\(|x^{2018}+|x-1||=x^{2018}+2404\)

\(\Leftrightarrow\orbr{\begin{cases}x^{2018}+|x-1|=-x^{2018}-2404\\x^{2018}+|x-1|=x^{2018}+2404\end{cases}\Leftrightarrow\orbr{\begin{cases}|x-1|=-\left(2x^{2018}+2404\right)\left(l\right)\\|x-1|=2404\left(n\right)\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=-2404\\x-1=2404\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2403\\x=2405\end{cases}}}\)

V...

\(\left|x^{2018}+\left|x-1\right|\right|=x^{2018}+2404\)

Ta thấy: \(x^{2018}\ge0\);\(\left|x-1\right|\ge0\)\(\Rightarrow x^{2018}+\left|x-1\right|\ge0\)

\(\Rightarrow\left|x^{2018}+\left|x-1\right|\right|=x^{2018}+2404\)

   \(\Leftrightarrow x^{2018}+\left|x-1\right|=x^{2018}+2404\) 

                          \(\left|x-1\right|=2404\)

\(\Rightarrow\orbr{\begin{cases}x-1=2404\\x-1=-2404\end{cases}}\Rightarrow\orbr{\begin{cases}x=2405\\x=-2403\end{cases}}\)

             Vậy \(x\in\left\{2405;-2403\right\}\)

\(\Leftrightarrow\frac{x+4}{9}+\frac{x+11}{8}+\frac{x+16}{7}+\frac{x+19}{6}=10\)

\(\Leftrightarrow\left(\frac{x+4}{9}-1\right)+\left(\frac{x+11}{8}-2\right)+\left(\frac{x+16}{7}-3\right)+\left(\frac{x+19}{6}-4\right)=0\)

\(\Leftrightarrow\frac{x+4-9}{9}+\frac{x+11-16}{8}+\frac{x+16-21}{7}+\frac{x+19-24}{6}=0\)

\(\Leftrightarrow\frac{x-5}{9}+\frac{x-5}{8}+\frac{x-5}{7}+\frac{x-5}{6}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+\frac{1}{6}\right)=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

V...

ta có tam giác BGD vuông tại G (BE ⊥ AD tại G )
=>BG^2+GD^2=BD^2
<=>BG^2+(AD/3)^2=AD^2(BD=AD=DC tính chất tam giác vuông )
<=>BG^2=8AD^2/9(1)
lại có tam giác ABG vuông tại G 
=>BG^2+AG^2=AB^2
<=>BG^2+(2AD/3)^2=6(2) 
từ (1) và (2) =>AD=3/căn 2
=>BC=2AD=6/căn2
tam giác ABC vuông tại A
=>AC^2=BC^2-AB^2
            =18-6
            =12
=>AC=2 căn 3

mình đang cần cm tam giác vuông bạn ạ

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

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

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

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

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

\(\Leftrightarrow x+2014=0\)

\(\Leftrightarrow x=-2014\)

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)

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