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

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:

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

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

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

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

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

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

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

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

\(B=1\)

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

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

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

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

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

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

\(\Rightarrow B=-1\)

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

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 :

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

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

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

\(B=-1\)

Đặt x -2006 = y 

pt <=>  \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)

<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)

<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)

<=> \(49y^2-49y+49=57y^2-57y+19\)

<=> \(8y^2-8y-30=0\)

<=> \(4y^2-4y+15=0\)

Giải tiếp nha 

Ta có nhận xét : \(a+b=1\) thì

\(f\left(a\right)+f\left(b\right)=\frac{4^a}{4^a+2}+\frac{4^b}{4^b+2}=\frac{4^a\left(4^a+2\right)+4^b\left(4^b+2\right)}{\left(4^a+2\right)\left(4^b+2\right)}\)

                 \(=\frac{4^{a+b}+2.4^a+4^{a+b}+2.4^b}{4^{a+b}+2.4^a+2.4^b+4}=\frac{2.4^a+2.4^b+8}{2.4^a+2.4^b+8}=1\)

Áp dụng kết quả trên ta có :

\(S=\left[f\left(\frac{1}{2007}\right)+f\left(\frac{2006}{2007}\right)\right]+\left[f\left(\frac{2}{2007}\right)+f\left(\frac{2005}{2007}\right)\right]+...+\left[f\left(\frac{1003}{2007}\right)+f\left(\frac{1004}{2007}\right)\right]\)

Vâyu \(S=1+1+1+...+1=1003\) (có 1003 số hạng)

khó quá ak

ừ, bạn bik làm thì giúp mình nha ^^

Theo mk thì hình như

=0 

Ko biết chắc !

\(\frac{1}{2007}.\left(\frac{1001}{2006}-2007\right)-\left(\frac{1}{2006}-2007\right).\frac{1001}{2007}\)

\(=\left(\frac{1001}{2007.2006}-\frac{2007}{2007}\right)-\left(\frac{1001}{2006.2007}-\frac{2007.1001}{2007}\right)\)

\(=\frac{1001}{2007.2006}-\frac{1001}{2006.2007}-1+1001\)

\(=-1+1001\)

\(=1000\)