b: Đặt g(x)=0

\(\Leftrightarrow x^2\left(x^2-x+1\right)+5x^2-5x+5=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+5\right)=0\)(vô lý)

Ta có : \(A=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)

\(=-24\)

Vậy ....

`A=4(x-6)-x^2(3x+2)+x(5x-4)+3x^2(x-1)`

`=4x-24-3x^3-2x^2+5x^2-4x+3x^3-3x^2`

`=-24` không phụ thuộc vào biến

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

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

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

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

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)

Ta có \(f\left(x\right)=x^4+x^3+4x^2+3x+3\)

\(=x^2\left(x+\frac{1}{2}\right)^2+\frac{15}{4}x^2+3x+3\)

\(=x^2\left(x+\frac{1}{2}\right)^2+\frac{15}{4}\left(x+\frac{2}{5}\right)^2+\frac{12}{5}>0\) với mọi \(x\inℝ\)

Vậy đa thức trên vô nghiệm

a, \(5x^2\)\(-3x-x^3+x^2+x^3-6x^2\)\(-10+3x\)=\(-10\)

b,\(x^3+x^2+x-x^3-x^2-x+5=5\)

OLM đang duyệt câu trả lời của mjk