câu 4: b, đề bài là tính giá trị của A tại x =-1/2;y=-1

Tk

Bài 2

a) F(x)-G(x)+H(x)= \(x^3-2x^2+3x+1-\left(x^3+x-1\right)+\left(2x^2-1\right)\)

= \(x^3-2x^2+3x+1-x^3-x+1+2x^2-1\)

=  \(x^3-x^3-2x^2+2x^2+3x-x+1+1-1\)

=  2x + 1

b) 2x + 1 = 0

 2x = -1

 x=\(\dfrac{-1}{2}\)

Tk

Bài 3

a)

f(x) + g(x)

\(x^3-2x+1+\left(2x^2-x^3+x-3\right)\)

\(x^3-2x+1+2x^2-x^3+x-3\)

\(x^3-x^3-2x+x+1-3+2x^2\)

\(-x-2+2x^2\)

f(x) - g(x)

\(x^3-2x+1-\left(2x^2-x^3+x-3\right)\)

\(x^3-2x+1-2x^2+x^3-x+3\)

\(x^3+x^3-2x-x+1+3-2x^2\)

\(2x^3-3x+4-2x^2\)

b)

Thay x = -1, ta có:

\(-\left(-1\right)-2+2\left(-1\right)^2\) = 1

x = -2, ta có

\(2\left(-2\right)^3-3\left(-2\right)+4-2\left(-2\right)^2\)

\(2\cdot\left(-8\right)+6+4-8\) = -14

\(#HaimeeOkk\)

\(a)\)

\(f ( x ) + g ( x ) = ( x ^3 − 2 x + 1 ) + ( 2 x ^2 − x ^3 + x − 3 ) \)

\(f ( x ) + g ( x ) = x ^3 − 2 x + 1 + 2 x ^2 − x ^3 + x − 3 \)

\(f ( x ) + g ( x ) = x ^3 − x ^3 + 2 x ^2 − 2 x + x + 1 − 3 \)

\(f ( x ) + g ( x ) = 2 x ^2 − x − 2\)

\(f ( x ) − g ( x ) = ( x ^3 − 2 x + 1 ) − ( 2 x ^2 − x ^3 + x − 3 ) \)

\(f ( x ) − g ( x ) =x^3- 2 x + 1 −2x^2+x^3-x+3\)

\(f ( x ) − g ( x ) = x ^3 + x ^3 − 2 x ^2 − 2 x − x + 1 + 3 \)

\(f ( x ) − g ( x ) = 2 x ^3 − 2 x ^2 − 3 x + 4\)

\(-----------------------------\)

\(b)\)

Thay \(x=-1\) vào \(f ( x ) + g ( x )\)

\(f ( x ) + g ( x ) = 2 x ^2 − x − 2\)

\(⇒ 2 ( − 1 ) ^2 − ( − 1 ) − 2 = 1\)

Thay \(x=-2\) vào \(f ( x ) + g ( x )\)

\(f ( x ) + g ( x ) = 2 x ^2 − x − 2\)

\(⇒ 2 ( − 2 ) ^2 − ( − 2 ) − 2 = 8\)

a: \(F\left(x\right)=x^3+2x^2+3x+4\)

\(G\left(x\right)=x^3-x^2+3x+1\)

b: \(F\left(x\right)+G\left(x\right)=2x^3+x^2+6x+5\)

\(F\left(x\right)-G\left(x\right)=3x^2+3\)

f(x)=x³ +2x²+3x+4

g(x)=x³ trừ x²+3x+1

a)

F(x)=x3+2x2+3x+4F(x)=x3+2x2+3x+4

G(x)=x3−x2+3x+1

b)

F(x)+G(x)=2x3+x2+6x+5F(x)+G(x)=2x3+x2+6x+5

F(x)−G(x)=3x2+3

Trừ biểu thức 2 cho biểu thức thứ 3 ta được: 

[g(x)+h(x)]-[f(x)+g(x)] = 2x²-2x+1-x²+4x-2

<=> h(x)-f(x) = x²+2x-1

Lại có: h(x)+f(x) = x²+2x+1

=> 2.f(x) = x²+2x+1-x²-2x+1 = 2

=> f(x) = 1

Đáp số: f(x) = 1 

Có \(f\left(x\right)=3x-2x+1\) và \(g\left(x\right)=2x2-3x+x-3\)

a) \(f\left(x\right)+g\left(x\right)=\left(3x-2x+1\right)+\left(2x2-3x+x-3\right)\)

