ĐK: \(x\ge\frac{2017}{2018}\)

\(pt\Leftrightarrow2017\sqrt{2017x-2016}-2017+\sqrt{2018x-2017}-1=0\)

\(\Leftrightarrow2017\frac{2017\left(x-1\right)}{\sqrt{2017x-2016}+1}+\frac{2018\left(x-1\right)}{\sqrt{2018x-2017}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{2017^2}{\sqrt{2017x-2016}+1}+\frac{2018}{\sqrt{2018x-2017}+1}\right)=0\)

Dễ thấy với \(x\ge\frac{2017}{2018}\Rightarrow\)\(\frac{2017^2}{\sqrt{2017x-2016}+1}+\frac{2018}{\sqrt{2018x-2017}+1}>0\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(\Delta=b^2-4ac=2017^2-2016.\left(-2018\right)=20341441>0\)

=> Phương trình có 2 nghiệm phân biệt

\(\orbr{\begin{cases}x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-2017-\sqrt{20341441}}{4032}\\x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-2017+\sqrt{20341441}}{4032}\end{cases}}\)

k mình nha bn thanks 

Pt tương đương:2015x-2014-2\(\sqrt{2017x-2016}\)=-X^2<=>2017x-2016-2\(\sqrt{2017x-2016}\)+1-2x+2-1=-X^2

<=>2017x-2016-2\(\sqrt{2017x-2016}\)+1=-x^2+2x-1

<=>(\(\sqrt{2017x-2016}\)-1)^2=-(x-1)^2

Rồi đánh giá(\(\sqrt{2017x-2016}\)-1)^2>=0

-(x-1)^2=<0 ( Ta thấy chỉ xảy ra khi bằng 0)

=>x-1=0<=>x=1

\(E\left(x\right)=x^{2018}-2019x^{2017}+2019x^{2016}-2019x^{2015}+...+2019x^2-2019x+1\)

Vì \(E\left(2018\right)\) nên :

\(\Rightarrow E\left(x\right)=2018^{2018}-2019.2018^{2017}+2019.2018^{2016}-2019.2018^{2015}+...+2019.2018^2-2019.2018+1\)

Tới đoạn này thì ghi dấu "=" rồi tính và làm tương tự

Lời giải

Ta có:

\(E(x)=x^{2018}-2019x^{2017}+2019x^{2016}-2019x^{2015}+...+2019x^2-2019x+1\)

\(E(x)=(x^{2018}-2018x^{2017})-(x^{2017}-2018x^{2016})+(x^{2016}-2018x^{2015})-....+(x^2-2018x)-x+1\)

\(E(x)=x^{2017}(x-2018)-x^{2016}(x-2018)+x^{2015}(x-8)-...+x(x-2018)-x+1\)

\(E(x)=(x-2018)(x^{2017}-x^{2016}+x^{2015}-...+x)-x+1\)

Suy ra \(E(2018)=-2018+1=-2017\)

Thay x = 2018 vào \(A=x^{2018}-2019x^{2017}+2019x^{2016}-2019x^{2015}+...+2019x^2-2019x-1\) ta được 

\(2018^{2018}-2019.2018^{2017}+2019.2018^{2016}-2019.2018^{2015}+...+2019.2018^2-2019.2018-1\)

\(=\)\(2018^{2018}-2019\left(2018^{2017}-2018^{2016}+2018^{2015}-...-2018^2+2018\right)-1\)

Đặt \(B=2018^{2017}-2018^{2016}+2018^{2015}-...-2018^2+2018\)

\(2018B=2018^{2018}-2018^{2017}+2018^{2016}-...-2018^3+2018^2\)

\(2018B+B=\left(2018^{2018}-2018^{2017}+...+2018^2\right)+\left(2018^{2017}-2018^{2016}+...+2018\right)\)

\(2019B=2018^{2018}-2018\)

\(B=\frac{2018^{2018}-2018}{2019}\)

\(\Rightarrow\)\(A=2018^{2018}-2019.B-1\)

\(\Rightarrow\)\(A=2018^{2018}-\frac{2019\left(2018^{2018}-2018\right)}{2019}-1\)

\(\Rightarrow\)\(A=2018^{2018}-\left(2018^{2018}-2018\right)-1\)

\(\Rightarrow\)\(A=2018^{2018}-2018^{2018}+2018-1\)

\(\Rightarrow\)\(A=2018-1\)

\(\Rightarrow\)\(A=2017\)

Vậy giá trị của \(A=2017\) tại \(x=2018\)

Chúc bạn học tốt ~