|-5| . |x| = | -20 |
= 5 . x = 20
=) x = 20 : 5 = 4
|x| < -5
=) x \(\in\left\{-6;-7;-8;-9;-10;....\right\}\)
đăng kí kênh của v-i-s nha !
12\(\ge\)x < 15
=) x \(\in\left\{12;13;14\right\}\)
a)\(\left|-5\right|.\left|x\right|=\left|-20\right|\)
\(5.\left|x\right|=20\)
\(\left|x\right|=4\)
\(x=\pm4\)
b) \(\left|x\right|< 5\)
\(\Rightarrow x\in\left\{0;\pm1;\pm2;\pm3;\pm4\right\}\)
c) \(12\le\left|x\right|< 15\)
\(\Rightarrow x\in\left\{\pm12;\pm13;\pm14;\right\}\)
d) \(\left|x-1\right|+\left(-3\right)=17\)
\(\left|x-1\right|=20\)
\(\Rightarrow\orbr{\begin{cases}x-1=20\\x-1=-20\end{cases}\Rightarrow\orbr{\begin{cases}x=21\\x=-19\end{cases}}}\)
e) \(\left|x-1\right|-\left(-4\right)=5\)
\(\left|x-1\right|=1\)
\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)
g)\(\left|x\right|-\left(-2\right)=-1\)
\(\left|x\right|=-3\left(vl\right)\)
12 ≥ x < 15
=> x ∈ {12;13;14}
Vậy........................
|-5| . |x| = | -20 |
= 5 . x = 20
=> x = 20 : 5 = 4
|x| < -5
=> x ∈ {−6; − 7; − 8; − 9; − 10;....}
|x-1 | + (-3) = 17
|x-1|=14
\(\Rightarrow\orbr{\begin{cases}x-1=14\Rightarrow x=14+1=15\\x-1=-14\Rightarrow x=-14+1=-13\end{cases}}\)
Vậy............................................................
|x+1| -(-4) = 5
|x+1|=5-4
|x+1|=1
\(\Rightarrow\orbr{\begin{cases}x+1=1\Rightarrow x=1-1=0\\x+1=-1\Rightarrow x=-1-1=-2\end{cases}}\)
Vậy.............................................
|x|-(-2) =(-1)
|x|+2=-1
|x|=-3
Vô lí vì |x| > 0
Vậy \(x\in\varnothing\)
a) \(\Rightarrow-5.x=20\)
\(\Rightarrow x=20:\left(-5\right)\Rightarrow x=-4\)
b)\(|x|< -5\Rightarrow x< 5\Rightarrow x=4;3;2;1;0;-1;-2;-3;....\)
c) chưa nghĩ ra
d) \(\Rightarrow|x-1|=20\)\(\Rightarrow\orbr{\begin{cases}x-1=20\\x-1=-20\end{cases}}\Rightarrow\orbr{\begin{cases}x=21\\x=-19\end{cases}}\)
e) \(\Rightarrow|x+1|+4=5\Rightarrow|x+1|=1\)\(\Rightarrow\orbr{\begin{cases}x+1=1\\x+1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
f) \(\Rightarrow\left|x\right|+2=-1\Rightarrow\left|x\right|=-3\)
Vì \(\left|x\right|\ge0\Rightarrow\)không có gt nào của x thỏa mãn để |x| = -3
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