A(x)2x3 2x3x2 1

B(x)2x2 3x3 x5

Dấu gì?

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\(a) A(x)=2x^3-3x^2+2x+1\\ B(x)=3x^3+2x^2-x-5\\ b) A(x)+B(x)=2x^3-3x^2+2x+1+3x^3+2x^2-x-5\\ =(2x^3+3x^3)+(-3x^2+2x^2)+(2x-x)+(1-5)\\ =5x^3-x^2+x-4\\ c) A(x)-B(x)=2x^3-3x^2+2x+1-(3x^3+2x^2-x-5)\\ = 2x^3-3x^2+2x+1-3x^3-2x^2+x+5\\ =(2x^3-3x^3)+(-3x^2-2x^2)+(2x+x)+(1+5)\\ = -x^3-5x^2+3x+6 \)

a: \(P\left(x\right)=5x^5-4x^4+2x^2+3x+6\)

\(Q\left(x\right)=-x^5+2x^4-2x^3+3x^2+x+\dfrac{1}{4}\)

b: \(P\left(x\right)+Q\left(x\right)=4x^5-2x^4-2x^3+5x^2+4x+\dfrac{25}{4}\)

a) Ta có : \(A\left(x\right)+B\left(x\right)\)

\(=2x^3+2x-3x^2+1+2x^2+3x^3-x-5\)

\(=\left(2x^3+3x^3\right)+\left(-3x^2+2x^2\right)+\left(2x-x\right)+\left(1-5\right)\)

\(=5x^3-x^2-x-4\)

b) Ta sẽ sắp xếp như sau :

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

\(B\left(x\right)=3x^3+2x^2-x-5\)

c) Ta có : \(A\left(x\right)-B\left(x\right)\)

\(=\left(2x^3+2x-3x^2+1\right)-\left(2x^2+3x^3-x-5\right)\)

\(=2x^3+2x-3x^2+1-2x^2-3x^3+x+5\)

\(=\left(2x^3-3x^3\right)+\left(-3x^2-2x^2\right)+\left(2x+x\right)+\left(1+5\right)\)

\(=-x^3-5x^2+3x+6\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`a)`

\(A(x) = 5x^5 + 2 - 7x - 4x^2 - 2x^5\)

`= (5x^5 - 2x^5) - 4x^2 - 7x + 2`

`= 3x^5 - 4x^2 - 7x + 2`

`b)`

`A(x)+B(x)`

`=`\((3x^5 - 4x^2 - 7x + 2)+(-3x^5 + 4x^2 + 3x - 7)\)

`= 3x^5 - 4x^2 - 7x + 2-3x^5 + 4x^2 + 3x - 7`

`= (3x^5 - 3x^5) + (-4x^2 + 4x^2) + (-7x + 3x) + (2-7)`

`= -4x - 5`

`b)`

`A(x) - B(x)`

`= 3x^5 - 4x^2 - 7x + 2 + 3x^5 - 4x^2 - 3x + 7`

`= (3x^5 + 3x^5) + (-4x^2 - 4x^2) + (-7x - 3x) + (2+7)`

`= 6x^5 - 8x^2 - 10x + 9`

`c)`

Thay `x=-1` vào đa thức `A(x)`

` 3*(-1)^5 - 4*(-1)^2 - 7*(-1) + 2`

`= 3*(-1) - 4*1 + 7 + 2`

`= -3 - 4 + 7 + 2`

`= -7+7 + 2`

`= 2`

Bạn xem lại đề ;-;.

`2,`

`M =` \(( 3 x - 2 )( 2 x + 1 )-( 3 x + 1 )( 2 x - 1 )\)

`= 3x(2x+1) - 2(2x+1) - [3x(2x-1) + 2x - 1]`

`= 6x^2 + 3x - 4x - 2 - (6x^2 - 3x + 2x - 1)`

`= 6x^2 - x - 2 - (6x^2 - x - 1)`

`= 6x^2 - x - 2 - 6x^2 + x + 1`

`= (6x^2 - 6x^2) + (-x+x) + (-2+1)`

`= -1`

Vậy, giá trị của biểu thức không phụ thuộc vào giá trị của biến.

