2 bài tìm x

P

phumanhpro

a. [TEX](x+3)^2 - (x-2)(x+2)= -5[/TEX]
<=> $x^2+6x+9-x^2+4+5=0$
<=> $6x+18=0$
<=> $x=-3$

b. [TEX](2x-1)^2 - 4 = 0[/TEX]
<=> $(2x-1)^2-2^2$
<=> $(2x-1-2)(2x-1+2)=0$
<=> $ 2x-3=0 <=>x=3/2$
$ 2x+1=0 <=> x=-1/2$

 
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N

nghgh97

Câu b

[tex](2x - 1){}^2 - 4 = 0[/tex]
[tex] \Leftrightarrow (2x - 1){}^2 = 4[/tex]
[tex] \Leftrightarrow \sqrt {{{(2x - 1)}^2}} = 2[/tex]
[tex] \Leftrightarrow \left| {2x - 1} \right| = 2[/tex]
+Nếu [tex]x \ge \frac{1}{2}[/tex] thì [tex] 2x - 1 = 2 \Leftrightarrow x = \frac{3}{2}[/tex]
+Nếu [tex]x < \frac{1}{2}[/tex] thì [tex]1 - 2x = 2 \Leftrightarrow x = - \frac{1}{2}[/tex]
 
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