I/ 1C 2C 3B 4D 5d

II/ 1c 2a 3d 4b

I:

1.D    2.D  3.B   4.D   5.D

II:

1.C   2.C   3.D   4.A

C

c

B-A-B-D-A-D-D-C-B-C

\(\Rightarrow x=\dfrac{5}{12}-\dfrac{1}{4}=\dfrac{1}{6}\)

\(0,25+x=\dfrac{5}{12}\Rightarrow\dfrac{1}{4}+x=\dfrac{5}{12}\)

\(\Rightarrow x=\dfrac{5}{12}-\dfrac{1}{4}=\dfrac{5-1.3}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)

x=1/6

\(a,\left(2x-1\right)^2-25=0.\\ \Leftrightarrow\left(2x-1\right)^2=25.\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=5.\\2x-1=-5.\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=\dfrac{5}{2}.\end{matrix}\right.\)

\(c,\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}-\dfrac{2\left(x^2+2\right)}{x^2-4}=0.\)

\(\left(x\ne\pm2\right).\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x^2-4}{\left(x-2\right)\left(x+2\right)}=0.\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x^2-4=0.\)

\(\Leftrightarrow x^2+3x+2+x^2-3x+2-2x^2-4=0.\)

\(\Leftrightarrow0=0\) (luôn đúng).

\(\dfrac{6}{5}\)

6/5

6/5

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