Ta thấy \(x^{2002}+x^{2000}+1\) có dạng \(x^{3m+1}+x^{3n+1}+1\)

Ta sẽ đi chứng minh \(x^{3m+1}+x^{3n+1}+1⋮x^2+x+1\)

Thật vậy,ta có:

\(x^{3m+1}+x^{3n+2}+1\)

\(=x^{3m+1}-x+x^{3n+2}-x^2+x^2+x+1\)

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

Mà \(x^{3m}-1⋮x^2+x+1;x^{3n}-1⋮x^2+x+1\) nên \(x^{3m+1}+x^{3n+2}+1⋮x^2+x+1\)

\(\frac{x^4-1}{x-1}=\frac{B}{x+1}\)

\(\Leftrightarrow\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x-1\right)}=\frac{B}{x+1}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)}=\frac{B}{x+1}\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=\frac{B}{x+1}\)

\(\Leftrightarrow\left(x+1\right)^2\left(x^2+1\right)=B\)

\(\Leftrightarrow B=\left(x^2+2x+1\right)\left(x^2+1\right)\)

\(\Leftrightarrow B=x^4+2x^3+x^2+x^2+2x+1\)

\(\Leftrightarrow B=x^4+2x^3+2x^2+2x+1\)

Em không biết đúng hay sai đâu nhé?!

Ta có: \(\frac{x^4-1}{x-1}=\frac{B}{x+1}\)

\(\Leftrightarrow B\left(x-1\right)=\left(x^4-1\right)\left(x+1\right)\)

\(\Leftrightarrow B\left(x-1\right)=\left(x^2-1\right)\left(x^2+1\right)\left(x+1\right)\)

\(\Leftrightarrow B\left(x-1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x+1\right)\)

\(\Leftrightarrow B\left(x-1\right)=\left(x-1\right)\left(x^2+1\right)\left(x+1\right)^2\)
\(\Leftrightarrow B=\frac{\left(x-1\right)\left(x^2+1\right)\left(x+1\right)^2}{x-1}\)

Vậy \(B=\left(x^2+1\right)\left(x+1\right)^2\)

$3^8{.5}^8-\left({15}^4-1\right)\left({15}^4+1\right)$

$={15}^8-\left({15}^8-1\right)$

$={15}^8-{15}^8+1=1$

$\mathrm{Học\; tốt}$

\(3^8.5^8-\left(15^4-1\right)\left(15^4+1\right)\)

\(=15^8-\left[\left(15^4\right)^2-1^2\right]\)

\(=15^8-\left[15^8-1\right]\)

\(=15^8-15^8+1=1\)

\(a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ca\) 

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)

\(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(=\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(a-c\right)^2\ge0\\\left(b-c\right)^2\ge0\end{cases}}\Rightarrow\)\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)

Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)

Vậy a = b = c (đpcm)

Tính nhanh :

126² - 152 x 126 + 5776

= 126 x 126 - 152 x 126 + 5776

= 126(126-152) + 5776

= -3276 + 5776

= 2500

\(126^2-152.126+5776\)

\(=126.126-152.126+5776\)

\(=126.\left(126-152\right)+5776\)

\(=126.\left(-26\right)+5776\)

\(=\left(-3276\right)+5776\)

\(=2500\)

Gọi khoảng cách từ A đến BM,DN lần lượt là h và k. Kẻ MH vuông góc AB.

Ta có \(S_{AMB}=\frac{MH.AB}{2}=\frac{S_{ABCD}}{2}\). Tương tự \(S_{AND}=\frac{S_{ABCD}}{2}\)

Do đó \(2S_{AMB}=2S_{AND}\) hay \(h.BM=k.DN\). Mà BM = DN nên \(h=k\)

Suy ra khoảng cách từ A đến 2 đường thẳng BM,DN là bằng nhau; BM cắt DN tại I

Vậy thì A nằm trên phân giác của ^DIB hay IA là phân giác góc DIB (đpcm).

ghiwyh

a) (x+y)^2: (x+y)=x+y