=> 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

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

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

\(\Rightarrow2014x+2029105=2015x\)

\(\Rightarrow2015x-2014x=2029105\)

\(\Rightarrow x=2029105\)

a) Nghiệm bằng 1 nha: 1^2016-1^2014=1-1=0

b)Không có nghiệm âm còn vì sao thì đợi lhi bạn k đug cho mk xog thì mk giải thick cho nha!

x²⁰¹⁶-x²⁰¹⁴=0

x²⁰¹⁴*(x²-1)=0

TH1:

x²⁰¹⁴=0

x=0

TH2

x²-1=0

x²=1

x=1

k mình nha

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

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

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

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

với x=2014

=> f(x)=x²⁰¹⁴-(x+1)x²⁰¹³+(x+1)x²⁰¹²-...-(x+1)x+(x+1)

=x²⁰¹⁴-x²⁰¹⁴-x²⁰¹³+x²⁰¹³+x²⁰¹²-...-x²-x+x+1

=1

cảm ơn nha

a) \(x-\dfrac{3}{5}=\dfrac{4}{-10}\)

\(x=\dfrac{4}{-10}+\dfrac{3}{5}\)

\(x=\dfrac{-4}{10}+\dfrac{6}{10}\)

\(x=\dfrac{1}{5}\)

b) \(\dfrac{3}{x}-2=\dfrac{4}{x}+4\)

\(\dfrac{3}{x}-2+2=\dfrac{4}{x}+4+2\)

\(\dfrac{3}{x}=\dfrac{4}{x}+4\)

\(\dfrac{3}{x}=\dfrac{4x+4}{x}\)

\(3x=\left(4x+4\right)x\)

\(3x=5x\cdot x+4x\)

\(3x=x\left(5x+4\right)\)

\(3=5x+4\)

\(5x=-1\)

\(x=\dfrac{-1}{5}\)

Toán này lớp 6 á???