Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

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

\(\Rightarrow a^2+b^2+c^2=0-2\cdot0\)

\(\Rightarrow a=b=c=0\)

Thế kết quả vào: \(\left(0-2017\right)^{2018}+\left(0-2017\right)^{2018}-\left(0+2017\right)^{2018}=2017^{2018}\)

Ps: \(\left(-2017\right)^{2018}=2017^{2018}\)

\(\frac{2018}{ab+2018a+2018}+\frac{b}{bc+a+2018}+\frac{c}{ac+c+1}\)

\(a.b.c=2018\Rightarrow a,b,c\ne0\)

Ta có \(\frac{2018}{ab+2018a+2018}\Rightarrow\frac{2018}{b+2018+bc}\)

\(\frac{c}{ac+c+1}=\frac{bc}{abc+bc+b}=\frac{bc}{2018+bc+b}\)

\(\Rightarrow S=\frac{2018}{b+2018+bc}+\frac{b}{bc+b+2018}+\frac{bc}{2018+bc+b}=\frac{2018+b+bc}{b+2018+bc}=1\)

để nghĩ tiếp

làm tiếp 

\(\frac{2013x+1}{2014x-2014}=\frac{2013\left(x-1\right)+2014}{2014\left(x-1\right)}=\frac{2013}{2014}+\frac{1}{x-1}\)

\(B_{max}\Leftrightarrow\frac{1}{x-1}max\)

+) Nếu x >1 thì x-1 >0 \(\Rightarrow\frac{1}{x-1}>0\)

+) Nếu x<1 thì x-1 <0 \(\Rightarrow\frac{1}{x-1}< 0\)

Xét x > 1 ta có 

\(\frac{1}{x-1}max\Rightarrow x-1\)là số nguyên dương nhỏ nhất 

\(\Rightarrow x-1=1\Rightarrow x=2\)

Vậy \(Bmax=1\frac{2018}{2019}\Leftrightarrow x=2\)

Dễ mà bạn

đưa x ra làm nhân tử chug

Do \(abc=2018,bc+b+1\ne0\) nên thay vào biểu thức A ta có :

  \(A=\frac{2018}{abc+bc+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+2018}\)

\(=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\)

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

\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)

\(=\frac{bc+b+1}{bc+b+1}=1\)

Vậy : \(A=1\) với a,b,c thỏa mãn đề.

\(A=\frac{2018}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+2018}\)

\(=\frac{abc}{abc+ab+a}+\frac{ab}{abc+ab+a}+\frac{a}{ab+a+abc}\)

\(=1\)

Vậy ...

Ta có: \(a+b+c=0\)

\(=>\left(a+b+c\right)^2=0\)

\(=>a^2+b^2+c^2+2ab+2bc+2ac=0\)

\(=>a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

\(=>a^2+b^2+c^2=0\)

\(=>a^2+b^2+c^2=ab+bc+ac\)

\(=>2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)

\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)(nhân phân phối, đổi qua bên kia dấu bằng, tách thành hằng đẳng thức)

\(=>\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

\(=>\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\)

\(=>a=b=c=0\)

***\(A=\left(a-1\right)^{22}+b^{12}+\left(c-1\right)^{2014}\)

\(A=\left(-1\right)^{22}+1+\left(-1\right)^{2014}\)

\(A=1+1+1\)

\(A=3\)

Ta có

a + b + c = 0

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

\(\Leftrightarrow\)a² + b² + c² = ab + bc + ca

Mà ta có a² + b² + c² \(\ge\) ab + bc + ca

Dấu = xảy ra khi a = b = c = 0

\(\Rightarrow\)(a - 1)²² + b¹² + (c - 1)²⁰¹⁴ = 1 + 0 + 1 = 2

Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

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

hay \(a^2+b^2+c^2=0\Rightarrow a=b=c=0\)

Thay a = b = c = 0 vào M rồi tính như bình thường nha bạn!

Ta có : 

\(a+b+c=0\)

\(\Leftrightarrow\)\(\left(a+b+c\right)^2=0\)

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

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

\(\Leftrightarrow\)\(a^2+b^2+c^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a^2=0\\b^2=0\\c^2=0\end{cases}\Leftrightarrow a=b=c=0}\)

\(\Rightarrow\)\(M=\left(a-2018\right)^{2019}+\left(b-2018\right)^{2019}-\left(c+2018\right)^{2019}\)

\(\Rightarrow\)\(M=-2018^{2019}-2018^{2019}-2018^{2019}\)

\(\Rightarrow\)\(M=-3.2018^{2019}\)

Chúc bạn học tốt ~