\(x^2-2015x+2014=0\)

\(x^2-2014x-x+2014=0\)

\(x\left(x-2014\right)-\left(x-2014\right)=0\)

\(\left(x-2014\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2014=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2014\\x=1\end{cases}}}\)

\(x^2-2015x+2014\)\(=0\)

\(\Rightarrow x^2-x-2014x+2014\)\(=0\)

\(\Rightarrow x\left(x-1\right)-2014\left(x-1\right)\)\(=0\)

\(\Rightarrow\left(x-1\right)\left(x-2014\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2014=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2014\end{cases}}\)

thanks

x^4-2014x^2+2015x-2014=0

<=>x⁴+x-2014x²+2014x-2014=0

<=>x.(x³+1)-2014.(x²-x+1)=0

<=>x.(x+1)(x²-x+1)-2014.(x²-x+1)=0

<=>(x²+x+1)[x.(x+1)-2014]=0

<=>x.(x+1)-2014=0 (vì x²+x+1 >0)

giải tiếp sao số xấu thế

Nhận xét: Tổng các hệ số của phương trình bằng 0 => phương trình có 1 nghiệm là 1

=> vế trái có nhân tử (x - 1)

pt <=> (x⁴ - 1 ) + (2015x³ - 2015x²) - (2015x - 2015)  = 0

<=> (x-1)(x+1).(x² + 1) + 2015x²(x - 1) - 2015.(x - 1) = 0

<=> (x - 1).[(x+1).(x² + 1) + 2015x² - 2015] = 0

<=> (x -1). [(x+1).(x² + 1) + 2015(x² - 1)] = 0

<=> (x -1). [(x+1).(x² + 1) + 2015(x - 1)(x+1)] = 0

<=> (x -1).(x+1).(x² + 1 + 2015x - 2015 ) = 0  

<=> x - 1 = 0 hoặc  x+ 1 = 0 hoặc x² + 1 + 2015x - 2015  = 0

+) x - 1 = 0 <=> x = 1

+) x + 1 = 0 <=> x = -1

+) x² + 1 + 2015x - 2015 = 0 <=> x² + 2015x - 2014 = 0 

<=> x² +2.x. \(\frac{2015}{2}\) + \(\left(\frac{2015}{2}\right)^2\) - \(\left(\frac{2015}{2}\right)^2\)   - 2015 = 0

<=> \(\left(x-\frac{2015}{2}\right)^2=\frac{2015^2+4030}{2}\)

<=>  \(x-\frac{2015}{2}=\sqrt{\frac{2015^2+4030}{2}}\) hoặc \(x-\frac{2015}{2}=-\sqrt{\frac{2015^2+4030}{2}}\)

<=> \(x=\frac{2015}{2}+\sqrt{\frac{2015^2+4030}{2}}\)hoặc \(x=\frac{2015}{2}-\sqrt{\frac{2015^2+4030}{2}}\)

Vậy pt có 4 nghiệm...

chính xác nè bạn nhớ sai ruj:

x⁴+2015x²+2014x+2015=0

<=>x⁴-x+2015x²+2015x+2015=0

<=>x(x³-1)+2015(x²+x+1)=0

<=>x(x-1)(x²+x+1)+2015(x²+x+1)=0

<=>(x²+x+1)[x(x-1)-2015]=0

<=>(x²+x+1)(x²-x-2015)=0

<=>x²+x+1=0 hoặc x²-x-2015=0

*x²+\(2x.\frac{1}{2}\)+\(\frac{1}{4}+\frac{3}{4}\)=0 

<=>(x+1/2)²+3/4=0(vô lí)

*x²-\(2x.\frac{1}{2}+\frac{1}{4}-\frac{8061}{4}\)

<=>(x-1/2)²-8061/4=0

<=>(x-1/2)²           =8061/4

<=>x-1/2              =\(\sqrt{\frac{8061}{4}}\)

<=>x                    =\(\sqrt{\frac{8061}{4}+}\frac{1}{2}\)

\(x^2-2015x+2014=0\)

\(x^2-x-2014x+2014=0\)

\(x\left(x-1\right)-2014\left(x-1\right)=0\)

\(\left(x-1\right)\left(x-2014\right)=0\)

TH1:x -1 = 0

=>x=1

TH2 : x-2014=0

=> x=2014

\(x^3-4x=0\)

\(x\left(x^2-4\right)=0\)

\(x\left(x-4\right)\left(x+4\right)=0\)

TH1: x=0

TH2:x-4=0

=> x= 4

TH3: x+4=0

=> x=(-4)

Hok tốt

Chính là 0

\(|2015x-2014|=|2015x+2014|\)

\(\Leftrightarrow\orbr{\begin{cases}-2015x+2014=|2015x+2014|\left(l\right)\\2015x-2014=|2015x+2014|\left(n\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2015x+2014=-2015x+2014\\2015x+2014=2015x-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}4030x=0\\0x=-4028\left(l\right)\end{cases}\Leftrightarrow}4030x=0\Leftrightarrow x=0}\)

=> x là số nguyên dượng

=> x+1+x+2+x+3+x+4+...+x+2014=|x+1|+|x+2|+|x+3|+|x+4|+...+|x+2014|=2015

=> 2014x + (1+2+3+4+...+2014)=2015x

=> 1+2+3+4+...+2014=x

P(x) = x²⁰¹⁶ - 2015x²⁰¹⁵ - 2015x²⁰¹⁴ - ... - 2015x² - 2015x 

<=> P(x) = x²⁰¹⁶ - 2016x²⁰¹⁵ + x²⁰¹⁵ - 2016x²⁰¹⁴ + x²⁰¹⁴ - ... - 2016x² + x² - 2016x + x 

<=> P(2016) = 2016²⁰¹⁶ - 2016.2016²⁰¹⁵ + 2016²⁰¹⁵ - 2016.2016²⁰¹⁴ + 2016²⁰¹⁴ -...- 2016.2016² + 2016² - 2016.2016 + 2016 

<=> P(2016)=2016²⁰¹⁶ - 2016²⁰¹⁶ + 2016²⁰¹⁵ - 2016²⁰¹⁵ + 2016²⁰¹⁴ - ... - 2016³ + 2016² - 2016² + 2016

<=> P(2016) = 2016

Vậy P(2016) = 2016

Ta có:

P(2016) = 2016²⁰¹⁶ - 2015 . 2016²⁰¹⁵ - 2015 . 2016²⁰¹⁴ -.....- 2015 . 2016² - 2015 . 2016 - 1

P(2016) = 2016²⁰¹⁶ - ( 2016 - 1 ) . 2016²⁰¹⁵ - ( 2016 -1 ) . 2016²⁰¹⁴ - ..... - ( 2016 - 1 ) . 2016² - ( 2016 - 1 ) . 2016 - 1

P(2016)= 2016²⁰¹⁶ - 2016²⁰¹⁶ + 2016²⁰¹⁵ - 2016²⁰¹⁵ + 2016²⁰¹⁴  - ..... - 2016³ + 2016² - 2016² + 2016 - 1

P(2016) = 2016 - 1

P(2016) = 2015.

cái chỗ bằng 1 là cộng 1 đấy

tek tức là nó = 2017

đúng không