\(f\left(x\right)+g\left(x\right)=\left(x+1\right)+\left(4x-3x+x-3\right)=\left(x+1\right)+\left(2x-3\right)\)

\(f\left(x\right)+g\left(x\right)=x+1+2x-3=\left(2x+x\right)\left(3-1\right)=3x-2\)

\(f\left(x\right)-g\left(x\right)=\left(3x-2x+1\right)-\left(2x2-3x+x-3\right)\)

\(f\left(x\right)-g\left(x\right)=\left(x+1\right)-\left(4x-3x+x-3\right)=\left(x+1\right)-\left(2x-3\right)\)

\(f\left(x\right)-g\left(x\right)=x+1-2x+3=\left(1+3\right)-\left(2x-x\right)=4-x\)

b) Có \(f\left(-1\right)=3\left(-1\right)-2\left(-1\right)+1=\left(-3\right)-\left(-2\right)+1=0\)

Và \(g\left(-1\right)=2.2\left(-1\right)-3\left(-1\right)+\left(-1\right)-3=\left(-4\right)-\left(-3\right)-1-3=-5\)

 \(x=-1\Leftrightarrow\left[f\left(x\right)+g\left(x\right)\right]=0+\left(-5\right)=0-5=-5\)

Có \(f\left(-2\right)=3\left(-2\right)-2\left(-2\right)+1=\left(-6\right)-\left(-4\right)+1=-1\)

\(g\left(-2\right)=2.2\left(-2\right)-3\left(-2\right)+\left(-2\right)-3=\left(-8\right)-\left(-6\right)-2-3=-3\)

\(x=-2\Leftrightarrow\left[f\left(x\right)+g\left(x\right)\right]=\left(-1\right)+\left(-3\right)=-4\)

Cho f(x) = x3 - 2x + 1

       g(x) = 2x2 - x3 + x - 3

a) f(x) + g(x) =3x -2

;f(x) - g(x) = 6x 

b) Tính f(x) + g(x) tại  x = - 1; x = - 2

Haizzz ...  chép đề quá ''ngu''

\(f\left(x\right)=x^3-2x+1\)

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

Ng ta đã ko bt viết mũ thì pk viết hộ ng ta chứ ? Tối cổ hả Minh ? 

a) Tính

 \(f\left(x\right)-g\left(x\right)+h\left(x\right)=\left(x^3-2x^2+3x+1\right)-\left(x^3+x-1\right)+\left(2x^2+2\right)\)

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

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

\(=2x+4\)

\(f\left(x\right)+g\left(x\right)+h\left(x\right)=\left(x^3-2x^2+3x+1\right)+\left(x^3+x-1\right)+\left(2x^2+2\right)\)

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

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

\(=2x^3+4x+2\)

\(f\left(x\right)-g\left(x\right)-h\left(x\right)=\left(x^3-2x^2+3x+1\right)-\left(x^3+x-1\right)-\left(2x^2+2\right)\)

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

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

\(=-4x^2+2x\)

b) Tìm x

\(f\left(x\right)-g\left(x\right)+h\left(x\right)=0\)

\(2x+4=0\)

\(2x=0-4=-4\)

\(x=\frac{-4}{2}=-2\)

\(f\left(x\right)-g\left(x\right)-h\left(x\right)=0\)

\(-4x^2+2x=0\)

\(-4x^2=-2x\)

\(x^2=\frac{-1}{2}x\)

\(\Leftrightarrow x^2+\frac{1}{2}x=0\)

\(x\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow x=0\)

Hoặc \(x+\frac{1}{2}=0\Leftrightarrow x=0-\frac{1}{2}=\frac{-1}{2}\)

a: \(f\left(x\right)+g\left(x\right)=2x^3-2x^2+4x\)

b: \(f\left(x\right)-g\left(x\right)=-2x^2+2x+2\)

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

\(f\left(x\right)+g\left(x\right)=2x^3-2x^2+4x\)

b)\(f\left(x\right)-g\left(x\right)=x^3-2x^2+3x+1-x^3-x+1\)

\(f\left(x\right)-g\left(x\right)=-2x^2+2x+2\)

`a,f(x)-g(x)+h(x)`

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

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

`=0+0+3x+1`

`=3x+1`

`b,f(x)-g(x)+h(x)=0`

`=>3x+1=0`

`=>x=-1/3`

đăng câu này lần 2

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