2 câu trả lời
Đáp án:
`a. sqrt{1-3x}=4`
`<=> 1-3x=16`
`<=> -3x=15`
`<=> x=-5`
Vậy `S={-5}`
`b. sqrt{x^2+4x+4}=2`
`<=> sqrt{(x+2)^2}=2`
`<=> |x+2|=2`
`<=>`\(\left[ \begin{array}{l}x+2=2\\x+2=-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=-4\end{array} \right.\)
Vậy `S={-4;0}`
a)$\sqrt{1-3x}$ $=$ $4$
$1-3x$ = $16$
$3x$ = $-15$
$x$ $=$ $-5$
b)$\sqrt{x^{2}+4x+4}$ $=$ $2$
$\sqrt{(x+2)^{2}}$ $=$ $2$
$|x+2|$ $=$ $2$
\(\left[ \begin{array}{l}x+2=2\\x+2=-2\end{array} \right.\)
\(\left[ \begin{array}{l}x=0\\x=-4\end{array} \right.\)
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