\(f\left(x\right)=x^3-2ax+a^2\)

\(\Rightarrow f\left(1\right)=1-2a+a^2\)

\(g\left(x\right)=x^3+\left(3a+1\right)x+a^2\)

\(\Rightarrow g\left(3\right)=27+\left(3a+1\right)3+a^2\)

Mà \(f\left(1\right)=g\left(3\right)\)

\(\Rightarrow1-2a+a^2=27+\left(3a+1\right)3+a^2\)

\(\Rightarrow1-2a=27+9a+3\)

\(\Rightarrow1-2a=30+9a\)

\(\Rightarrow-29=11a\)

\(\Rightarrow a=\dfrac{-29}{11}\)

Vậy \(a=\dfrac{-29}{11}\) thì \(f\left(1\right)=g\left(3\right)\)

tra loi qua hay

Các bạn ơi mình làm đc rồi nhé!

Làm như nào vậy

a: f(a)=g(a)

=>5a-3=-1/2a+1

=>5,5a=4

=>\(a=\dfrac{4}{5.5}=\dfrac{8}{11}\)

b: f(b-2)=g(2b+4)

=>\(5\left(b-2\right)-3=-\dfrac{1}{2}\left(2b+4\right)+1\)

=>\(5b-13=-b-2+1=-b-1\)

=>6b=12

=>b=2

f(a) = g(a)

⇔ 5a - 3 = -a/2 + 1

⇔ 5a + a/2 = 1 + 3

⇔ 11a/2 = 4

⇔ 11a = 8

⇔ a = 8/11

Vậy a = 8/11 thì f(a) = g(a)

b) f(b - 2) = g(2b + 4)

⇔ 5.(b - 2) - 3 = -(2b + 4)/2 + 1

⇔ 5b - 10 - 3 = -b - 2 + 1

⇔ 5b + b = 1 + 13

⇔ 6b = 14

⇔ b = 7/3

Vậy b = 7/3 thì f(b - 2) = g(2b + 4)

f(x)=x^3-2x^2+3x+1

g(x)=x^3+x^2-5x+3

a: f(-1/3)=-1/27-2/9-1+1=-1/27-6/27=-7/27

g(-2)=-8+4+10+3=17-8=9

b: f(x)-g(x)=x^3-2x^2+3x+1-x^3-x^2+5x-3

=x^2+8x-2

f(x)+g(x)

=x^3-2x^2+3x+1+x^3+x^2-5x+3

=2x^3-x^2-2x+4

VT = (3a + 2b - 1)(a + 5) - 2b(a - 2)

= 3a² + 2ab - a + 15a + 10b - 5 - 2ab + 4b

= 3a² + 14a + 14b - 5 

= 3a² + 9a + 5a + 15 + 14b - 20

= 3a(a + 3) + 5(a + 3) + 2(7b - 10)

= (3a + 5)(a + 3) + 2(7b - 10)

= VP (đpcm)

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