\(\frac{x+4}{2010}+\frac{x+3}{2011}=\frac{x+2}{2012}+\frac{x+1}{2013}\)

\(\Leftrightarrow\left(\frac{x+4}{2010}+1\right)+\left(\frac{x+3}{2011}+1\right)=\left(\frac{x+2}{2012}+1\right)+\left(\frac{x+1}{2013}+1\right)\)

\(\Leftrightarrow\frac{x+2014}{2010}+\frac{x+2014}{2011}=\frac{x+2014}{2012}+\frac{x+2014}{2013}\)

\(\Leftrightarrow\frac{x+2014}{2010}+\frac{x+2014}{2011}-\frac{x+2014}{2012}-\frac{x+2014}{2013}=0\)

\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)

\(\Leftrightarrow x+2014=0\)

\(\Leftrightarrow x=-2014\)

V...

Ta có  : \(B=\frac{x^2+3+12}{x^2+3}=1+\frac{12}{x^2+3}.\)Do \(x^2\ge0\)với mọi x nên \(x^2+3\ge3\Rightarrow\frac{12}{x^2+3}\le4\Rightarrow\frac{12}{x^2+3}+1\le4+1\)hay \(B\le5.\)Vậy \(maxB=37\)đạt được khi \(x=0.\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)suy ra đpcm.

tam giác OAH và OBHcó

góc OAH=OBH=90 độ, góc AOH=BOH (phân giác), cạnh OH chung

=>tam giac OAH=OBH(cạnh huyền góc nhọn)

=>OA=OB; AH=BH

tam giác ABO và BAH có

OA=OB,AH=BH,AB chung

=>tam giác ABO=BAH (c.c.c)

x y H A B

a) ta có : (x-5)\(^2\) =x-5

\(\Rightarrow\)(x-5)\(^2\) - (x-5)=0

\(\Rightarrow\)(x-5)(x-6)=0

\(\Rightarrow\)\(\orbr{\begin{cases}x-5=0\\x-6=0\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=5\\x=6\end{cases}}\)

a)\(\left(x-5\right)^2=x-5\Leftrightarrow\left(x-5\right)^2-\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)