c. \(\left(x+2\right)^4-6\left(x+2\right)^2+5=0\)

\(\Leftrightarrow\left(x+2\right)^4-\left(x+2\right)^2-5\left(x+2\right)^2+5=0\)

\(\Leftrightarrow\left(x+2\right)^2\left[\left(x+2\right)^2-1\right]-5\left[\left(x+2\right)^2-1\right]=0\)

\(\Leftrightarrow\left[\left(x+2\right)^2-1\right]\left[\left(x+2\right)^2-5\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+1\right)\left(x+2+\sqrt{5}\right)\left(x+2-\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+1=0\\x+2+\sqrt{5}=0\\x+2-\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\\x=-\sqrt{5}-2\\x=\sqrt{5}-2\end{matrix}\right.\)

Vậy: Phương trình có tập nghiệm \(S=\left\{-3;-1;-\sqrt{5}-2;\sqrt{5}-2\right\}\)

lm cho em câu 4 nữa đc ko

Bài 1: - \(\dfrac{5}{7}\) x \(\dfrac{31}{33}\) + \(\dfrac{-5}{7}\) x \(\dfrac{2}{33}\) + 2\(\dfrac{5}{7}\)

         = - \(\dfrac{5}{7}\) \(\times\) ( \(\dfrac{31}{33}\) + \(\dfrac{2}{33}\)) + 2 + \(\dfrac{5}{7}\)

         =  - \(\dfrac{5}{7}\)  + 2 + \(\dfrac{5}{7}\)

          = 2

2, \(\dfrac{3}{14}\): \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\): \(\dfrac{1}{28}\) - 8

   = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) : \(\dfrac{1}{28}\) - 8

   =  \(\dfrac{2}{7}\) x 28 - 8

   = 8 - 8

    = 0 

3, \(\dfrac{37}{43}\) . \(\dfrac{17}{29}\) - \(\dfrac{21}{41}\).\(\dfrac{1}{2}\) + \(\dfrac{9}{58}\): 1\(\dfrac{6}{37}\) - \(\dfrac{6}{29}\): 1\(\dfrac{20}{21}\)

=    \(\dfrac{37}{43}\). \(\dfrac{17}{29}\) - \(\dfrac{21}{41}\) . \(\dfrac{1}{2}\) + \(\dfrac{9}{58}\).\(\dfrac{37}{43}\)  - \(\dfrac{6}{29}\).\(\dfrac{21}{41}\)

=    (\(\dfrac{37}{43}\).\(\dfrac{17}{29}\) + \(\dfrac{9}{58}\).\(\dfrac{37}{43}\)) - (\(\dfrac{21}{41}\).\(\dfrac{1}{2}\) + \(\dfrac{6}{29}\).\(\dfrac{21}{41}\))

=     \(\dfrac{37}{43}\).( \(\dfrac{17}{29}\) + \(\dfrac{9}{58}\)) - \(\dfrac{21}{41}\).( \(\dfrac{1}{2}\) + \(\dfrac{6}{29}\))

=       \(\dfrac{37}{43}\).\(\dfrac{43}{58}\) - \(\dfrac{21}{41}\).\(\dfrac{41}{58}\)

=         \(\dfrac{37}{58}\) - \(\dfrac{21}{58}\)

=          \(\dfrac{16}{58}\)

= \(\dfrac{8}{29}\)

Câu 9:

d) Xét ΔABC có 

AH là đường trung tuyến ứng với cạnh BC

BD là đường trung tuyến ứng với cạnh AC

AH cắt BD tại G

Do đó: G là trọng tâm của ΔABC(đpcm)

Đề bài yêu cầu cụ thể điều gì vậy bạn?