Vì \(x=2018\Rightarrow x+1=2019\)

Thay x+1=2019 vào biểu thức A  ta được :

\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-...-\left(x+1\right)x+x+1\)

\(=x^6-x^6-x^5+x^5+x^4-...-x^2-x+x+1\)

\(=1\)

\(A=x^6-2019x^5+2018x^4-2019x^3+2019x^2-2019x+2019\)

\(=x^6-2018x^5-x^5+2018x^4+x^4-2018x^3-x^3+2018x^2+x^2\)

\(-2018x-x+2019\)

\(=x^5\left(x-2018\right)-x^4\left(x-2018\right)-x^3\left(x-2018\right)+x^2\left(x-2018\right)\)

\(+x\left(x-2018\right)-\left(x-2018\right)+1\)

= 1

Vì \(x=2018\Rightarrow x+1=2019\)

Thay \(x+1=2019\) vào biểu thức \(A\) ta được :

\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x-1\right)\)

\(=x^6-x^6-x^5+x^5-x^4+x^4-x^3+x^3-x^2+x^2-x+x+1\)

\(=1\)

C

C

C

\(\dfrac{x-2017}{2019}+\dfrac{x-2019}{2017}=\dfrac{x+6}{2021}\)

\(\Rightarrow\dfrac{x-2017}{2019}-1+\dfrac{x-2019}{2017}-1=\dfrac{x+6}{2021}-2\)

\(\Rightarrow\dfrac{x-2017}{2019}-\dfrac{2019}{2019}+\dfrac{x-2019}{2017}-\dfrac{2017}{2017}=\dfrac{x+6}{2021}-\dfrac{4042}{2021}\)

\(\Rightarrow\dfrac{x-2017-2019}{2019}+\dfrac{x-2019-2017}{2017}=\dfrac{x+6-4042}{2021}\)

\(\Rightarrow\dfrac{x-4036}{2019}+\dfrac{x-4036}{2017}=\dfrac{x-4036}{2021}\)

\(\Rightarrow\dfrac{x-4036}{2021}-\dfrac{x-4036}{2019}-\dfrac{x-4036}{2017}=0\)

\(\Rightarrow\left(x-4036\right)\left(\dfrac{1}{2021}-\dfrac{1}{2019}-\dfrac{1}{2017}\right)=0\)

=> x - 4036 = 0

=> x = 4036

−

=> x - 4036 = 0

=> x = 4036

\(x^2+y^2=6\left(x-y-3\right)\)\(\Rightarrow x^2+y^2-6\left(x-y-3\right)=0\)

\(\Leftrightarrow x^2+y^2-6x+6y+18=0\)\(\Leftrightarrow\left(x^2-6x+9\right)+\left(y^2+6x+9\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+3\right)^2=0\)(1)

Vì \(\hept{\begin{cases}\left(x-3\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-3\right)^2+\left(y+3\right)^2\ge0\forall x,y\)(2)

Từ (1) và (2) \(\Rightarrow\left(x-3\right)^2+\left(y+3\right)^2=0\Leftrightarrow\hept{\begin{cases}x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)

\(\Rightarrow M=3^{2019}+\left(-3\right)^{2019}+\left(3-3\right)^{2020}=0\)

\(Ta \) \(có : \) \(x ^2 + y^2 = 6. ( x - y - 3 )\)

\(\Leftrightarrow\)\(x^2 + y^2 - 6. ( x - y - 3 ) = 0\)

\(\Leftrightarrow\)\(x^2 + y^2 - 6x + 6y + 18 = 0\)

\(\Leftrightarrow\)\(( x^2 - 6x + 9 ) + ( y^2 + 6y + 9 ) = 0\)

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

\(\Leftrightarrow\)\(( x - 3 )^2 = 0 \) \(và \) \(( y - 3 )^2 = 0\)

\(\Leftrightarrow\)\(x - 3 = 0 \) \(và \) \(y + 3 = 0\)

\(\Leftrightarrow\)\(x = 3 \) \(và \) \(y = - 3\)

\(Thay\) \(x = 3 ; y = - 3 \) \(vào \) \(M \)\(ta \) \(được :\)

\(M = 3\)^\(2019\) \(+ (- 3 )\)^\(2019\) \(+ [ 3 + ( - 3 ) ]\)^\(2020\)

\(M = 0 \)

Đặt \(x+6=a;y-7=b;z-9=c\)

\(\Rightarrow\hept{\begin{cases}a+b+c=0\\a^3+b^3+c^3=0\end{cases}}\)

Bạn hiểu chưa :))

Đặt x+6=a, y-7=b, z-9=c

Vì x+y+z=10 nên a+b+c=0

Xét \(a^3+b^3+c^3=0\Leftrightarrow a^3+b^3+c^3-3abc=-3abc\)(1)

Ta có đẳng thức (bạn nên học đẳng thức này nhé vì nó cực kì thông dụng trong toán nâng cao):

\(a^3+b^3+c^3-3abc=\frac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2}\)(2)

Vì a+b+c=0 nên từ (1), (2) suy ra \(\hept{\begin{cases}-3abc=0\\a+b+c=0\end{cases}\Rightarrow a=b=c=0}\)

Vậy M = a²⁰¹⁹+b²⁰¹⁹+c²⁰¹⁹=0

ĐKXĐ:....

Lấy pt trên cộng 2 lần pt dưới ta được:

\(\frac{x+2019}{x+2018}+\frac{10}{x+2016}=12\)

Số to quá, đặt \(x+2016=a\Rightarrow\frac{a+3}{a+2}+\frac{10}{a}=12\)

\(\Leftrightarrow12a\left(a+2\right)=a\left(a+3\right)+10\left(a+2\right)\)

\(\Leftrightarrow12a^2+24a-a^2-3a-10a-20=0\)

\(\Leftrightarrow11a^2+11a-20=0\)

Nghiệm rất xấu, bạn tự giải tiếp