5 là số nguyên tố. Theo định lý Fermat nhỏ

\(5^{2017}-5\equiv0\) (mod 2017)

\(\Rightarrow5^{2021}=5^{2017}.5^4=\left(5^{2017}-5+5\right).5^4=\)

\(=5^4\left(5^{2017}-5\right)+5^5=5^4\left(5^{2017}-5\right)+3125=\)

\(=5^4\left(5^{2017}-5\right)+2017+1108\)

Ta có

\(5^4\left(5^{2017}-5\right)+2017⋮2017\)

\(\Rightarrow5^{2021}\equiv1108\) (mod 2017)

\(x\left(x-8\right)=9\)

\(\Leftrightarrow x^2-8x-9=0\)

\(\Leftrightarrow x\left(x-9\right)+\left(x-9\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)

\(\Leftrightarrow x=9\) hoặc \(x=-1\)

\(x\left(x+14\right)=15\)

\(\Leftrightarrow x^2+14x-15=0\)

\(\Leftrightarrow x\left(x+15\right)-\left(x+15\right)=0\)

\(\Leftrightarrow\left(x+15\right)\left(x-1\right)=0\)

\(\Leftrightarrow x=-15\) hoặc \(x=1\)

\(4x\left(x-3\right)=7\)

\(\Leftrightarrow4x^2-12x-7=0\)

\(\Leftrightarrow2x\left(2x+1\right)-7\left(2x+1\right)=0\)

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

\(\Leftrightarrow x=-\dfrac{1}{2}\) hoặc \(x=\dfrac{7}{2}\)

\(9x\left(x-4\right)=13\)

\(\Leftrightarrow9x^2-36x-13=0\)

\(\Leftrightarrow3x\left(3x+1\right)-13\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(3x-13\right)=0\)

\(\Leftrightarrow x=-\dfrac{1}{3}\) hoặc \(x=\dfrac{13}{3}\)

\(\left(x+6\right)^2-x\left(x-3\right)=81\)

\(\Leftrightarrow x^2+12x+36-x^2+3x=81\)

\(\Leftrightarrow15x=45\)

\(\Leftrightarrow x=3\)

\(\left(2x+3\right)^2-x\left(x-3\right)=9\)

\(\Leftrightarrow4x^2+12x+9-x^2+3x=9\)

\(\Leftrightarrow3x^2+15x+9=9\)

\(\Leftrightarrow3x\left(x+5\right)=0\)

\(\Leftrightarrow x=0\) hoặc \(x=-5\)

\(\left(x-7\right)^2-x\left(x+4\right)=121\)

\(\Leftrightarrow x^2-14x+49-x^2-4x=121\)

\(\Leftrightarrow-18x=72\)

\(\Leftrightarrow x=-4\)

\(\left(3x-2\right)^2-9x\left(x-4\right)=100\)

\(\Leftrightarrow9x^2-12x+4-9x^2+36x=100\)

\(\Leftrightarrow24x=96\)

\(\Leftrightarrow x=4\)

\(\left(\dfrac{1}{2}a-\dfrac{2}{3}b\right)^3=\left(\dfrac{1}{2}a\right)^3-3.\left(\dfrac{1}{2}a\right)^2.\dfrac{2}{3}b+3.\dfrac{1}{2}a.\left(\dfrac{2}{3}b\right)^2-\left(\dfrac{2}{3}b\right)^3\)

                        \(=\dfrac{1}{8}a^3-\dfrac{1}{2}a^2b+\dfrac{2}{3}ab^2-\dfrac{8}{27}b^3\)

\(\left(\dfrac{1}{2}a-\dfrac{2}{3}b\right)^2\)

\(=\left(\dfrac{1}{2}a\right)^2-2.\left(\dfrac{1}{2}a\right).\left(\dfrac{2}{3}b\right)+\left(\dfrac{2}{3}b\right)^2\)

\(=\dfrac{1}{4}a^2-\dfrac{2}{3}ab+\dfrac{4}{9}b^2\)

cứu với

Ta có

\(a^{2020}+b^{2020}=a^{2021}+b^{2021}\)

\(\Leftrightarrow a^{2021}-a^{2020}=b^{2020}-b^{2021}\)

\(\Leftrightarrow a^{2020}\left(a-1\right)=b^{2020}\left(1-b\right)\)

\(\Leftrightarrow\dfrac{a-1}{1-b}=\dfrac{b^{2020}}{a^{2020}}=\left(\dfrac{b}{a}\right)^{2020}\) (1) 

Ta có

\(a^{2021}+b^{2021}=a^{2022}+b^{2022}\)

\(\Leftrightarrow\dfrac{a-1}{1-b}=\left(\dfrac{b}{a}\right)^{2021}\) (2)

Từ (1) và (2) \(\Rightarrow\left(\dfrac{b}{a}\right)^{2020}=\left(\dfrac{b}{a}\right)^{2021}\)

\(\Rightarrow\dfrac{b}{a}=1\Rightarrow a=b\)

\(\Rightarrow2.a^{2020}=2.a^{2021}\Leftrightarrow a^{2020}=a^{2021}\Rightarrow a=b=1\)

\(\Rightarrow S=a^{2021}+b^{2021}=1+1=2\)

Are you missed something? \(C\) is a first degree polynomial so we can't find a minimum value of \(C\)

I think you mean \(C=12x^2+6x-10\) 

Am I correct?

Ta có: \(x+y+z+t=0\)

\(\Rightarrow t=-\left(x+y+z\right)\)

\(VT=x^3+y^3+z^3+t^3\)

\(=x^3+y^3+z^3-\left(x+y+z\right)^3\)

\(=x^3+y^3+z^3-\left[x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\right]\)

\(=-3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(VP=3\left[xy+z\left(x+y+z\right)\right]\left(z-x-y-z\right)\)

\(=3\left(xy+yz+zx+z^2\right)\left(-x-y\right)\)

\(=-3\left(y+z\right)\left(x+z\right)\left(x+y\right)\)

\(\Rightarrow VT=VP\)

x+y+z+t=0

<=> t= - (x+y+z)

<=> t³ = - (x+y+z)³

<=> t³ = - x³- y³- z³ - 3(x+y)(y+z)(z+x)

=>  x³+y³+z³+t³ = x³+y³+z³ + (- x³- y³- z³ - 3(x+y)(y+z)(z+x))

=> 3(y+z)(xt-yz) = -3(x+y)(y+z)(z+x)

=>xt-yz= (x+y)(z+x)

=> x²+xy+xz+xt=0

=> x(x+y+z+t)=0 luôn đúng => đpcm