Không có j đâu

\(25-y^2=8\left(x-2009\right)^2\)

Ta có \(8\left(x-2009\right)^2=25-y^2\)

        \(8\left(x-2009\right)^2+y^2=25\) (*)

Vì \(y^2\ge0\) nên \(\left(x-2009\right)^2\le\dfrac{25}{8}\) , suy ra \(\left(x-2009\right)^2=0\) hoặc \(\left(x-2009\right)^2=1\)

Với \(\left(x-2009\right)^2=1\) thay vào (*) ta có \(y^2=17\) (loại)

Với \(\left(x-2009\right)^2=0\) thay vào (*) ta có \(y^2=25\) suy ra \(y=5\) (do \(y\in N\) )

Từ đó tìm đuợc (x=2009 ; y = 5)

9 nguời

Thay x = 206 , y = 1 vào biểu thức C

\(C=\left(1-\dfrac{206}{2}\right)^3-6\left(y-\dfrac{206}{2}\right)^2+12\left(y-\dfrac{206}{2}\right)-8\)

\(C=\left(\dfrac{-204}{2}\right)^3-6.\left(\dfrac{-204}{2}\right)^2+12.\left(\dfrac{-204}{2}\right)-8\)

\(C=-102^3-6.\left(-102\right)^2+12\left(-102\right)-8\)

\(C=-1061208-6.\left(-10404\right)+12.\left(-102\right)-8\)

\(C=-1061208-\left(-62424\right)+\left(-1224\right)\)

\(C=-998784+\left(-1224\right)\)

\(C=-1000008\)

a, góc IAJ kè với góc JAK

b, góc EBF kề vơi góc FBG ; góc FBG kề với góc GBH ; góc EBG kề với góc GBH

a, \(\left|x+\dfrac{1}{5}\right|-4=-2\)

\(\left|x+\dfrac{1}{5}\right|=\left(-2\right)+4\)

\(\left|x+\dfrac{1}{5}\right|=2\)\(\left\{{}\begin{matrix}x+\dfrac{1}{5}=2\\x+\dfrac{1}{5}=-2\end{matrix}\right.\)

với \(x+\dfrac{1}{5}=2\Rightarrow x=2-\dfrac{1}{5}\)\(\) hay \(x=\dfrac{9}{5}\)

với \(x+\dfrac{1}{5}=-2\Rightarrow x=-2-\dfrac{1}{5}\) hay \(x=-\dfrac{11}{5}\)

b, \(-\dfrac{15}{12}x+\dfrac{3}{7}=\dfrac{6}{5}x-\dfrac{1}{2}\)

\(\dfrac{6}{5}x+\dfrac{5}{4}x=\dfrac{3}{7}+\dfrac{1}{2}\)

\(\left(\dfrac{6}{5}+\dfrac{5}{4}\right)x=\dfrac{6}{14}+\dfrac{7}{14}=\dfrac{13}{14}\)

\(\dfrac{49}{20}x=\dfrac{13}{14}\)

\(x=\dfrac{13}{14}:\dfrac{49}{20}\)

\(x=\dfrac{13}{14}.\dfrac{20}{49}=\dfrac{260}{686}\)

\(x=\dfrac{130}{343}\)

\(\left[18\dfrac{1}{6}-\left(0,06:7\dfrac{1}{2}+3\dfrac{2}{5}.0,38\right)\right]:\left(19-2\dfrac{2}{3}.4\dfrac{2}{4}\right)\)

\(=\left[\dfrac{109}{6}-\left(\dfrac{6}{100}:\dfrac{15}{2}+\dfrac{17}{5}.\dfrac{38}{100}\right)\right]:\left(19-\dfrac{8}{3}.\dfrac{19}{4}\right)\)

\(=\left[\dfrac{109}{6}-\left(\dfrac{3}{50}.\dfrac{2}{15}+\dfrac{17}{5}.\dfrac{19}{50}\right)\right]:\left(19-\dfrac{38}{3}\right)\)

\(=\left[\dfrac{109}{6}-\left(\dfrac{2}{250}+\dfrac{323}{250}\right)\right]:\dfrac{19}{3}\)

\(=\left(\dfrac{109}{6}-\dfrac{13}{10}\right).\dfrac{3}{19}\)

\(=\dfrac{506}{30}.\dfrac{3}{19}\)

\(=\dfrac{253}{95}\)

a,a > b

b, a >b

d