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Bài 1:
a)
\(F+G+H=(x^3-2x^2+3x+1)+(x^3+x-1)+(2x^2-1)\)
\(=2x^3+4x-1\)
b)
\(F-G+H=0\)
\(\Leftrightarrow (x^3-2x^2+3x+1)-(x^3+x-1)+(2x^2-1)=0\)
\(\Leftrightarrow 2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Bài 2:
a)
\(A=-4x^5-x^3+4x^2-5x+9+4x^5-6x^2-2\)
\(=(-4x^5+4x^5)-x^3+(4x^2-6x^2)-5x+(9-2)\)
\(=-x^3-2x^2-5x+7\)
\(B=-3x^4-2x^3+10x^2-8x+5x^3\)
\(=-3x^4+(5x^3-2x^3)+10x^2-8x\)
\(=-3x^4+3x^3+10x^2-8x\)
b)
\(P=A+B=(-x^3-2x^2-5x+7)+(-3x^4+3x^3+10x^2-8x)\)
\(=-3x^4+(3x^3-x^3)+(10x^2-2x^2)-(8x+5x)+7\)
\(=-3x^4+2x^3+8x^2-13x+7\)
\(P(-1)=-3.(-1)^4+2(-1)^3+8(-1)^2-12(-1)+7=23\)
\(Q=A-B=(-x^3-2x^2-5x+7)-(-3x^4+3x^3+10x^2-8x)\)
\(=3x^4-(x^3+3x^3)-(2x^2+10x^2)+(8x-5x)+7\)
\(=3x^4-4x^3-12x^2+3x+7\)
a) Q(x) - P(x)
= x4 + 4x3 + 2x2 - 4x + 1 - (x4 + 2x2 + 1)
= x4 + 4x3 + 2x2 - 4x + 1 - x4 - 2x2 - 1
= (x4 - x4) + 4x3 + (2x2 - 2x2) - 4x + (1 - 1)
= 4x3 - 4x
b) Để Q(x) - P(x) = 0 thì 4x3 - 4x = 0
=> 4x (x2 - 1) = 0
=> 4x = 0 hoặc x2 - 1 = 0
=> x = 0 hoặc x2 = 1
=> x thuộc {-1 ; 0 ; 1}
a, Q(x) - P(x) = \(x^4+2x^2+1-x^4-4x^3-2x^2+4x-1\)
\(=-4x^3+4x\)
Vậy Q(x) - P(x) = \(-4x^3+4x\)
b, Để Q(x) - P(x) = 0 thì \(-4x^3+4x=0\)
\(\Leftrightarrow-4x\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{1}\end{cases}}}\)
Vậy x = ...
a: \(f\left(x\right)=6x^4-3x^2-5\)
\(g\left(x\right)=x^4+3x^3-5x^2-4x+2\)
\(f\left(x\right)+g\left(x\right)=7x^4+3x^3-8x^2-4x-3\)
b: \(f\left(x\right)-g\left(x\right)=5x^4-3x^3+2x^2+4x-7\)
c: \(f\left(1\right)=6-3-5=-2\)
\(g\left(2\right)=16+3\cdot8-5\cdot4-4\cdot2+2\)
=16+24-20-16+2
=40-20-16+2
=20-16+2
=4+2=6
a) ( 3x - 1 )2 - 4
= ( 3x - 1 ) - 22
= ( 3x - 1 - 2 )( 3x - 1 + 2 )
= ( 3x - 3 )( 3x + 1 )
= 3( x - 1 )( 3x + 1 )
b) ( x + y )2 - x2
= ( x + y - x )( x + y + x )
= y( 2x + y )
c) 100 - ( 2x - y )2
= 102 - ( 2x - y )2
= [ 10 - ( 2x - y ) ][ 10 + ( 2x - y ) ]
= ( 10 - 2x + y )( 10 + 2x - y )
d) ( 2x - 1 )2 - ( x - 1 )2
= [ ( 2x - 1 ) - ( x - 1 ) ][ ( 2x - 1 ) + ( x - 1 ) ]
= ( 2x - 1 - x + 1 )( 2x - 1 + x - 1 )
= x( 3x - 2 )
e) 4( x + 6 )2 - 9( 1 + x )2
= 22( x + 6 )2 - 32( 1 + x )2
= ( 2x + 12 )2 - ( 3 + 3x )2
= [ ( 2x + 12 ) - ( 3 + 3x ) ][ ( 2x + 12 + ( 3 + 3x ) ]
= ( 2x + 12 - 3 - 3x )( 2x + 12 + 3 + 3x )
= ( 9 - x )( 5x + 15 )
= 5( 9 - x )( x + 3 )
a. Ta có: P=\(-3x^2+x-5x^3+2x+x^2+3x^3-4\)4
\(P=-2x^3-2x^2+3x-5\)
\(Q=7x^3-3-x-3x^2-4x-4x^3+x^2-3x^3\)
\(\Rightarrow Q=-2x^2-5x-3\)