\(\frac{1}{25}-x^2\)

\(=\left(\frac{1}{5}\right)^2-x^2\)

\(=\left(\frac{1}{5}-x\right)\left(\frac{1}{5}+x\right)\)

#H

x=565>2 nhé

Đáp án C 

\(\frac{CD}{BD}=\frac{AC}{AB}=\frac{3}{4}\)(định lí)

\(\Rightarrow C\)

\(1^2-2^2+3^2-4^2+...+2009^2-2010^2+2011^2\)

\(=1^2+\left(3^2-2^2\right)+\left(5^2-4^2\right)+...+\left(2011^2-2010^2\right)\)

\(=1+\left(3-2\right)\left(3+2\right)+\left(5-4\right)\left(5+4\right)+...+\left(2011-2010\right)\left(2011+2010\right)\)

\(=1+2+3+...+2010+2011\)

\(=\frac{2011.2012}{2}=2023066\)

a) \(-5x^2-2xy-2y^2+14x-10y-1\)

\(=-x^2-y^2-9-2xy+6x+6y-4x^2+8x-4-y^2+4y-4+16\)

\(=-\left(x+y-3\right)^2-4\left(x-1\right)^2-\left(y-2\right)^2+16\le16\)

Dấu \(=\)khi \(\hept{\begin{cases}x+y-3=0\\x-1=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\).

b) \(-8x^2-3y^2-26x+6y+100\)

\(=-8\left(x+\frac{13}{8}\right)^2-3\left(y-1\right)^2+\frac{993}{8}\le\frac{993}{8}\)

Dấu \(=\)khi \(\hept{\begin{cases}x+\frac{13}{8}=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{13}{8}\\y=1\end{cases}}\)

Đặt A = -x² - y² + xy + 2x + 2y

=> 4A = -4x² - 4y² + 4xy + 8x + 8y

         = -(4x² - 4xy + y²) + 4(2x - y) - 4 - 3y² + 12y - 12 + 16 

         = -(2x - y)² + 4(2x - y) - 4 - 3(y² - 4y + 4) + 16

         = -(2x - y - 2)² - 3(y - 2)² + 4 \(\le16\)

=> A \(\le4\)

=> Max A = 4

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-y-2=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=2\end{cases}}\)

Vậy Max A = 4 <=> x = y = 2

c) \(O\)là trung điểm \(BC\)suy ra \(OA=OB\)

\(\Rightarrow\Delta OAB\)cân tại \(O\)nên \(\widehat{OAB}=\widehat{OBA}\). 

\(\Delta AMN~\Delta ACB\Rightarrow\widehat{AMN}=\widehat{ACB}\)

\(\Rightarrow\widehat{OAB}+\widehat{AMN}=\widehat{ABC}+\widehat{ACB}=90^o\)

\(\Rightarrow\widehat{AIM}=90^o\)

suy ra đpcm. 

d) \(\frac{P_{AMN}}{P_{ABC}}=\frac{12}{24}=\frac{1}{2}\)suy ra hệ số đồng dạng của hai tam giác \(AMN\)và \(ACB\)là \(\frac{1}{2}\).

\(\Rightarrow\frac{MN}{CB}=\frac{1}{2}\)mà \(MN=AH,BC=2OA\)nên \(\frac{AH}{OA}=1\)

do đó tam giác \(ABC\)vuông cân tại \(A\)nên \(\widehat{ABC}=45^o\).

Đặt \(d=11...1\)(\(n\)chữ số \(1\)) suy ra \(10^n=9d+1\).

\(a=10^n.d+d=\left(9d+1\right).d+d=9d^2+2d\)

\(b=10d+1\)

\(c=6d\)

\(a+b+c+8=9d^2+2d+10d+1+6d+8\)

\(=9d^2+18d+9=\left(3d+3\right)^2\)là số chính phương.