Ta có: x=100

nên x+1=101

Ta có: \(f\left(x\right)=x^8-101x^7+101x^6-101x^5+...+101x^2-101x+25\)

\(=x^8-x^7\left(x+1\right)+x^6\left(x+1\right)-x^5\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+25\)

\(=x^8-x^7-x^7+x^7+x^6-x^6-x^5+x^5-x^4+...+x^3+x^2-x^2-x+25\)

\(=-x+25\)

\(=-100+25=-75\)

Ta có: x=100

\(\Leftrightarrow x+1=101\)

Ta có: \(f\left(x\right)=x^{10}-101x^9+101x^8-101x^7+...+101x+2021\)

\(=x^{10}-x^9\cdot\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...+x\left(x+1\right)+2021\)

\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^2+x+2021\)

\(=x+2021\)

\(=100+2021=2121\)

f(100)=x⁸-(100+1)x⁷+(100+1)x⁶-(100+1)x⁵+....+(100+1)x²-(100+1)x+25

=x⁸-(x+1)x⁷+(x+1)x⁶-(x+1)x⁵+....+(x+1)x²-(x+1)x+25

=x⁸-x⁸-x⁷+x⁷+x⁶-x⁶-x⁵+...+x³+x²-x²-x+25

=25

vậy f(100)=25

Mk ko hiểu lắm! Hoàng Phúc

-x+25=25?

ai ket bn ko

giải hộ

1. Ta có :

f(x) = ( m - 1 ) . 1² - 3m . 1 + 2 = 0

f(x) = m - 1 - 3m + 2 = -2m + 1 = 0

\(\Rightarrow m=\frac{1}{2}\)

2.

a) M(x) = -2x² + 5x = 0 

\(\Rightarrow-2x^2+5x=x.\left(-2x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\-2x+5=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{2}\end{cases}}\)

b) N(x) = x . ( x - 1/2 ) + 2 . ( x - 1/2 ) = 0

N(x) = ( x + 2 ) . ( x - 1/2 ) = 0 

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-\frac{1}{2}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2}\end{cases}}\)

c) P(x) = x² + 2x + 2015 = x² + x + x + 1 + 2014 = x . ( x + 1 ) + ( x + 1 ) + 2014 = ( x + 1 ) . ( x + 1 ) + 2014 = ( x + 1 )² + 2014

vì ( x + 1 )² + 2014 > 0 nên P(x) không có nghiệm

bài 3 . 

tham khảo ở đây :  Câu hỏi của Trần Hà Mi - Toán lớp 7 - Học toán với OnlineMath

bài 4 . 

Ta có : 2n - 3 = 2n + 2 - 5 = 2 . ( n + 1 ) - 5

Để 2n - 3 \(⋮\)n + 1 thì 2 . ( n + 1 ) - 5 \(⋮\)n + 1 mà 2 . ( n + 1 ) \(⋮\)n + 1 nên 5 \(⋮\)n + 1

\(\Rightarrow\)n + 1 \(\in\)Ư ( 5 ) = { 1 ; -1 ; 5 ; -5 }

\(\Rightarrow\)n \(\in\){ 0 ; -2 ; 4 ; -6 }

\(f\left(100\right)\Rightarrow x=100\)

\(\Rightarrow x+1=101\)

Thay x + 1 = 101 ta được:

\(f\left(100\right)-x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-\left(x+1\right)x^5+...+\left(x+1\right)x^2-\left(x+1\right)x+25\)

\(=x^8-\left(x^8+x^7\right)+\left(x^7+x^6\right)-\left(x^6+x^5\right)+...+\left(x^3+x^2\right)-\left(x^2+x\right)+25\)

\(=x^8-x^8-x^7+x^7+x^6-x^6-x^5+...+x^3+x^2-x^2-x+25\)

\(=-x+25\)

\(=-100+25\)

\(=-75\)

Thank you very much