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\(\Rightarrow x=\dfrac{5}{12}-\dfrac{1}{4}=\dfrac{1}{6}\)
\(0,25+x=\dfrac{5}{12}\Rightarrow\dfrac{1}{4}+x=\dfrac{5}{12}\)
\(\Rightarrow x=\dfrac{5}{12}-\dfrac{1}{4}=\dfrac{5-1.3}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
\(a,\left(2x-1\right)^2-25=0.\\ \Leftrightarrow\left(2x-1\right)^2=25.\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=5.\\2x-1=-5.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=-2.\end{matrix}\right.\)
\(b,\left(x-3\right)\left(5-2x\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0.\\5-2x=0.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=\dfrac{5}{2}.\end{matrix}\right.\)
\(c,\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}-\dfrac{2\left(x^2+2\right)}{x^2-4}=0.\)
\(\left(x\ne\pm2\right).\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x^2-4}{\left(x-2\right)\left(x+2\right)}=0.\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x^2-4=0.\)
\(\Leftrightarrow x^2+3x+2+x^2-3x+2-2x^2-4=0.\)
\(\Leftrightarrow0=0\) (luôn đúng).
26 different
27 pronunciation
28 comfortable
29 performers
30 windy
31 unhealthy
32 childhood
33
34 flight
35 suddenly
IV.
26. different
27. pronunciation
28. comfortable
29. performers
30. windy
31. unhealthy
32. childhood
33. successfully
34. flights
35. suddenly
Lời giải:
a.
$3\leq x< 10$
$\Rightarrow x\in\left\{3; 4; 5;6 ;7; 8;9\right\}$
b.
$24< x< 27$
$\Rightarrow x\in\left\{25; 26\right\}$
c.
$x\vdots 2; x<10$
$\Rightarrow x\in\left\{0; 2; 4; 6; 8\right\}$
d.
$x\in\left\{1;2;4;5;7;8;;10;11;13;14\right\}$