Ta có:\(\hept{\begin{cases}\left(y-3\right)^{2014}\ge0\\\left|2x+1\right|^{2015}\ge0\end{cases}}\)\(\Rightarrow\left(y-3\right)^{2014}+\left|2x+1\right|^{2015}\ge0\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(y-3\right)^{2014}=0\\\left|2x+1\right|^{2015}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}y-3=0\\2x+1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}y=3\\x=-\frac{1}{2}\end{cases}}\)

Ta có: (y-3)²⁰¹⁴ \(\ge\)0 và |2x+1|²⁰¹⁵ \(\ge\)0

Mà (y-3)²⁰¹⁴ + |2x+1|²⁰¹⁵ = 0 => (y-3)²⁰¹⁴ = 0 và |2x+1|²⁰¹⁵ = 0

=> y - 3 = 0 và 2x + 1 = 0

=> y = 3 và 2x = -1

=> y = 3 và x = -1/2.

Vậy y = 3 và x = -1/2.

Vì (y-3)²⁰¹⁴lớn hơn hoặc =0 với mọi y

|2x+1|²⁰¹⁵ lớn hơn hoặc =0 với mọi x

=>  (y-3)²⁰¹⁴+|2x+1|²⁰¹⁵ lớn hơn hoặc =0 với mọi x,y

Mà (y-3)²⁰¹⁴+|2x+1|²⁰¹⁵=0

=>  (y-3)²⁰¹⁴=0 và |2x+1|²⁰¹⁵=0

<=> y-3=0         và |2x+1|=0

<=>y=3             và 2x+1=0

<=>y=3             và 2x=-1

<=>y=3             và x=-1/2

Vậy y=3; x=-1/2

a: \(\left(2x-3\right)^{2012}+\left(y-\dfrac{2}{5}\right)^{2014}+\left|x+y-z\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\y-\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{2}{5}\\z=\dfrac{19}{10}\end{matrix}\right.\)

b: 2015-|x-2015|=x

=>|x-2015|=2015-x

=>x-2015<=0

hay x<=2015

d: |x-999|+|1998-2x|=0

=>x-999=0

hay x=999

Ta có:/x-3/^2014>=0;/6+2y/^2015>=0

=>/x-3/^2014+/6+2y/^2015>=0

mà theo đề bài, /x-3/^2014+/6+2y/^2015<=0

=>/x-3/^2014=/6+2y/^2015=0

=>/x-3/=0;       /6+2y/=0

=>x-3=0           =>6+2y=0

=>x=3              =>2y=-6=>y=-3

vậy x=3; y=-3

3 hoặc -3 đúng zồi đó

3;-3 đúng 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000%

f)

\(A=\sqrt{\frac{\left(x+1\right)}{x-3}}=\sqrt{1+\frac{4}{x-3}}\)

x-3={-4)=> x=-1

x=3

y=-3

T..i..c..k mk nha