Ta có \(x^2-5x+10=\left(x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{15}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}>0\) với mọi \(x\)

\(x^2-5x+10=x^2-\dfrac{2.5}{2}x+\dfrac{25}{4}-\dfrac{25}{4}+10=\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}>0\)

Vậy bth luôn dương với mọi biến 

1, \(x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)

\(\Leftrightarrow x^3\left(x-2\right)+4x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^3+4x^2+4x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^3+3x^2+x^2+3x+x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2+x+1>0\right)\left(x+3\right)\left(x-2\right)=0\Leftrightarrow x=-3;x=2\)

2, \(2\left(x^3-1\right)-7x\left(x-1\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(x^2+x+1\right)-7x\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^2-5x+2\right)=0\Leftrightarrow x=1;x=\dfrac{1}{2};x=2\)

`(-5x).(x-3).(2x-4)-(x-7).(x-3)+(5x-2).(3x-4)`

`=-10x^3 +50x^2-60x - x^2 +10x -21+15x^2 -26x +8`

`=-10x^3 + (50x^2 -x^2 + 15x^2) -(60x +10x - 26x) -(21+8)`

`=-10x^2 + 64x^2 -76x -13`

\(\left(x-3\right)\left(-10x^2+20x-x+7\right)+15x^2-20x-6x+8\)

\(=\left(x-3\right)\left(-10x^2+19x+7\right)+15x^2-26x+8\)

\(=-10x^3+19x^2-21+30x^2-57x-21+15x^2-26x+8\)

\(=-10x^3+64x^2-83x-34\)

13) \(\left(x-3\right)^2-16=\left(x-3-4\right)\left(x-3+4\right)=\left(x-7\right)\left(x+1\right)\)

\(1...1\) (2n chữ số)

\(=1+10+10^2+...+10^{2n-1}\)

\(=\dfrac{10^{2n}-1}{9}\)

 \(4...4\) (n chữ số) 

\(=4.\left(1+10+10^2+...+10^{n-1}\right)\)

\(=4.\dfrac{10^n-1}{9}\)

\(=\dfrac{4.10^n-4}{9}\)

Viết lại: \(D=\dfrac{10^{2n}-1}{9}+\dfrac{4.10^n-4}{9}+1=\dfrac{\left(10^n\right)^2-1+4.10^n-4+9}{9}=\dfrac{\left(10^n\right)^2+4.10^n+4}{9}=\left(\dfrac{10^n+2}{3}\right)^2\)

Vì \(\left\{{}\begin{matrix}10^nmod3=1\\2mod3=2\end{matrix}\right.\Rightarrow\left(10^n+2\right)⋮3\)

\(\Rightarrow\dfrac{10^n+2}{3}\) là số tự nhiên

Vậy D là số chính phương.

(x+y)² + (x-y)² = x² +2xy +y² + x² - 2xy + y² = 2x² + 2y²

\(\left(x+y\right)^2+\left(x-y\right)^2\)

\(=x^2+2.x.y+y^2+x^2-2x.y+y^2\)

\(=x^2+2xy+y^2+x^2-2xy+y^2\)

\(=2x^2+y^2+y^2\)

\(=2x^2+2y^2\)

\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\\ =\left[\left(x^2+1\right)^2-x^2\right]\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\\ =\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\\ =\left[\left(x^4+1\right)^2-x^4\right]\left(x^8-x^4+1\right)\\ =\left(x^8+x^4+1\right)\left(x^8-x^4+1\right)\\ =\left(x^8+1\right)^2-x^8\\ =x^{16}+x^8+1\)

\(x^2+2x-8\)

\(=x^2-2x+4x-8\)

\(=x\left(x-2\right)+4\left(x-2\right)\)

\(=\left(x-2\right)\left(x+4\right)\)