Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)

Vì a, b, c là các số dương \(\Rightarrow a=b=c=0\) ( loại )

\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Rightarrow a=b=c\) ( tự chứng minh )

\(\Rightarrow M=\left(\dfrac{a}{b}-1\right)+\left(\dfrac{b}{c}-1\right)+\left(\dfrac{c}{a}-1\right)=0\)

Vậy M = 0

\(2^{2^{6n+2}}+3⋮19\)

\(\Leftrightarrow4^{2\left(3n+1\right)}+3⋮19\)

\(\Leftrightarrow16^{3n+1}+3⋮19\)

\(\Leftrightarrow\left(16+3\right)\left(16^{3n}-...+1\right)⋮19\) ( luôn đúng )

\(\Rightarrowđpcm\)

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

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

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

\(\Leftrightarrow x=2\)

Vậy x = 2

\(A=x^4-6x^3+27x^2-54x+32\)

\(=x^4-2x^3-4x^3+8x^2+19x^2-38x-16x+32\)

\(=x^3\left(x-2\right)-4x^2\left(x-2\right)+19x\left(x-2\right)-16\left(x-2\right)\)

\(=\left(x^3-4x^2+19x-16\right)\left(x-2\right)\)

\(x^2-xy+y^2-4=0\)

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

Để \(x\left(x-y\right)+\left(y-2\right)\left(y+2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x\left(x-y\right)=0\\\left(y-2\right)\left(y+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\end{matrix}\right.\)

Vậy x = y = 2 hoặc x = y = -2

Bổ sung bài làm bạn dưới thêm 1 trường hợp:

TH2: \(2x+3y-1=0\)

\(\Rightarrow\left\{{}\begin{matrix}2x+1=0\\3y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Giải:
Kẻ Bz // Ax \(\Rightarrow\)Ax // Bz // Cy

Ta có: Ax // Bz \(\Rightarrow\widehat{A}=\widehat{B_1}=40^o\left(slt\right)\)

Bz // Cy \(\Rightarrow\widehat{C}=\widehat{B_2}=30^o\left(slt\right)\)

\(\widehat{ABC}=\widehat{B_1}+\widehat{B_2}=70^o\)

Vậy...