2:

M=6x^2+3x-4x-2-6x^2+3x-2x+1

=-1

1;

a: A(x)=3x^5-4x^2-7x+2

b: B(x)=-3x^5+4x^2+3x-7

B(x)+A(x)

=-3x^5-4x^2-7x+2+3x^5+4x^2+3x-7

=-4x-5

A(x)-B(x)

=-3x^5-4x^2-7x+2-3x^5-4x^2-3x+7

=-6x^5-8x^2-10x+9

Bài `1`

\(a,A\left(x\right)=5x^5+2-7x-4x^2-2x^5\\ =\left(5x^5-2x^5\right)-4x^2-7x+2\\ =3x^5-4x^2-7x+2\)

\(b,A\left(x\right)+B\left(x\right)=3x^5-4x^2-7x+2+\left(-3x^5+4x^2+3x-7\right)\\ =3x^5-4x^2-7x+2-3x^5+4x^2+3x-7\\ =\left(3x^5-3x^5\right)+\left(-4x^2+4x^2\right)+\left(-7x+3x\right)+\left(2-7\right)\\ =-4x-5\)

\(A\left(x\right)-B\left(x\right)=\left(3x^5-4x^2-7x+2\right)-\left(-3x^5+4x^2+3x-7\right)\\ =3x^5-4x^2-7x+2+3x^5-4x^2-3x+7\\ =\left(3x^5+3x^5\right)+\left(-4x^2-4x^2\right)+\left(-7x-3x\right)+\left(2+7\right)\\ =6x^5-8x^2-10x+9\)

`c,` Thay `x=-1` Vào từng biểu thức ta có :

\(A\left(x\right)=3x^5-4x^2-7x+2\\=3\left(-1\right)^5-4.\left(-1\right)^2-7.\left(-1\right)+2\\ =3.\left(-1\right)-4.1-\left(-7\right)+2\\ =-3-4+7+2\\ =2\)

Cậu xem lại đề ạa

\(2,\\ M=\left(3x-2\right)\left(2x+1\right)-\left(3x+1\right)\left(2x-1\right)\\ =6x^2+3x-4x-2-\left(6x^2-3x+2x-1\right)\\=6x^2-x-2-6x^2+3x-2x+1\\ =\left(6x^2-6x^2\right)+\left(-x+3x-2x\right)+\left(-2+1\right)\\ =-1\)

`->` Vậy biểu thức không phụ thuộc vào biến `x`

a: A(x)=2x^3+x^2+4x+1

B(x)=-2x^3+x^2+3x+2

b: M(x)=A(x)+B(x)

=2x^3+x^2+4x+1-2x^3+x^2+3x+2

=2x^2+7x+3

c: M(x)=0

=>2x^2+7x+3=0

=>2x^2+6x+x+3=0

=>(x+3)(2x+1)=0

=>x=-3 hoặc x=-1/2

\(a,A\left(x\right)=2x^3-3x^2+2x+1\\ B\left(x\right)=3x^3+2x^2-x-5\\ b,A\left(x\right)+B\left(x\right)=\left(2x^3+3x^3\right)+\left(2x^2-3x^2\right)+\left(2x-x\right)+\left(1-5\right)=5x^3-x^2+x-4\\ A\left(x\right)-B\left(x\right)=\left(2x^3-3x^3\right)+\left(-3x^2-2x^2\right)+\left(2x+x\right)+\left(1-5\right)=-x^3-5x^2+3x-4\)

a: A(x)=3x^5+x^4-x^2+x

B(x)=3x^5-x^4+x^2+x-2

b: M(x)=B(x)-A(x)

=3x^5-x^4+x^2+x-2-3x^5-x^4+x^2-x

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

A(x) + B(x) = x⁴ - 3x + 3 + x⁴ - x + 128

A(x) +B(x) = (x⁴ + x⁴) - (3x+x) +( 3 +128)

A(x) + B(x) = 2x⁴ - 4x + 131

A(x) -B(x) = x⁴ - 3x + 3 - (x⁴ - x + 128)

A(x) -B(x) = x⁴ - 3x + 3 - x⁴ + x - 128

A(x) - B(x) = (x⁴  - x⁴) - (3x - x)  - ( 128 - 3)

A(x) - B(x) = 0 - 2x - 125

A(x) - B(x) = -2x - 125

 A(x) =  x⁴ + 3 - 3x

   A(x) = x⁴ - 3x + 3

 B(x) = 5³ + 3 - 3x² + x⁴ - 2x + 3x² + x

   B(x) = (125 + 3) - ( 3x² - 3x²) + x⁴ -( 2x - x)

   B(x) = 128 - 0 + x⁴ - x

B(x) = x⁴ - x + 128 

b, A(2) = 2⁴ - 3 \(\times\) 2 + 3

   A(2) = 16 - 6 + 3

  A(2) = 10 + 3

  A(2) = 13