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a: \(F\left(x\right)=x^5-3x^2+x^3-x^2-2x+5\)
\(=x^5+x^3-4x^2-2x+5\)
\(G\left(x\right)=x^5-x^4+x^2-3x+x^2+1\)
\(=x^5-x^4+2x^2-3x+1\)
b: Ta có: \(H\left(x\right)=F\left(x\right)+G\left(x\right)\)
\(=x^5+x^3-4x^2-2x+5+x^5-x^4+2x^2-3x+1\)
\(=2x^5-x^4+x^3-2x^2-5x+6\)
a: f(x)=x^3-2x^2+2x-5
g(x)=-x^3+3x^2-2x+4
b: Sửa đề: h(x)=f(x)+g(x)
h(x)=x^3-2x^2+2x-5-x^3+3x^2-2x+4=x^2-1
c: h(x)=0
=>x^2-1=0
=>x=1 hoặc x=-1
\(a) f ( x ) = 2 x ^4 + 3 x ^2 − x + 1 − x ^2 − x ^4 − 6 x ^3\)
\(= ( 2 x ^4 − x ^4 ) − 6 x ^3 + ( 3 x ^2 − x ^2 ) − x + 1\)
\(= x ^4 − 6 x ^3 + 2 x ^2 − x + 1\)
\(g ( x ) = 10 x ^3 + 3 − x ^4 − 4 x ^3 + 4 x − 2 x ^2\)
\(= − x ^4 + ( 10 x ^3 − 4 x ^3 ) − 2 x ^2 + 4 x + 3\)
\(= − x ^4 + 6 x ^3 − 2 x ^2 + 4 x + 3\)
\(b) f ( x ) + g ( x ) = x ^4 − 6 x ^3 + 2 x ^2 − x + 1 − x ^4 + 6 x ^3 − 2 x ^2 + 4 x + 3\)
\(= ( x ^4 − x ^4 ) + ( − 6 x ^3 + 6 x ^3 ) + ( 2 x ^2 − 2 x ^2 ) + ( − x + 4 x ) + ( 1 + 3 )\)
\(= 3 x + 4\)
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 \(h ( x ) \)
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)
\(P\left(-1\right)=2\cdot\left(-1\right)-5=-2-5=-7\)
\(P\left(3\right)=2\cdot3-5=6-5=1\)
\(a)\)
\(f\left(x\right)=-x^3+3x^2+x-3+2^3-x^2\)
\(\Leftrightarrow f\left(x\right)=-x^3+\left(3x^2-x^2\right)+x-3+2^3\)
\(\Leftrightarrow f\left(x\right)=-x^3+2x^2+x-3+8\)
\(\Leftrightarrow f\left(x\right)=-x^3+2x^2+x+5\)
\(g\left(x\right)=-3x^3-x^2+2x^3+5x-3-4x\)
\(\Leftrightarrow g\left(x\right)=\left(-3x^3+2x^3\right)-x^2+\left(5x-4x\right)-3\)
\(\Leftrightarrow g\left(x\right)=-x^3-x^2+x-3\)
\(b)\)
Theo đề ra: \(h\left(x\right)=f\left(x\right)+g\left(x\right)\)
\(f\left(x\right)=-x^3+2x^2+x+5\)
\(g\left(x\right)=-x^3-x^2+x-3\)
\(\Rightarrow h\left(x\right)=x^2+2x+2\)
1:
a: f(x)=2x^4+2x^3+2x^2+5x+6
g(x)=x^4-2x^3-x^2-5x+3
c: h(x)=2x^4+2x^3+2x^2+5x+6+x^4-2x^3-x^2-5x+3=3x^4+x^2+9
K(x)=f(x)-2g(x)-4x^2
=2x^4+2x^3+2x^2+5x+6-2x^4+4x^3+2x^2+10x-6-4x^2
=6x^3+15x
c: K(x)=0
=>6x^3+15x=0
=>3x(2x^2+5)=0
=>x=0
d: H(x)=3x^4+x^2+9>=9
Dấu = xảy ra khi x=0
\(a)\)
\(f\left(x\right)=2x.\left(x^2-3\right)-4.\left(1-2x\right)+x^2.\left(x-2\right)+\left(5x+3\right)\)\(=2x^3-6x-4+8x+x^3-2x^2+5x+3=3x^3+7x-1-2x^2=3x^3-2x^2+7x-1\)\(g\left(x\right)=-3.\left(1-x^2\right)-2.\left(x^2-2x-1\right)=-3+3x^2-2x^2+4x+2=-1+x^2+4x=x^2+4x-1\)
\(b)\)
\(h\left(x\right)=f\left(x\right)-g\left(x\right)=\left(3x^3-2x^2+7x-1\right)-\left(-1+x^2+4x\right)=x^2+4x-1=3x^3-2x^2+7x-1+1-x^2-4x=3x^3-3x^2+3x\)
\(\text{Xét}:\)
\(3x^3-3x^2+3x=0\)
\(\rightarrow3x.\left(x^2-x+1\right)=0\)
\(\rightarrow x.\left(x^2-x+1\right)=0\)
\(\rightarrow\orbr{\begin{cases}3x.\left(x^2-x+1\right)=0\\x.\left(x^2-x+1\right)=0\end{cases}}\) \(\rightarrow\orbr{\begin{cases}x=0\\x^2-x+1=0\end{cases}}\)
\(\rightarrow\orbr{\begin{cases}x=0\\x\notinℝ\end{cases}}\) \(\rightarrow x=0\)
\(\text{Vậy nghiệm của}\)\(h\left(x\right)\)\(\text{là}:\)\(0\)