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}

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\(a)3-\left(17-x\right)=-24+12\)

\(\Leftrightarrow3-17+x=-12\)
\(\Leftrightarrow-14+x=-12\)

\(\Leftrightarrow x=-12+14=2\)

a)  3 - (17 - x) = -24 + 12

=> 3 - 17 + x = -12

=> -14 + x = -12

=> x = -12 + 14

=> x = 2

b) -5 + (16 - x) = 2².3 - 5²

=> -5 + 16 - x = 12 - 25

=> 11 - x = -13

=> x = 11 + 13

=> x = 24

c) -26 - (x - 7) = 3⁴ - 9²

=> -26 - x + 7 = 81 - 81

=> -19 - x = 0

=> x = -19 - 0

=> x = -19

d) -36 - (4 - x) = 37 + (-5)

=> -36 -  4 + x = 32

=> -40 + x = 32

=> x = 32 + 40

=> x = 72

\(b)-5+\left(16-x\right)=2^2.3-5^2\)

\(\Leftrightarrow-5+16-x=4.3-25\)

\(\Leftrightarrow11-x=-13\Leftrightarrow x=24\)

a)9.x + 1=73

 9x=73-1

9x=72

  x=72:9

  x=8

b)2.x - 5 = -17 - 12

2x-5=-29

2x=-29+5

2x=-24

  x=-24:2

  x=-12

c)10 - x - 5 = - 5 - 7 -11

10-x-5=-12-11

10-x-5=-23

 10-x=-23+5 

  10-x=18

       x=10-18

       x=-8

d)(-9) . x + 3 = (-2) . (-7) +16

-9x+3=14+16

-9x+3=30

-9x=30-3

-9x=27

   x=27:(-9)

   x=-3

(-12) . x - 34 =2

-12x=2+34

-12x=36

     x=36:(-12)

     x=-3

(-11).x + 9 =130

-11x=130-9

-11x=121

     x=121:(-11)

     x=11 

(-5) .x + 5 = (-15) .(-4) -12

-5x+5=60-12

-5x+5=48

-5x=48-5

-5x=43

   x=43:(-5)

  x=-8,6

IxI -3=0

|x|=3

=>x=+3

(7 - IxI).(2.x - 4) =0

*7-|x|=0            * 2x-4=0

|x|=7                   2x=4

=>x=+7                 x=4:2

                             x=2

280-(x-140):35 =270

(x-140):35=280-270

(x-140):35=10

x-140=10.35

x-140=350

       x=350+140

       x=490

(1900 - 2.x ) : 35- 32 =16

1900-2x=(16+32).35

1900-2x=1680

         2x=1900-1680

         2x=220

           x=220:2

           x=110

720 :[41-(2x -5 )] =2³ .5

720:[41-(2x-5)]=40

41-(2x-5)=720:40

41-(2x-5)=18

2x-5=41-18

2x-5=23

2x=23+5

2x=28

  x=28:2

  x=14

(x - 5).(x² - 4 ) =0

* x-5=0           * x²-4=0

      x=0+5         x²=4

      x=5             x²=2²

                    => x=+2

3(x+1) + 2(x+5)

= 3x + 3 + 2x + 10

= 5x + 13

Ta có: \(3\left(x+1\right)+2\left(x+5\right)\)

\(=3x+3+2x+10\)

\(=5x+13\)