Ta có với a,b,c,d là các số thực khác 0 

\(\Rightarrow\frac{a-b+c+d}{b}=\frac{a+b-c+d}{c}=\frac{a+b+c-d}{d}=\frac{b+c+d-a}{a}\)

\(\Rightarrow\frac{a-b+c+d}{b}+1=\frac{a+b-c+d}{c}+1=\frac{a+b+c-d}{d}+1=\frac{b+c+d-a}{a}+1\)

\(\Rightarrow\frac{a+c+d}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}=\frac{b+c+d}{a}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có: 

\(\Rightarrow\frac{a+c+d}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}=\frac{b+c+d}{a}=\frac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)

Ta có M= \(\left(\frac{a+c+d}{b}\right)\left(\frac{a+b+d}{c}\right)\left(\frac{a+b+c}{d}\right)\left(\frac{b+c+d}{a}\right)\)

=> M= 3.3.3.3 

=> M =81

Áp dụng TC cuae DTSBN ta có:

a-b+c+d/b = a+b-c+d/c = a+b+c-d/d = b+c+d-a/a = \(\frac{a-b+c+d+a+b-c+d+a+b+c-d+b+c+d-a}{b+c+d+a}=\frac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)

=> a-b+c+d/b = 3 => a-b+c+d = 3b => a+c+d = 4b

a+b-c+d/c = 3 => a+b-c+d = 3c => a+b+d = 4c

a+b+c-d/d = 3 => a+b+c-d = 3d => a+b+c = 4d

b+c+d-a/a = 3 => b+c+d-a = 3a => b+c+d = 4a

=> M = \(\frac{\left(a+b+c\right)\left(a+b+d\right)\left(b+c+d\right)\left(c+d+a\right)}{abcd}=\frac{4d.4c.4a.4b}{abcd}=\frac{256abcd}{abcd}=256\)

Vậy M = 256

cho 2012=x+1

B=x²⁰¹² - (x+1)x^2010+(x+1)x^2009-...+(x+1)x+1

B=x^2012-x^2012-x^2011+x^2011+x^2010-...+x^2+x+1

B=x+1=2012

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 ...