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a)f(x)+g(x)=\(x^5-4x^4-2x^2-7-2x^5+6x^4-2x^2+6.\)
=\(-x^5+2x^4-4x^2-1\)
f(x)-g(x)=\(x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
=\(3x^5-10x^4-13\)
b)f(x)+g(x)=\(5x^4+7x^3-6x^2+3x-7-4x^4+2x^3-5x^2+4x+5\)
=\(x^4+9x^3-11x^2+7x-2\)
f(x)-g(x)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)
=\(9x^4+5x^3-x^2-x-12\)
a )
\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)
\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
\(f\left(x\right)=x^5-4x^4-2x^2-7\)
\(g\left(x\right)=-2x^5+6x^4-2x^2+6\)
\(f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)
\(f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)
Câu a nhé: 2x . x^2 - 2x . 7x - 2x . 3 = 2x^3 - 14x^2 - 6x
a)\(f\left(x\right)=\left(3x+4\right)\cdot\left(5x-1\right)+\left(5x+2\right)\cdot\left(1-3x\right)+2\)
\(=15x^2-3x+20x-4+5x-15x^2+2-6x+2\)
\(=16x\)
b)\(g\left(x\right)=\left(5x-1\right)\cdot\left(2x+3\right)-3\cdot\left(3x-1\right)\)
\(=10x^2+15x-2x-3-9x+3\)
\(=10x^2+4x\)
a) \(f\left(x\right)=x^3+2x^2+x+2=0\)
\(\Leftrightarrow\)\(\left(x+2\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x+2=0\)
\(\Leftrightarrow\)\(x=-2\)
Vậy...
b) \(g\left(x\right)=\left(x-1\right)x^2+\left(x-1\right)x+4x+4=0\)
\(\Leftrightarrow\)\(x\left(x-1\right)\left(x+1\right)+4\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-x+4\right)=0\)
\(\Leftrightarrow\)\(x+1=0\)
\(\Leftrightarrow\)\(x=-1\)
Vậy...
c) \(h\left(x\right)=x^4-x^3+5x^2-5x=0\)
\(\Leftrightarrow\)\(x\left(x-1\right)\left(x^2+5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy...