`a)|6x-3|=15`
`<=>|2x-1|=5`
`+)2x-1=5<=>x=3`
`+)2x-1=-5<=>x=-2`
Vậy x=-2 hoặc `x=3`
`b)(|x|-2)(6-2|x|)=0`
`<=>(|x|-2)(3-|x|)=0`
`<=>` \(\left[ \begin{array}{l}|x|=2\\|x|=3\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-3\\x=3\\x=-3\end{array} \right.\)
a)|6x−3|=15a)|6x-3|=15
⇔|2x−1|=5⇔|2x-1|=5
+)2x−1=5⇔x=3+)2x-1=5⇔x=3
+)2x−1=−5⇔x=−2+)2x-1=-5⇔x=-2
Vậy x=-2 hoặc x=3x=3
b)(|x|−2)(6−2|x|)=0b)(|x|-2)(6-2|x|)=0
⇔(|x|−2)(3−|x|)=0⇔(|x|-2)(3-|x|)=0
⇔⇔ [|x|=2|x|=3[|x|=2|x|=3
⇔⇔ ⎡⎢ ⎢ ⎢⎣x=2x=−3x=3x=−3[x=2x=−3x=3x=−3
Phần a bn tự làm nha
b)\(\left(\left|x\right|-2\right).\left(6-2.\left|x\right|\right)=0\)
\(\Rightarrow\left(\left|x\right|-2\right)=0\) hoặc \(\left(6-2.\left|x\right|\right)=0\)
1)\(\left|x_1\right|-2=0\) 2)\(6-2.\left|x_2\right|=0\)
\(\left|x_1\right|=2\) \(2.\left|x_2\right|=6\)
\(\Rightarrow x_1\in\left\{\pm2\right\}\) |x2|=3
\(\Rightarrow x_2\in\left\{\pm3\right\}\)
Vậy \(x_1\in\left\{\pm2\right\};x_2\in\left\{\pm3\right\}\)
Chúc bn học tốt
Phần a bn tự làm nha
b)(|x|−2).(6−2.|x|)=0(|x|−2).(6−2.|x|)=0
⇒(|x|−2)=0⇒(|x|−2)=0 hoặc (6−2.|x|)=0(6−2.|x|)=0
1)|x1|−2=0|x1|−2=0 2)6−2.|x2|=06−2.|x2|=0
|x1|=2|x1|=2 2.|x2|=62.|x2|=6
⇒x1∈{±2}⇒x1∈{±2} |x2|=3
⇒x2∈{±3}⇒x2∈{±3}
Vậy x1∈{±2};x2∈{±3}
