Tại x= - 3

=> \(B=\left[\left(-3\right)^{2017}+3\left(-3\right)^{2016}-1\right]^{2017}\)

=> \(B=\left[\left(-3\right)^{2017}+3^{2017}-1\right]^{2017}\)

=> \(B=\left(-1\right)^{2017}\)

=> B = - 1

Ta có:

\(B=\left(x^{2007}+3x^{2006}-1\right)^{2007}\)

\(B=\left(\left(-3\right)^{2007}+3\left(-3\right)^{2006}-1\right)^{2007}\)

\(B=\left(\left(-3\right)^{2007}+3\left(3\right)^{2006}-1\right)^{2007}\)

\(B=\left(\left(-3\right)^{2007}+3^1\left(3\right)^{2006}-1\right)^{2007}\)

\(B=\left(\left(-3\right)^{2007}+3^{1+2006}-1\right)^{2007}\)

\(B=\left(\left(-3\right)^{2007}+3^{2007}-1\right)^{2007}\)

\(B=\left(0-1\right)^{2007}\)

\(B=\left(-1\right)^{2007}\)

\(B=1\)

\(B=\left(x^{2007}+3x^{2006}-1\right)^{2007}\)

\(\Rightarrow B=\left[\left(-3\right)^{2007}+3.\left(-3\right)^{2006}-1\right]^{2007}\)

\(\Rightarrow B=\left[\left(-3\right)^{2006}.\left(3+\left(-3\right).1\right)-1\right]^{2007}\)

\(\Rightarrow B=\left[\left(-3\right)^{2006}.0-1\right]^{2007}\)

\(\Rightarrow B=\left(0-1\right)^{2007}\)

\(\Rightarrow B=-1^{2007}\)

\(\Rightarrow B=-1\)

Khi x=-3 thì biểu thức:

 \(\Rightarrow B=\left(-3^{2007}+3.\left(-3\right)^{2006}-1\right)^{2007}\)

\(\Rightarrow B=.............\)

máy tính tính cũng không ra nha bạn

Thay \(x=-3\) vào biểu thức B ta được :

\(B=\left(-3^{2007}+3.\left(-3\right)^{2006}-1\right)^{2007}\)

\(=\left(-3^{2007}+3^{2007}-1\right)^{2007}\)

\(=-1^{2007}\)

\(=-1\)

Thay x =\(-3\)vào biểu thức B ta được :

\(B=\left(-3^{2017}+3.\left(-3\right)^{2016}-1\right)^{2007}\)

\(B=\left(-3^{2007}+3^{2007}-1\right)^{2007}\)

\(B=\left(-1\right)^{2007}\)

\(B=-1\)

a) \(A=x^{15}+3x^{14}+5\)

\(=x^{14}\left(x+3\right)+5\)

\(=x^{14}.0+5\)

= 5

b) x = -3 => x + 3 = 0

\(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)

\(=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)

\(=\left(x^{2006}.0+1\right)^{2007}\)

\(=1^{2007}=1\)

\(A=x^{15}+3.x^{14}+5\text{ biết x+3=0}\)

\(A=x^{14}.\left(x+3\right)+5\)

\(\text{Do x+3=0}\Rightarrow A=x^{14}.0+5\)

\(A=0+5\)

\(A=5\)        \(\text{Vậy }A=5\text{ với x+3=0}\)

\(B=\left(x^{2007}+3.x^{2006}+1\right)^{2007}\text{ biết x=-3}\)

\(B=\left[x^{2006}.\left(x+3\right)+1\right]^{2007}\)

\(\text{Do x=-3}\Rightarrow B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)

\(B=\left(x^{2006}.0+1\right)^{2007}\)

\(B=\left(0+1\right)^{2007}\)

\(B=1^{2007}\)

\(B=1\)           \(\text{Vậy }B=1\text{ với x=-3}\)

Ngu thì câm mồm đi

Tại \(x=-3\) thì:

\(B=\left[-3^{2007}+3.\left(-3\right)^{2006}-1\right]^{2007}\)

\(\Rightarrow B=\left[-3^{2006}\left(-1+3\right)-1\right]^{2007}\)

\(\Rightarrow B=\left[-3^{2006}.2-1\right]^{2007}\)

Thay \(x=3\) vào B ta có:

\(B=\left[\left(-3\right)^{2007}+3.\left(-3\right)^{2006}-1\right]^{2007}\)

\(=\left[\left(-3\right)^{2007}+3^{2007}-1\right]^{2007}\)

\(=\left(-1\right)^{2007}=-1\)

B=\(\left(x^{2007}+3x^{2006}-1\right)^{2007}\)

\(B=\left[\left(-3^{2007}\right)+3\left(-3^{2006}\right)-1\right]^{2007}\)

\(B=\left[\left(-3^{2007}\right)+3.3^{2006}-1\right]^{2007}\)

\(B=\left[\left(-3^{2007}\right)+3^{2007}-1\right]^{2007}\)

\(B=\left(-1\right)^{2007}=\left(-1\right)\)

Ta có: \(B=\left(x^{2007}+3x^{2006}-1\right)^{2007}\)

Thay x = -3 vào B ta có:

\(B=\left(\left(-3\right)^{2007}+3\left(-3\right)^{2006}-1\right)^{2007}\)

=>\(B=\left(\left(-3\right)^{2007}+3\cdot3^{2006}-1\right)^{2007}\)

=>\(B=\left(\left(-3\right)^{2007}+3^{2007}-1\right)^{2007}\)

=>\(B=\left(0-1\right)^{2007}\)

\(=>B=\left(-1\right)^{2007}=-1\)

1) 1

2) \(a\in\left\{-4;-2;0;2\right\}\)

3)\(\dfrac{7}{9}\)

tick nha!!!

Ta có:B=\(\left(x^{2007}+3x^{2006}-1\right)^{2007}\)

\(\Rightarrow\)B=\(\left(\left(-3\right)^{2007}+3\times\left(-3\right)^{2006}-1\right)^{2007}\)

B=\(\left(\left(-3\right)\times\left(-3\right)^{2006}+3\times\left(-3\right)^{2006}-1\right)^{2007}\)

B=\(\left(\left(\left(-3\right)+3\right)\times\left(-3\right)^{2006}-1\right)^{2007}\)

B=\(\left(0\times\left(-3\right)^{2006}-1\right)^{2007}\)

B=\(\left(0-1\right)^{2007}\)

B=\(\left(-1\right)^{2007}\)

B=\(\left(-1\right)\)

Mk sửa lại đề nha tìm GTNN

a) B=|x- 2006| -|2007- x|

           Vì |x- 2006|\(\ge\)0

                |2007- x|\(\ge\)0

       Suy ra:|x- 2006| -|2007- x|\(\ge\)0

Dấu = xảy ra khi x-2006=0;x=2006

                           2007-x=0;x=2007

             Vậy Min B=0 khi x=2006

                                        x=2007

\(B=\left|x-2006\right|-\left|2007-x\right|=\left|x-2006\right|-\left|x-2007\right|\)

Áp dụng bđt: \(\left|A\right|-\left|B\right|\le\left|A-B\right|\)

=>\(B\le\left|x-2006-x+2007\right|=1\)

Vậy GTLN của B là 1 khi \(2006\le x\le2007\)

lm dưới rùi