\(\widehat{E}=53^0\)

Áp dụng tslg trong tam giác DEF vuông tại D:

\(tanE=\dfrac{DF}{ED}=\dfrac{4}{3}\Rightarrow\widehat{E}\approx53^0\)

Vì P(x) có nghiệm bằng 2 nên:

P(2) = 0

=> m.2 + 3 = 0

     2m        = -3

       m        = \(\frac{-3}{2}\)

Ta có:

\(A=\dfrac{\sqrt{x}-x}{\sqrt{x}-1}=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-\sqrt{x}\)

Vậy \(A=-\sqrt{x}\)

\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-\sqrt{x}\)

\(=\sqrt{x}\)

B

CHọn B

B

\(\Leftrightarrow-2^{3x+1}=-128\\ \Leftrightarrow-2^{3x+1}=-2^7\\ \Leftrightarrow3x+1=7\\ \Leftrightarrow x=2\)

\(\Leftrightarrow41-2\left(x-2\right)=-1\\ \Leftrightarrow2\left(x-2\right)=42\\ \Leftrightarrow x-2=21\\ \Leftrightarrow x=23\)

\(\dfrac{2}{7}\cdot\dfrac{14}{15}\cdot\dfrac{5}{6}:\dfrac{3}{8}:\dfrac{2020}{2021}\)

\(=\dfrac{2}{7}\cdot\dfrac{14}{15}\cdot\dfrac{5}{6}\cdot\dfrac{8}{3}\cdot\dfrac{2021}{2020}\)

\(=\dfrac{14}{7}\cdot\dfrac{5}{15}\cdot\dfrac{8}{6}\cdot\dfrac{2}{3}\cdot\dfrac{2021}{2020}\)

\(=2\cdot\dfrac{1}{3}\cdot\dfrac{4}{3}\cdot\dfrac{2}{3}\cdot\dfrac{2021}{2020}\)

\(=\dfrac{8084}{13635}\)

Gọi các điểm như hình vẽ:

Ta có: góc ACD kề bù với góc DCB => góc DCB=180-150=30

Ta có: góc ACE đối đỉnh góc DCB => góc ACE=góc DCB=30

Lại có góc ACD đối đỉnh với góc BCE => góc ACD=góc BCE=150

28.

\(\Leftrightarrow2x+1=x-1\)

\(\Leftrightarrow x=-2\)

29.

\(\Leftrightarrow x^3-2x^2-3x^2+6x=0\)

\(\Leftrightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2-3x=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=3\end{matrix}\right.\)

30.\(\Leftrightarrow3x^2+3x+2x+2=0\)

\(\Leftrightarrow3x\left(x+1\right)+2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(28,\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=-x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\\ 29,\\ \Leftrightarrow x\left(x^2-5x+6\right)=0\\ \Leftrightarrow x\left(x^2-2x-3x+6\right)=0\\ \Leftrightarrow x\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\\ \Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\) 

\(30,\\ \Leftrightarrow3x^2+3x+2x+2=0\\ \Leftrightarrow3x\left(x+1\right)+2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=-1\end{matrix}\right.\)