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\(f\left(x\right)=x^3-2x^2+3x+2\)
\(g\left(x\right)=-x^3-3x^2+2\)
Bài 1:
a) Ta có: \(P\left(x\right)=3x^4+2x^2-3x^4-2x^2+2x-5\)
\(=\left(3x^4-3x^4\right)+\left(2x^2-2x^2\right)+2x-5\)
\(=2x-5\)
Bài 1:
b)
\(P\left(-1\right)=2\cdot\left(-1\right)-5=-2-5=-7\)
\(P\left(3\right)=2\cdot3-5=6-5=1\)
\(a,\)
\(\Rightarrow f\left(x\right)=x^4-x^3+3x-1\)
\(\Rightarrow g\left(x\right)=x^4+4x^3+x-5\)
\(b,\)
\(A\left(x\right)=f\left(x\right)-g\left(x\right)=x^4-x^3+3x-1-x^4-4x^3-x+5\)
\(=-5x^3-x+4\)
\(B\left(x\right)=f\left(x\right)+g\left(x\right)=x^4-x^3+3x-1+x^4+4x^3+x-5\)
\(=2x^4+3x^3+4x-6\)
\(c,\)
Thay \(x=-2\) vào \(A\left(x\right)\) , ta được :
\(A\left(x\right)=-5.\left(-2\right)^3+2+4=46\)
Thay \(x=2\) vào \(A\left(x\right)\) , ta được :
\(A\left(x\right)=-5.2^3-2+4=-38\)
\(a) f ( x ) = 2 x ^4 + 3 x ^2 − x + 1 − x ^2 − x ^4 − 6 x ^3\)
\(= ( 2 x ^4 − x ^4 ) − 6 x ^3 + ( 3 x ^2 − x ^2 ) − x + 1\)
\(= x ^4 − 6 x ^3 + 2 x ^2 − x + 1\)
\(g ( x ) = 10 x ^3 + 3 − x ^4 − 4 x ^3 + 4 x − 2 x ^2\)
\(= − x ^4 + ( 10 x ^3 − 4 x ^3 ) − 2 x ^2 + 4 x + 3\)
\(= − x ^4 + 6 x ^3 − 2 x ^2 + 4 x + 3\)
\(b) f ( x ) + g ( x ) = x ^4 − 6 x ^3 + 2 x ^2 − x + 1 − x ^4 + 6 x ^3 − 2 x ^2 + 4 x + 3\)
\(= ( x ^4 − x ^4 ) + ( − 6 x ^3 + 6 x ^3 ) + ( 2 x ^2 − 2 x ^2 ) + ( − x + 4 x ) + ( 1 + 3 )\)
\(= 3 x + 4\)
c)Có \(h ( x ) = f ( x ) + g ( x ) = 3 x + 4\)
\(Cho h ( x ) = 0 ⇒ 3 x + 4 = 0\)
\(⇒ 3 x = − 4\)
\(⇒ x = − \frac{4 }{3} \)
Vậy \(x=-\frac{4}{3}\) là nghiệm của \(h ( x ) \)
a: \(F\left(x\right)=x^3+2x^2+3x+4\)
\(G\left(x\right)=x^3-x^2+3x+1\)
b: \(F\left(x\right)+G\left(x\right)=2x^3+x^2+6x+5\)
\(F\left(x\right)-G\left(x\right)=3x^2+3\)
Bài 1:
a) Ta có: \(f\left(x\right)=x^2+2x^4+10x-3x^2+x^2-x+5\)
\(=2x^4-x^2+9x+5\)
Ta có: \(g\left(x\right)=x-5x-x^2-x^4+3x+x^2-2x^2-2x^3-3x\)
\(=-x^4-2x^3-2x^2-4x\)
b) Ta có: \(f\left(x\right)+g\left(x\right)\)
\(=2x^4-x^2+9x+5-x^4-2x^3-2x^2-4x\)
\(=x^4-2x^3-3x^2+5x+5\)
Ta có: \(f\left(x\right)-g\left(x\right)\)
\(=2x^4-x^2+9x+5+x^4+2x^3+2x^2+4x\)
\(=3x^4+2x^3+x^2+13x+5\)
c) Ta có: \(f\left(x\right)+g\left(x\right)=x^4-2x^3-3x^2+5x+5\)
nên khi x=-1 thì \(f\left(-1\right)+g\left(-1\right)=\left(-1\right)^4-2\cdot\left(-1\right)^3-3\cdot\left(-1\right)^2+5\cdot\left(-1\right)+5\)
\(=1+2-3-5+5\)
\(=0\)
Ta có: \(f\left(x\right)-g\left(x\right)=3x^4+2x^3+x^2+13x+5\)
nên khi x=-1 thì \(f\left(-1\right)-g\left(-1\right)=3\cdot\left(-1\right)^4+2\cdot\left(-1\right)^3+\left(-1\right)^2+12\cdot\left(-1\right)+5\)
\(=3+2+1-12+5\)
\(=-1\)