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\(a,f\left(x\right)+g\left(x\right)=5x^2-2x+5+5x^2-6x-\dfrac{1}{3}\\ =10x^2-8x+\dfrac{14}{3}\\ b,f\left(x\right)-g\left(x\right)=5x^2-2x+5-5x^2+6x+\dfrac{1}{3}\\ =4x+\dfrac{16}{3}\\ c,f\left(x\right)-g\left(x\right)=4x+\dfrac{16}{3}=0\\ \Leftrightarrow4x=-\dfrac{16}{3}\Leftrightarrow x=-\dfrac{4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) f(x) + g(x) = \(5x^2-2x+5+5x^2-6x-\dfrac{1}{3}=10x^2-8x+\dfrac{14}{3}\)
b) f(x) - g(x) = \(5x^2-2x+5-5x^2+6x+\dfrac{1}{3}=4x+\dfrac{16}{3}\)
c) Ngiệm của f(x) - g(x) chính là nghiệm của \(4x+\dfrac{16}{3}\)
Ta có: \(4x+\dfrac{16}{3}=0\Leftrightarrow4x=-\dfrac{16}{3}\Leftrightarrow x=-\dfrac{4}{3}\)
Vậy nghiệm của f(x) - g(x) là \(-\dfrac{4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(f\left(x\right)+g\left(x\right)=5x^2-4x+13+9x-7-5x^2=5x+6\)
\(f\left(x\right)-g\left(x\right)=5x^2-4x+13-9x+7+5x^2=10x^2-13x+20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) f(x) = x(x - 5) + 2(x - 5)
x(x - 5) + 2(x - 5) = 0
<=> (x - 5)(x - 2) = 0
x - 5 = 0 hoặc x - 2 = 0
x = 0 + 5 x = 0 + 2
x = 5 x = 2
=> x = 5 hoặc x = 2
a, f(x) có nghiệm
\(\Leftrightarrow x\left(x-5\right)+2\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)
->tự kết luận.
b1, để g(x) có nghiệm thì:
\(g\left(x\right)=2x\left(x-2\right)-x^2+5+4x=0\)
\(\Rightarrow2x^2-4x-x^2+5+4x=0\)
\(\Rightarrow x^2+5=0\)
Do \(x^2\ge0\forall x\)nên\(x^2+5\ge5\forall x\)
suy ra: k tồn tại \(x^2+5=0\)
Vậy:.....
b2,
\(f\left(x\right)=x\left(x-5\right)+2\left(x-5\right)\)
\(=x^2-5x+2x-10\)
\(=x^2-3x-10\)
\(f\left(x\right)-g\left(x\right)=x^2+5-\left(x^2-3x-10\right)\)
\(=x^2+5-x^2+3x-10=3x-5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(f\left(-2\right)=5\cdot4-8-8=4\)
b: \(f\left(x\right)+g\left(x\right)=6x^2+2x-8\)
c: Đặt G(x)=0
=>x(x-2)=0
=>x=0 hoặc x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
a: f(x)=3x^4+2x^3+6x^2-x+2
g(x)=-3x^4-2x^3-5x^2+x-6
b: H(x)=f(x)+g(x)
=3x^4+2x^3+6x^2-x+2-3x^4-2x^3-5x^2+x-6
=x^2-4
f(x)-g(x)
=3x^4+2x^3+6x^2-x+2+3x^4+2x^3+5x^2-x+6
=6x^4+4x^3+11x^2-2x+8
c: H(x)=0
=>x^2-4=0
=>x=2 hoặc x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(f\left(x\right)-g\left(x\right)=\left[x\left(x^2-2x+7\right)-1\right]-\left[x\left(x^2-2x-1\right)-1\right]\)
\(f\left(x\right)-g\left(x\right)=x^3-2x^2+7x-1-x^3+2x^2+x+1\)
\(f\left(x\right)-g\left(x\right)=8x\)
\(f\left(x\right)+g\left(x\right)=x\left(x^2-2x+7\right)-1+x\left(x^2-2x-1\right)-1\)
\(f\left(x\right)+g\left(x\right)=x^3-2x^2+7x-1+x^3-2x^2-x-1\)
\(f\left(x\right)+g\left(x\right)=2x^3-4x^2+6x-2\)
b) 8x=0
=> x=0
=> Nghiệm đa thức f(x)-g(x)
c) Thay \(x=-\frac{3}{2}\)vào BT f(x)+g(x) ta được :
\(2.\left(-\frac{3}{2}\right)^3-4\left(-\frac{3}{2}\right)^2+6\left(-\frac{3}{2}\right)-2\)
\(=6,75+9-9-2\)
\(=4,75\)
#H
![](https://rs.olm.vn/images/avt/0.png?1311)
a) cho f(x )=0
\(=>2x^2-x=0=>x\left(2x-1\right)=0\)
\(=>\left[{}\begin{matrix}x=0\\2x=1\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)cho \(f\left(2\right)+g\left(2\right)=0\)
\(=>2.2^2-2+m.2^2+2m+1=0\)
\(8-2+4m+2m+1=0\)
\(6+2m\left(2+1\right)+1=0\)
\(6+6m=-1\)
\(6m=-7=>m=-\dfrac{7}{6}\)
a)f(x)+g(x)=10xmũ2-8x+ 14/3
b)f(x)-g(x)=10x mũ 2 +4x+16/3
nghiệm chưa tính ddcj nha
a;\(f\left(x\right)+g\left(x\right)=\left(5x^2-2x+5\right)+\left(5x^2-6x-\frac{1}{3}\right)=25x^2-8x+\frac{1}{4}\)
b'\(f\left(x\right)-g\left(x\right)=\left(5x^2-2x+5\right)-\left(5x^2-6x-\frac{1}{3}\right)=4x+\frac{16}{3}\)
c;\(f\left(x\right)-g\left(x\right)=0\Leftrightarrow4x+\frac{16}{3}=0\)
\(\Leftrightarrow4x=-\frac{16}{3}\)
\(\Leftrightarrow x=-\frac{4}{3}\)
Vậy nghiệm của đa thức f(x)-g(x) là : x=-4/3