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I

iceghost

Sr mình nhầm -_-
2)
$b)-3x^2+2x=0 \\
\iff x(-3x+2)=0 \\
\iff x=0 \; hay \; -3x+2=0 \iff x=\dfrac23 \\
c)-3x+5x^2=0 \\
\iff x(-3+5x)=0 \\
\iff x=0 \; hay \; -3+5x=0 \iff x=\dfrac35 \\
\\~\\
d) x^2+5x+4=0 \\
\iff x^2+x+4x+4=0 \\
\iff x(x+1)+4(x+1)=0 \\
\iff (x+4)(x+1)=0 \\
\iff x+4=0 \iff x=-4 \; hay \; x+1=0 \iff x=-1$
 
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Q

quynhphamdq

2. d .
[TEX]x^2+5x+4=0 [/TEX]
[TEX]\Leftrightarrow x^2 +4x +x +4 = 0[/TEX]
[TEX]\Leftrightarrow x(x+4)+(x+4)=0[/TEX]
[TEX]\Leftrightarrow (x+1)(x+4)=0[/TEX]
[TEX]\Rightarrow x+1=0 \Rightarrow x=-1[/TEX]
Hoặc[TEX] x+4= 0 \Rightarrow x=-4.[/TEX]
Vậy [TEX]x=-1 [/TEX]và[TEX] x=-4 .[/TEX]
2c.
[TEX] -3x+5x^2=0[/TEX]
\Leftrightarrow[TEX]x(5x-3)=0[/TEX]
[TEX]\Rightarrow x=0 [/TEX]Hoặc [TEX]5x-3 =0 \Rightarrow x=\frac{3}{5}[/TEX]
Vậy $x=0$ và [TEX]x=\frac{3}{5}[/TEX]
 
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P

pinkylun

2a) x/3=y/4 va xy=12

$\dfrac{x}{3}=\dfrac{y}{4}=> \dfrac{x.y}{3.4}=\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{12}{12}=1$

$=>x=\pm 3$ và $y= \pm 4$

b) $-3x^2+2x=0$

$=>x(-3x+2)=0$

$=>x=0$ hoặc $x=\dfrac{2}{3}$
 
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