x=7

nên x+1=8

\(A=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-5\)

\(=x-5=7-5=2\)

thay x=7

ta có:7^15-8*7^14+887^13-8*7^12+...-8*7^2+8*7-2015

f(7)=7^15-8.7^14+8.7^13-8.7^12+... 
-8.7^2+8.7-5= 
= -7^14+8.7^13-8.7^12+...-8.7^2+8.7-5= 
=7^13-8.7^12+...-8.7^2+8.7-5= 
= -7^12+...-8.7^2+8.7-5= 
=...= -7^2+8.7-5=7-5=2 
Kết quả là 2

sao mà 2 câu giống nhau thế

f(7)=7^15-8.7^14+8.7^13-8.7^12+... 
-8.7^2+8.7-5= 
= -7^14+8.7^13-8.7^12+...-8.7^2+8.7-5= 
=7^13-8.7^12+...-8.7^2+8.7-5= 
= -7^12+...-8.7^2+8.7-5= 
=...= -7^2+8.7-5=7-5=2 
Kết quả là 2

vào câu hỏi tương tự

tích nha

x=7 nên x+1=8

B=x^15-x^14(x+1)+x^13(x+1)-...-x^2(x+1)+x(x+1)-5

=x^15-x^15-x^14+x^14+...-x^3-x^2+x^2+x-5

=x-5

=7-5

=2

\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)

\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)

\(=2\)

\(\dfrac{7}{8x}+\dfrac{5-x}{4x^2-8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}\)
ĐKXĐ: x ≠ 0; x ≠ 2

\(< =>\dfrac{14x-28+20-4x}{16x\left(x-2\right)}=\dfrac{8x-8+2x}{16x\left(x-2\right)}\)
Suy ra: 14x - 28 + 20 - 4x = 8x - 8 + 2x
<=> 14x - 8x - 2x - 4x = 28 - 20 - 8
<=> 0x = 0
Vậy: S = { x | x ≠ 0;2 }

          \(P\left(x\right)=8x^3\)         +    5x   -1 

       +  \(Q\left(x\right)\)=          \(4x^2\)  -    3x  + 7

       + \(R\left(x\right)=8x^3+8x^2+7x\)

Tổng :            16x^3 + 12x^2 +9x + 6

\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\) ( ĐKXĐ: \(x\ne0;x\ne2\) )

\(\Leftrightarrow\dfrac{x-1}{2x\left(x-2\right)}-\dfrac{7}{8x}=\dfrac{5-x}{4x\left(x-2\right)}-\dfrac{1}{8\left(x-2\right)}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)4}{8x\left(x-2\right)}-\dfrac{7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{2\left(5-x\right)}{8x\left(x-2\right)}-\dfrac{1x}{8x\left(x-2\right)}\)

\(\Rightarrow4x-4-7x+14=10-2x-x\)

\(\Leftrightarrow-3x+2x+x=10+4-14\)

\(\Leftrightarrow0=0\)

          Vậy pt đã cho có nghiệm đúng với mọi x

khó thế

Vì x = 7

\(\Rightarrow\)\(x+1=8\)

\(\Rightarrow\)\(A=x^{15}\)\(-\)\(8x^{14}\)\(+\)\(8x^{13}\)\(-\)\(8x^{12}\)\(+\)... \(-\)\(8x^2\)\(+8x-5\)

\(=\)\(x^{15}\)\(-\left(x+1\right)x^{14}\)\(+\left(x+1\right)x^{13}\)\(-\left(x+1\right)x^{12}\)\(+\)... \(-\)\(\left(x+1\right)x^2\)\(+\left(x+1\right)x-5\)

\(=\)\(x^{15}\)\(-\)\(x^{15}\)\(-\)\(x^{14}\)\(+\)\(x^{14}\)\(+\)\(x^{13}\)\(-\)\(x^{13}\)\(-\)\(x^{12}\)\(+\)... \(-\)\(x^3\)\(-\)\(x^2\)\(+\)\(x^2\)\(+\)\(x\)\(-\)\(5\)

\(\Rightarrow\)\(x=-5\)

\(\Rightarrow\)\(A=7-5=2\)

Vậy  \(A=2\) khi  \(x=7\)