Question

1, Chứng tỏ rằng các đa thức sau ko phụ thuộc vào biến:

a,x(2x+1)-x²(x+2)+(x³-x+3)

b,4(x-6)-x²(2+3x)+x(5x-4)+3x²(x-1)

2,Tìm x

a,5x(12x+7)-3x(20x-5)=-100

b,0,6x(x-0,5)-0,3x(2x+1,3)=0,138

Answer 1

1)

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

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

=>đpcm

b) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)

\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2=-24\)

=>đpcm

2,

a) \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)

\(\Leftrightarrow60x^2+35x-60x^2+15x=-100\)

\(\Leftrightarrow50x=-100\)

\(\Leftrightarrow x=-2\)

b) \(0,6x\left(x-0,5\right)-0,3x\left(2x+1,3\right)=0,138\)

\(\Leftrightarrow0,6x^2-0,3x-0,6x^2-0,39x=0,138\)

\(\Leftrightarrow-0,69x=0,138\)

\(\Leftrightarrow x=-0,2\)

Answer 2

Câu 1:

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

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

\(=x^3+3\)(ko thể CM)

b)\(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)

\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)

\(=-24\)(đpcm)

Answer 3

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

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

\(=3\)

=> đpcm

 b ) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)

\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)

\(=-24\)

=> đpcm

2 ) Tìm x :

a ) \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)

\(\Leftrightarrow60x^2+35x-60x^2+15x=100\)

\(\Leftrightarrow50x=100\)

\(\Leftrightarrow x=2\)

b ) \(0,6c\left(x-0,5\right)-0,3x\left(2x+1,3\right)=0,138\)

\(\Leftrightarrow0,6x^2-0,3x-0,6x^2-0,39x=0,138\)

\(\Leftrightarrow-0,69x=0,138\)

\(\Leftrightarrow-0,2\)

Answer 4

Chết r mk ghi nhầm đề

 

Answer 5

Câu 2:

a)\(5.\left(12x+7\right)-3.\left(20x-5\right)=-100\)

   \(60x+35-60x+15=-100\)

   \(35+15=-100\)

        Do đó ko tìm được x

b)\(0,6.\left(x-0,5\right)-0,3.\left(2x+1,3\right)=0,138\)

   \(0,6x-0,3-0,6x-0,39=0,138\)

   \(-0,3-0,39=0,138\)

         Do đó ko tìm được x