giai dum tui bai tim x:

0

0915549009

[TEX]a) 3-36xmu2=0[/TEX]
[TEX]b)xmu2+x+1=0[/TEX]
[TEX]c)x(x-4)4-2x+8=0[/TEX]
[TEX]d)x(x-5)-3x+15=0[/TEX]
[TEX]a) 3-36x^2=0 \Leftrightarrow 36x^2=3\Leftrightarrow x=\pm \ \frac{\sqrt{3}}{6}[/TEX]
[TEX]b)x^2+x+1=0[/TEX]
Vì [TEX]x^2+x+1>0[/TEX] nên PT vô nghiệm
[TEX]c)x(x-4)-2x+8=0\Leftrightarrow x(x-4)-2(x-4)=0\Leftrightarrow (x-2)(x-4)=0\Leftrightarrow \left[\begin{x=2}\\{x=4} [/TEX]
[TEX]d)x(x-5)-3x+15=0\Leftrightarrow x(x-5)-3(x-5)=0\Leftrightarrow (x-3)(x-5)=0\Leftrightarrow \left[\begin{x=3}\\{x=5} [/TEX]
 
B

bang8h

a) 3-36x^2=0
<=>x^2=3/36
=>x=căn bậc 2 của 1/12
b)x^2+x+1=0
=>(x^2+2.x.1/2+1/4)+3/4=0
=>(x+1/2)^2+3/4=0
=>pt trên vô nghiệm vì (x+1/2)^2 luôn lớn hơn hôặc bằng 0
c)x(x-4)4-2x+8=0
4x^2-16x-2x+8=0
4x^2-2x-16x+8=0
2x(x-1)-8(x-1)=0
(x-1)(2x-8)=0
th1) x-1=0
=>x=1
th2) 2x-8=0
=>x=4
d) x^2-5x-3x+15=0
=>x^2-3x-5x+15=0
=>x(x-3)-5(x-3)=0
=>(x-3))(x-5)=0
th1)x-3=0
=>x=3
th2)x-5=0
=>x=5
 
T

thaonguyenpham

[TEX]a) 3-36x^2=0 \Leftrightarrow 36x^2=3\Leftrightarrow x=\pm \ \frac{\sqrt{3}}{6}[/TEX]
[TEX]b)x^2+x+1=0[/TEX]
Vì [TEX]x^2+x+1>0[/TEX] nên PT vô nghiệm
[TEX]c)x(x-4)-2x+8=0\Leftrightarrow x(x-4)-2(x-4)=0\Leftrightarrow (x-2)(x-4)=0\Leftrightarrow \left[\begin{x=2}\\{x=4} [/TEX]
[TEX]d)x(x-5)-3x+15=0\Leftrightarrow x(x-5)-3(x-5)=0\Leftrightarrow (x-3)(x-5)=0\Leftrightarrow \left[\begin{x=3}\\{x=5} [/TEX]
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