Question

[↭ﾟ°☆Şαƙυɾα☆°︵✰ﾟঔ]

Bài 46:

11, - 4|x-2| = -8

12, 5|x+2| = -10(-2)

13, 6|x-2| = 18: (- 3)

14, -7|x+4| = 21 : (-3)

15, 4|x+1| = 8(-2) - 8(-5)

16, 3|x+5| = -9

17, -8|x-3| = 24 - 16: 2

18, -3|x+6| = 6.2 -9

19, 5-|x+7| = 4

20, 12-|x+8| = 10

Giúp mk nha. Mk càn ngay đó

Answer 1

[💋Amanda💋]

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 46:

11: Ta có: \(-4\left|x-2\right|=-8\)

\(\Leftrightarrow\left|x-2\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: x∈{0;4}

12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)

\(\Leftrightarrow5\left|x+2\right|=20\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy: x∈{-6;2}

13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)

\(\Leftrightarrow6\left|x-2\right|=-6\)(1)

Ta có: \(\left|x-2\right|\ge0\forall x\)

\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)

Ta có: -6<0(3)

Từ (1), (2) và (3) suy ra x∈∅

Vậy: x∈∅

14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)

\(\Leftrightarrow-7\left|x+4\right|=-7\)

\(\Leftrightarrow\left|x+4\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: x∈{-5;-3}

15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)

\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)

\(\Leftrightarrow4\left|x+1\right|=24\)

\(\Leftrightarrow\left|x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Vậy: x∈{-7;5}

16: Ta có: \(3\left|x+5\right|=-9\)(4)

Ta có: |x+5|≥0∀x

⇒3|x+5|≥0∀x(5)

Ta có: -9<0(6)

Từ (4), (5) và (6) suy ra x∈∅

Vậy: x∈∅

17: Ta có: \(-8\left|x-3\right|=24-16:2\)

\(\Leftrightarrow-8\left|x-3\right|=16\)

\(\Leftrightarrow\left|x-3\right|=-2\)

mà |x-3|≥0>-2∀x

nên x∈∅

Vậy: x∈∅

18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)

\(\Leftrightarrow-3\left|x+6\right|=3\)

\(\Leftrightarrow\left|x+6\right|=-1\)

mà |x+6|≥0>-1∀x

nên x∈∅

Vậy: x∈∅

19: Ta có: \(5-\left|x+7\right|=4\)

\(\Leftrightarrow\left|x+7\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)

Vậy: x∈{-8;-6}

20: Ta có: \(12-\left|x+8\right|=10\)

\(\Leftrightarrow\left|x+8\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)

Vậy: x∈{-10;-6}