Khi x = 100 

<=> x + 1 = 101

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

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

= - x + 25 =  -75 

Lời giải:
a. $MN\parallel BC$ nên theo định lý Talet:

$\frac{MN}{BC}=\frac{AN}{AC}=\frac{AM}{AB}=\frac{1}{2}(1)$

$\Rightarrow N$ là trung điểm $AC$

$NP\parallel AB$ nên theo định lý Talet:
$\frac{NP}{AB}=\frac{CP}{CB}=\frac{CN}{CA}=\frac{1}{2}(3)$

$\Rightarrow  P$ là trung điểm $BC$

$\Rightarrow \frac{BP}{BC}=\frac{1}{2}(2)$

Từ $(1); (2)\Rightarrow \frac{MN}{BC}=\frac{BP}{BC}=\frac{1}{2}\Rightarrow MN=BP$

Từ $(1); (3)\Rightarrow \frac{NP}{AB}=\frac{AM}{AB}=\frac{1}{2}$

$\Rightarrow NP=AM$. Mà $AM=BM$ nên $NP=BM$

b.

$MN\parallel BC$ nên $\widehat{ANM}=\widehat{NCP}$ (đồng vị) 

$AN=NC$ (do $N$ là trung điểm $AC$)

$MN=PC$ (cùng = BP)

$\Rightarrow \triangle AMN=\triangle NPC$ (c.g.c)

Hình vẽ:

bạn phải nêu rõ 2 đáy của h thang mình mới tính đc

a)

X có CTHH là AB₂

=> NTK_A + 2.NTK_B = 120 (đvC)

Mà NTK_A : NTK_B = 7 : 4

=> NTK_A = 56 (đvC); NTK_B = 32 (đvC)

=> A là Fe, B là S

b) PTK_Y = x.1 + 31.1 + 16.4 = 98 (đvC)

=> x = 3

c) 

Z có CTHH là A₂B₅

PTK_Z = 2.NTK_A + 5.NTK_B = 6,75.\(PTK_{CH_4}\) = 108 (đvC)

Mà NTK_A : NTK_B = 7 : 8 

=> NTK_A = 14 (đvC); NTK_B = 16 (đvC)

=> A là N, B là O

- Ta có: \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

\(\Rightarrow\left(x^3-1\right)⋮\left(x^2+x+1\right)\)

- Áp dụng hệ quả nhị thức Newton ta có: \(x^n-1⋮x-1\) với \(n\in N\).

- Vì \(n\) không chia hết cho \(3\) \(\Rightarrow n\) có dạng \(3k+1\) hoặc \(3k+2\) \(\left(k\in N\right)\)

- Với \(n=3k+1\) thì:

\(x^{2n}+x^n+1=x^{2\left(3k+1\right)}+x^{3k+1}+1=x^{6k+2}+x^{3k+1}+1=x^{3k+2}\left(x^{3k}-1\right)+x^{3k}\left(x^2+x+1\right)-\left(x^{3k}-1\right)\)

- Do \(\left\{{}\begin{matrix}x^{3k+2}\left(x^{3k}-1\right)⋮\left(x^{3k}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\\x^{3k}\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\\\left(x^{3k}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\end{matrix}\right.\) 

\(\Rightarrow x^{3k+2}\left(x^{3k}-1\right)+x^{3k}\left(x^2+x+1\right)-\left(x^{3k}-1\right)⋮\left(x^2+x+1\right)\)

hay \(x^{2n}+x^n+1⋮x^2+x+1\) khi \(n=3k+1\left(k\in N\right)\) (1).

- Với \(n=3k+2\) thì:

\(x^{2n}+x^n+1=x^{2\left(3k+2\right)}+x^{3k+2}+1=x^{6k+4}+x^{3k+2}+1=x^{3k+4}\left(x^{3k}-1\right)+x^{3k+2}\left(x^2+x+1\right)-\left(x^{3k+3}-1\right)\)- Do \(\left\{{}\begin{matrix}x^{3k+4}\left(x^{3k}-1\right)⋮\left(x^{3k}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\\x^{3k+2}\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\\\left(x^{3\left(k+1\right)}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\end{matrix}\right.\) 

\(\Rightarrow x^{3k+4}\left(x^{3k}-1\right)+x^{3k+2}\left(x^2+x+1\right)-\left(x^{3k+3}-1\right)⋮\left(x^2+x+1\right)\)

 hay  \(x^{2n}+x^n+1⋮x^2+x+1\) khi \(n=3k+2\left(k\in N\right)\) (2).

- Từ (1), (2) ta suy ra đpcm

D , E , F lần lượt là trung điểm của AB , AC , BC 

=> DE , DF và EF sẽ lần lượt là các đường trung bình ứng với BC , AC , AB 

\(\Rightarrow\left\{{}\begin{matrix}DE=\dfrac{1}{2}BC=\dfrac{1}{2}.14=7\left(cm\right)\\DF=\dfrac{1}{2}AC=\dfrac{1}{2}.10=5\left(cm\right)\\\text{EF}=\dfrac{1}{2}AB=\dfrac{1}{2}.6=3\left(cm\right)\end{matrix}\right.\)

\(\left(x-3\right)^4+\left(x-1\right)^4-16\)

\(=\left(x-3\right)^4+\left[\left(x-1\right)^2-4\right]\left[\left(x-1\right)^2+4\right]\)

\(=\left(x-3\right)^4+\left(x-1-2\right)\left(x-1+2\right)\left(x^2-2x+5\right)\)

\(=\left(x-3\right)^4+\left(x-3\right)\left(x+1\right)\left(x^2-2x+5\right)\)

\(=\left(x-3\right)\left(x^3-9x^2+27x-27\right)+\left(x-3\right)\left(x^3-2x^2+5x+x^2-2x+5\right)\)

\(=\left(x-3\right)\left(x^3-9x^2+27x-27\right)+\left(x-3\right)\left(x^3-x^2+3x+5\right)\)

\(=\left(x-3\right)\left(x^3-9x^2+27x-27+x^3-x^2+3x+5\right)\)

\(=\left(x-3\right)\left(2x^3-10x^2+30x-22\right)\)

\(=2\left(x-3\right)\left(x^3-5x^2+15x-11\right)\)

\(=2\left(x-3\right)\left(x^3-x^2-4x^2+4x+11x-11\right)\)

\(=2\left(x-3\right)\left[x^2\left(x-1\right)-4x\left(x-1\right)+11\left(x-1\right)\right]\)

\(=2\left(x-3\right)\left(x-1\right) \left(x^2-4x+11\right)\)