là sao?

\(a^3+b^3=2021c^3\\ \Leftrightarrow a^3+b^3+c^3=2022c^3⋮6\left(2022⋮6\right)\left(1\right)\)

Mặt khác: \(\left(a^3+b^3+c^3\right)-\left(a+b+c\right)=\left(a-1\right)a\left(a+1\right)+\left(b-1\right)b\left(b+1\right)+\left(c-1\right)c\left(c+1\right)\)

Có \(\left(a-1\right)a\left(a+1\right);\left(b-1\right)b\left(b+1\right);\left(c-1\right)c\left(c+1\right)\) là 3 cặp số nguyên liên tiếp nên chia hết cho 6

Do đó \(\left(a^3+b^3+c^3\right)-\left(a+b+c\right)⋮6\)

Kết hợp (1) ta được đpcm

\(A=x^2-x+xy-y+2x-x^2-xy+1=x-y+1\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}y=2x-3\\x+2mx-3m=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x-3\\x\left(2m+1\right)=3m+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3m+1}{2m+1}\\y=\dfrac{6m+2-6m-3}{2m+1}=\dfrac{-1}{2m+1}\end{matrix}\right.\)

Ta có \(mx+3y=1\Leftrightarrow\dfrac{3m^2+m}{2m+1}-\dfrac{3}{2m+1}=1\Leftrightarrow3m^2+m-3=2m+1\)

\(\Leftrightarrow3m^2-m-4=0\\ \Leftrightarrow\left[{}\begin{matrix}m=\dfrac{4}{3}\\m=-1\end{matrix}\right.\)

\(=\left(\dfrac{1}{2}x\right)^2-2\cdot\dfrac{1}{2}x\cdot1+1=\dfrac{1}{4}x^2-x+1\)

\(a,\Rightarrow\dfrac{\left(-3\right)^x}{\left(-3\right)^4}=\left(-3\right)^3\\ \Rightarrow\left(-3\right)^{x-4}=\left(-3\right)^3\\ \Rightarrow x-4=3\Rightarrow x=7\\ b,Sửa:\left(x-\dfrac{1}{2}\right)^2=25\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=5\\x-\dfrac{1}{2}=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{5}\\x=-\dfrac{9}{5}\end{matrix}\right.\)

PT có VSN

\(\Leftrightarrow\dfrac{m}{1}=\dfrac{1}{m}=\dfrac{2m-1}{m}\\ \Leftrightarrow\left\{{}\begin{matrix}m^2=1\\2m^2-m=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=\pm1\\\left[{}\begin{matrix}m=0\\m=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow m=1\)

Áp dụng định lí Ta lét đảo ta có:

\(\dfrac{OD}{OA}=\dfrac{OE}{OB}=\dfrac{OF}{OC}=\dfrac{1}{4}\Rightarrow DE\text{//}AB;EF\text{//}BC;DF\text{//}AC\\ \Rightarrow\dfrac{DE}{AB}=\dfrac{EF}{BC}=\dfrac{DF}{AC}=\dfrac{OD}{OA}=\dfrac{1}{4}\\ \Rightarrow\Delta ABC\sim\Delta DEF\left(c.c.c\right)\)

Tỉ số đồng dạng là: \(\dfrac{DE}{AB}=\dfrac{1}{4}\)