=> 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

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

Ta có :

|x+1| \(\ge\)0

|x+2| \(\ge\)0

|x+3| \(\ge\)0

..........

|x+2014| \(\ge\)0

=> |x+1| + |x+2| + |x+3| +..........+ |x+2014| \(\ge\)0

=> 2015x \(\ge\)0

Mà 2015 \(\ge\)0

=> x \(\ge\)0 

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

= x + 1 + x + 2 + x + 3 +.................... + x + 2014 = 2015x

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

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

=> x = 2029105

toán bồi dưởng à

Ko bài kiểm tra

[ ] là giá trị tuyệt đối à

Ta thấy :

\(\left|x+1\right|\ge0\)

\(\left|x+2\right|\ge0\)

............

|x + 2014| \(\ge0\)

Cộng vế với vế ta được :

\(\left|x+1\right|+\left|x+2\right|+....+\left|x+2014\right|\ge0\)

Mà \(\left|x+1\right|+\left|x+2\right|+....+\left|x+2014\right|=2015x\Rightarrow2015x\ge0\Rightarrow x\ge0\)\(\Rightarrow x+1+x+2+....+x+2014=2015x\)

\(\Rightarrow2014x+\frac{2014.2015}{2}=2015x\)

\(\Rightarrow2014x+2029105=2015x\)

\(\Rightarrow2015x-2014x=2029105\)

\(\Rightarrow x=2029105\)

Ta có:

\(\left|x+1\right|\ge0,\left|x+2\right|\ge0,...,\left|x+2014\right|\ge0\)

\(\Rightarrow\)\(\left|x+1\right|+\left|x+2\right|+...+\left|x+2014\right|\ge0\)

\(\Rightarrow2015x\ge0\)

\(\Rightarrow x\ge0\)

Khi đó :\(\left|x+1\right|=x+1,\left|x+2\right|=x+2,...,\left|x+2014\right|=x+2014\)\(\Rightarrow x+1+x+2+...+x+2014=2015x\)

\(\Rightarrow2014x+1+2+...+2014=2015x\)

\(\Rightarrow1+2+..+2014=x\)

\(\Rightarrow x=\dfrac{\left(1+2014\right)2014}{2}=2029105\)

\(x^4-2015x^3+2015x^2-2015x+2015\)

\(=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)(vì x=2014 nên 2015=x+1)

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

\(=1\)

a=x⁴-2223x³+2223x²-2223x+2223

=x³(x-2223)+x(x-2223)+2222x²+2003(*)

thay x=2222,ta co:

(*)<=>-2222³-2222+2222³+2223=1

dung thi chon nha