4n}=\left(\dfrac{1}{2}\right)^6\)
\(\Rightarrow4n=6\)
\(\Rightarrow n=\dfrac{6}{4}=\dfrac{3}{2}\)
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M+N=(3/2x6-7x+4x^5+2,5x^2)+(-3x^6+1/2^5-13/2x^2+4x)
M+N=3/2x6-7x+4x^5+2,5x^2+-3x^6+1/2^5-13/2x^2+4x
= (3/2x^6-3x^6)+(7x+4x)+(4x^5+1/2^5)+(2,5x^2-13/2x^2)
=-1,5x^6+11x+4,5x^5-4x^2
M-N=(3/2^6-7x+4x^5+2,5x^2)-(-3x^6+1/2^5-13/2x^2+4x)
=3/2^6-7x+4x^5+2,5x^2+3x^6-1/2^5+13/2x^2-4x
= (3/2x^6+3x^6)+(-7x-4x)+(4x^5-1/2^5)+(2,5x^2+13/2x^2)
= 4,5x^6-11x+3,5x^5+9x^2
N-M=(-3x^6+1/2^5-13/2x^2+4x)-(3/2^6-7x+4x^5+2,5x^2)
= -3x^6+1/2^5-13/2x^2+4x-3/2^6-7x-4x^5-2,5x^2
= (-3x^6-3/2x^6)+(1/2x^5-4x^5)+(-13/2x^2-2,5x^2)+(4x-7x)
= -4,5x^6-3,5x^5-9x^2-3x
Đáp án :
\(\frac{2^{10}.9^4}{6^4.8^2}=\frac{2^{10}.9^4}{6^4.\left(2^3\right)^2}=\frac{2^{10}.9^4}{6^4.2^6}=2^4.\left(\frac{3}{2}\right)^4=16.\frac{81}{16}=81\)
Hok tốt