\(x^2-6x+10\)

\(=x^2-6x+9+1\)

\(=\left(x-3\right)^2+1\ge1\forall x\)

Mà 1>0 

\(\Rightarrow x^2-6x+10\) luôn dương \(\forall x\left(đpcm\right)\)

ta có

(a+b)(a-b)=a²-b²

áp dụng hằng đẳng thức đáng nhớ

nó là (a+B) nó là B chứ ko phải b

\(2^3.19-2^3.14+1^{2018}\)

= \(8.19-8.14+1\)

= \(8.\left(19-14\right)+1\)

= \(8.5+1\)

= \(40+1\)

= \(41\)

Chúc bạn học tốt !!!

\(ab\left(a-b\right)-ac\left(a+c\right)+bc\left(2a-b+c\right)\)

\(=a^2b-ab^2-a^2c-ac^2+2abc-b^2c+bc^2\)

\(=a^2b-ab^2-a^2c-ac^2+abc+abc-b^2c+bc^2\)

\(=\left(bc^2-ac^2+abc-a^2c\right)-\left(b^2c-abc-ab^2+a^2b\right)\)

\(=c\left(bc-ac+ab-a^2\right)-b\left(bc-ac-ab+a^2\right)\)

\(=\left(c-b\right)\left(bc-ac+ab-a^2\right)\)

\(=\left(c-b\right)\left[c\left(b-a\right)+a\left(b-a\right)\right]\)

\(=\left(c-b\right)\left(c+a\right)\left(b-a\right)\)

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

\(\Leftrightarrow3\left|x\right|=x^2-4\)

\(\Leftrightarrow3x=\pm\left(x^2-4\right)\)

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

Ta giải 2 phương trình này được \(s=\left\{-4;4\right\}\)

Chúc bạn học tốt !!!

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-3x=4\\x^2-3\left(-x\right)=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-3x-4=0\\x^2+3x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-4\right)\left(x+1\right)=0\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm4\left(tm\right)\\x=\pm1\left(ktm\right)\end{cases}}\)

\(\Rightarrow x=\pm4\)

\(B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=5^{32}-1< 5^{32}\)

Vậy \(B< A\)