A,-16+23+x=-16

<=>x=-16+16-23

<=>x=-23

B,10-2(4-3x)=-4

<=>10-8+6x=-4

<=>6x=-4-10+8

<=>6x=-6

<=>x=-1

C,-12+3(-x+7)=-18

<=>-12-3x+21=-18

<=>-3x=-18+12-21

<=>-3x=-27

<=>x=9

D,24:(3x-2)=-3

<=>24/(3x-2)=-3

<=>-3(3x-2)=24

<=>3x-2=-8

<=>3x=-6

<=>x=-2

E,|x+8|-7=8

<=>|x+8|=15

<=>x+8=+-15

Chia 2 TH:

TH1:x+8=15

<=>x=7

TH2:x+8=-15

<=>x=-23

F,-45:5(-3-2x)=3

<=>-9(-3-2x)=3

<=>27+18x=3

<=>18x=-24

<=>x=-4/3

G,|x-1|=0

<=>x-1=0

<=>x=1

H,-13|x|=-26

<=>|x|=2

<=>x=+-2

a) -16 + 23 + x = -16

x = -16 + 16 - 23

x = -23

Vậy x = -23

b) 10 - 2. (4 - 3x) = -4

10 - 2 . 4 + 2. 3x = -4

10 - 8 + 6x = -4

6x = -4 - 10 + 8

6x = -6

x = (-6) : 6

x = -1

Vậy x = -1

c) -12 + 3. (-x + 7) = -18

-12 + 3. (-x) + 3. 7 = -18

-12 + (-3x) + 21 = -18

-3x = -18 + 12 - 21

-3x = -27

x = (-27) : (-3)

x = -9

Vậy x = -9

d) 24 : (3x - 2) = -3

3x - 2 = 24 : (-3)

3x - 2 = -8

3x = -8 + 2

3x = -6

x = (-6) : 3

x = -2

Vậy x = -2

e) lx + 8l - 7 = 8

lx + 8l = 8 + 7

lx + 8l = 15

\(\Rightarrow\hept{\begin{cases}x+8=15\\x=18-5\\x=13\end{cases}hay\hept{\begin{cases}x+8=-15\\x=-15-8\\x=-23\end{cases}}}\)

Vậy \(x\in\left\{13;-23\right\}\)

f) (-45) : 5. (-3 - 2x) = 3

(-9). (-3 - 2x) = 3

(-9). (-3) + 9. 2x = 3

27 + 18x = 3

18x = 3 - 27

18x = -24

x = (-24) : 18

x =\(\frac{-4}{3}\)

Vậy x =\(\frac{-4}{3}\)

g) lx - 1l = 0

\(\Rightarrow x-1=0\)

x = 0 + 1

x = 1

Vậy x = 1

h) (-13). lxl = -26

lxl = (-26) : (-13)

lxl = 2

\(\Rightarrow x=2\)

Vậy x = 2

Chúc bạn học tốt!!!

a, 440 + 2 . (125 - x) = 546

2 . (125 - x) = 546 - 440

2 . (125 - x) = 106

125 - x = 106 : 2

125 - x = 53

x = 125 - 53

x = 72

Các câu kia tương tự

Chúc bạn học tốt

còn câu e làm sao?

g: =>3x-10=200

hay x=70

c) (7x + x – 28) :100 – 38 = 9

<=> 8x - 28 = 4700

<=> x = 591

d) 936 – (4x + 5x ) = 900

<=> 9x = 36

<=> x = 4

dài thế

dai the giai bao gio xong chac mua quyt nam sau moi giai xong

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}

bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\) 

     \(x\times\dfrac{11}{16}=2\) 

     \(x=2:\dfrac{11}{16}\) 

    \(x=\dfrac{32}{11}\)

Bài 1 : 

 \(\dfrac{x}{16}\times\left(2017-1\right)=2\)

          \(\dfrac{x}{16}\times2016=2\)

                      \(\dfrac{x}{16}=\dfrac{2}{2016}\)

                         \(x=\dfrac{2}{2016}\times16\)

                         \(x=\dfrac{1}{63}\)

1- (5\(\dfrac{4}{9}\) +x+7\(\dfrac{7}{18}\)) : 15\(\dfrac{3}{4}\) = 0 

1- (\(\dfrac{49}{9}+x+\dfrac{133}{18}\)) : \(\dfrac{63}{4}=0\) 

 (\(\dfrac{49}{9}+\dfrac{133}{18}\)+\(x\) ) : \(\dfrac{63}{4}\) = 1 - 0 

       (\(\dfrac{77}{6}\) + \(x\) ) : \(\dfrac{63}{4}\) = 1

         \(\dfrac{77}{6}+x\)         = 1 x \(\dfrac{63}{4}\) 

          \(\dfrac{77}{6}\) + \(x\)          = \(\dfrac{63}{4}\)

                  \(x\)            = \(\dfrac{63}{4}\) - \(\dfrac{77}{6}\)

                  \(x=\) \(\dfrac{35}{12}\)