8⁷-2¹⁸=2²¹-2¹⁸=2¹⁸(2³-1)=2¹⁸.7=14.2^{17 chia hết cho} 14 nha

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\(B=\left|x-1\right|+2\left|x-2\right|+3\left|x-3\right|+4\left|x-4\right|+5\left|x-5\right|\)

\(=\left(\left|x-1\right|+\left|x-5\right|\right)+2\left(\left|x-2\right|+\left|x-5\right|\right)+2\left(\left|x-3\right|+\left|x-5\right|\right)+\left(\left|x-3\right|+\left|x-4\right|\right)+3\left|x-4\right|\)

\(B\ge\left|x-1+5-x\right|+2\left|x-2+5-x\right|+2\left|x-3+5-x\right|+\left|x-3+4-x\right|+3\left|x-4\right|\)

\(=4+2.3+2.2+1+3.0\)

= 15

B min = 15 khi x =4

\(\left\{{}\begin{matrix}\left(x+2y\right)^2-4xy=5\\4xy\left(x+2y\right)+5\left(x+2y\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2-4b=5\\4ab+5a=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}4b=a^2-5\\a\left(a^2-5\right)+5a=1\end{matrix}\right.\)

\(\Rightarrow a^3=1\)=> a=1 => 4b= 1 -5 =4=> b = -1

=>\(\left\{{}\begin{matrix}x+2y=1\\xy=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=1\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=2\end{matrix}\right.\end{matrix}\right.\)

\(1=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+\left(b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{a}{bc}\right)\)

\(\Leftrightarrow\left(b+c\right)\left(\dfrac{bc+ac+ab+a^2}{abc}\right)=0\)

\(\dfrac{\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(a+c\right)}{abc}=0\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)

Xét 3 TH

=> P=0 ( đề bài BT ở giữa có 1 số mũ sai nha )

\(\dfrac{ab}{\left|a-b\right|}=c\Rightarrow ab=c\left|a-b\right|\)

Vì c là số nguyên tố => \(a⋮c\) hoặc \(b⋮c\)

=> c thuộc { 2;3;5;7}

+ c =2 => ab=2|a-b|

Nếu a >b => b =\(\dfrac{2a}{a+2}=2-\dfrac{4}{a+2}\in N^{\cdot}\)=>a=2

=> b =1 (TM)

Nếu a <b => \(a=\dfrac{2b}{b+2}\) tưng tụ trên => b =2

=> a =1 ( TM)

+ Nếu c =3 ; 5;7 bạn tự làm nha.

Vì ax + by =2c

ax + cz =2b

by + cz = 2a

=>Ta có ax + by + cz =a+b+c

=> ax + 2a=a+b+c

và 2c + cz =a+b+c

và 2b+ by =a+b+c

=> \(x=\dfrac{b+c-a}{a}\); \(y=\dfrac{a+c-b}{b}\);\(z=\dfrac{b+a-c}{c}\)

=> \(x+2=\dfrac{b+c+a}{a}\); \(y+2=\dfrac{a+c+b}{b}\);\(z+2=\dfrac{b+a+c}{c}\)

=>\(M=\dfrac{1}{x+2}+\dfrac{1}{y+2}+\dfrac{1}{z+2}=\dfrac{a+b+c}{a+b+c}=1\)