Ta có :

A = | x + 1 | + | x - 6 |

A = | x + 1 | + | 6 - x | \(\ge\)| x + 1 + 6 - x | = 7

\(\Rightarrow\)GTLN của A là 7 khi ( x + 1 ) . ( 6 - x ) \(\ge\)0 hay -1 \(\le\)x \(\le\)6

Ta có :

H = 2²⁰¹⁷ - 2²⁰¹⁶ - ... - 2 - 1

H = 2²⁰¹⁶ - ... - 2 - 1

...

H = 1

\(\Rightarrow\)2017^H = 2017¹ = 2017

giải chi tiết đi

vì ( x - 2014 )²⁰¹⁴ \(\ge\)0 \(\forall\)x

( y - 2015 )^{2014 \(\ge\)}0 \(\forall\)y

\(\Rightarrow\)( x - 2014 )²⁰¹⁴ + ( y - 2015 )²⁰¹⁴ \(\ge\)0 \(\forall\)x,y

Mà ( x - 2014 )²⁰¹⁴ + ( y - 2015 )²⁰¹⁴ = 0 

\(\Rightarrow\)\(\hept{\begin{cases}\left(x-2014\right)^{2014}=0\\\left(y-2015\right)^{2014}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=2014\\y=2015\end{cases}}\)

Vậy ( x ; y ) = ( 2014 ; 2015 )

Vì (x-2014)²⁰¹⁴ \(\ge\) 0

(y-2015)²⁰¹⁴ \(\ge\)0

=> (x-2014)²⁰¹⁴ + (y-2015)²⁰¹⁴ \(\ge\) 0

Mà (x-2014)²⁰¹⁴ + (y-2015)²⁰¹⁴  = 0

=> \(\hept{\begin{cases}\left(x-2014\right)^{2014}=0\\\left(y-2015\right)^{2015}=0\end{cases}\Rightarrow\hept{\begin{cases}x-2014=0\\y-2015=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2014\\y=2015\end{cases}}}\)

Xét hiệu :

H = \(\frac{a^2+b^2}{2}-\left(\frac{a+b}{2}\right)^2=\frac{2.\left(a^2+b^2\right)}{4}-\frac{\left(a+b\right)^2}{4}\)

\(=\frac{2a^2+2b^2-a^2-b^2-2ab}{4}=\frac{\left(a-b\right)^2}{2^2}=\left(\frac{a-b}{2}\right)^2\ge0\)\(\forall\)a,b

Dấu " = " xảy ra khi \(\left(\frac{a-b}{2}\right)^2=0\Leftrightarrow a=b\)

\(\Rightarrow\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2\)

Vậy ...

(x-2)⁸ = (x-2)¹⁰

=> (x-2)⁸ - (x-2)¹⁰ = 0

=> (x-2)⁸ - (x-2)².(x-2)⁸ = 0

=> (x-2)⁸.[1 - (x-2)² ] = 0

=> \(\orbr{\begin{cases}\left(x-2\right)^8=0\\1-\left(x-2\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x-2=0\\x-2=\pm1\end{cases}\Rightarrow}x\in\left\{1;2;3\right\}}\)