d, \(\left(3x-2^4\right).7^3=2.7^4\)

\(\Rightarrow3x-2^4=2.7^4:7^3\)

\(\Rightarrow3x-16=2.7\\ \Rightarrow3x=14+16\\ \Rightarrow3x=30\Rightarrow x=10\)

Vậy.....

e, \(x-\left[42+\left(-28\right)\right]=-8\)

\(\Rightarrow x-14=-8\\ \Rightarrow x=6\)

Vậy.....

g, \(x-7=-5\)

\(\Rightarrow x=-5+7\Rightarrow x=2\)

Vậy.....

h, \(15-5\left(x+4\right)=-12-3\)

\(\Rightarrow15-5x-20=-15\)

\(\Rightarrow-5x=-15-15+20\)

\(\Rightarrow-5x=-10\Rightarrow x=2\)

Vậy.....

Chúc bạn học tốt!!!

d/ \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

\(\Rightarrow3x-16=\dfrac{2\cdot7^4}{7^3}=14\)

\(\Rightarrow3x=14+16=30\)

\(\Rightarrow x=\dfrac{30}{3}=10\)

e/ Đễ ==> tự lm thì tốt hơn nhé

g/ Đễ ==> tự lm thì tốt hơn nhé

h/ \(15-5\left(x+4\right)=-12-3\)

\(\Rightarrow15-5x-20=-15\)

\(\Rightarrow-5x=-15+20-15=-10\)

\(\Rightarrow x=\dfrac{-10}{-5}=2\)

i/ \(\left(7-x\right)-\left(25+7\right)=-25\)

\(\Rightarrow7-x-25-7=-25\)

\(\Rightarrow-x=-25-7+7+25\)

\(\Rightarrow-x=0\Rightarrow x=0\)

k/ \(\left|x+2\right|=0\Rightarrow x+2=0\Rightarrow x=-2\)

l/ \(\left|x-3\right|=7-\left(-2\right)\)

\(\Rightarrow\left|x-3\right|=9\)

\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)

m/ \(\left|x-5\right|=\left|-7\right|\Rightarrow\left|x-5\right|=7\)

\(\Rightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)

i, \(\left(7-x\right)-\left(25+7\right)=-25\)

\(\Rightarrow7-x-25-7=-25\)

\(\Rightarrow-x=-25-7+25+7\)

\(\Rightarrow x=0\)

Vậy.....

k, \(\left|x+2\right|=0\)

\(\Rightarrow x+2=0\Rightarrow x=-2\)

Vậy.....

l, \(\left|x-3\right|=-7-\left(-2\right)\)

\(\Rightarrow\left|x-3\right|=-5\)

Với mọi giá trị của \(x\in Z\) ta có:

\(\left|x-3\right|\ge0\) mà \(-5< 0\) nên không tìm được giá trị nào của x thoả mãn \(\left|x-3\right|=-5\).

Vậy \(x\in\varnothing\)

m, \(\left|x-5\right|=\left|-7\right|\)

\(\Rightarrow\left|x-5\right|=7\\ \Rightarrow\left\{{}\begin{matrix}x-5=-7\\x-5=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-2\\x=12\end{matrix}\right.\)

Vậy...,..

Chúc bạn học tốt!!!

\(c)\) \(\left|2x-1\right|-2x=3\)

\(\Leftrightarrow\)\(\left|2x-1\right|=2x+3\)

Ta có : \(\left|2x-1\right|\ge0\)

\(\Rightarrow\)\(2x+3\ge0\)\(\Rightarrow\)\(2x\ge-3\)\(\Rightarrow\)\(x\ge\frac{-3}{2}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=2x+3\\2x-1=-2x-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x-2x=3+1\\2x+2x=-3+1\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}0=4\\4x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}0=4\left(loai\right)\\x=\frac{-1}{2}\left(tm\right)\end{cases}}}\)

Vậy \(x=\frac{-1}{2}\)

Chúc bạn học tốt ~ 

\(b)\) \(3\left(2x-1\right)-\left|x-5\right|=7\)

\(\Leftrightarrow\)\(3\left(2x-1\right)-7=\left|x-5\right|\)

\(\Leftrightarrow\)\(6x-3-7=\left|x-5\right|\)

\(\Leftrightarrow\)\(\left|x-5\right|=6x-10\)

Ta có : \(\left|x-5\right|\ge0\)

\(\Rightarrow\)\(6x-10\ge0\)\(\Rightarrow\)\(6x\ge10\)\(\Rightarrow\)\(x\ge\frac{5}{3}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=6x-10\\x-5=10-6x\end{cases}\Leftrightarrow\orbr{\begin{cases}6x-x=-5+10\\x+6x=10+5\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}5x=5\\7x=15\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(loai\right)\\x=\frac{15}{7}\left(tm\right)\end{cases}}}\)

Vậy \(x=\frac{15}{7}\)

Chúc bạn học tốt ~ 

a: Trường hợp 1: x<-2

Pt sẽ là -x-2+3-2x=5

=>-3x+1=5

=>-3x=4

hay x=-4/3(loại)

Trường hợp 2: -2<=x<3/2

Pt sẽ là x+2+3-2x=5

=>5-x=5

hay x=0(nhận)

Trường hợp 2: x>=3/2

Pt sẽ là x+2+2x-3=5

=>3x-1=5

hay x=2(nhận)

b: \(\Leftrightarrow\left|x-5\right|=6x-3-7=6x-10\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{3}\\\left(6x-10-x+5\right)\left(6x-10+x-5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{3}\\\left(5x-5\right)\left(7x-15\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{15}{7}\)

c: \(\Leftrightarrow\left|2x-1\right|=2x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{2}\\\left(2x+3+2x-1\right)\left(2x+3-2x+1\right)=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{2}\)

d: =>|3x-2|=3x-2

=>3x-2>=0

hay x>=2/3

Bài 1:

1/6x+1/10x-4/15x+1=0

(1/6+1/10-4/15)x+1=0

0x+1=0

0x=0-1

0x=-1

x=-1/0

ko tim đc x thỏa mãn yêu cầu

Bài 2

a/ l5/3xl=l-1/6l

    l5/3xl=1/6

\(\Rightarrow\)5/3x=1/6 hoặc 5/3x=-1/6

TH1:5/3x=1/6                      TH2:5/3x=-1/6

           x=1/6:5/3                            x=-1/6:5/3  

           x=1/10                                x=-1/10

Vậy x\(\in\) {1/10 và -1/10}

a; |\(x+2\)| = 0

    \(x+2=0\)

   \(x\)         = - 2

Vậy \(x\)   = - 2

b; |\(x-5\)| = |-7|

   | \(x-5\) |  =  7

    \(\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

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

   \(\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)

Vậy \(x=12\)

      \(x=-2\)

a) x - 7 = -5

x = -5 + 7 

x = 2

b) 128 - 3 . ( x + 4 ) = 23

3 . ( x + 4 ) = 128 - 23

3 . ( x + 4 ) =  105

x + 4 = 105 : 3 

x + 4 = 35

x = 35 - 4

x = 31

d) | x - 3 | = 7 - ( -2 )

| x - 3 | = 9

\(\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}}\)

\(\orbr{\begin{cases}x=9+3\\x=-9+3\end{cases}}\)

\(\orbr{\begin{cases}x=12\\x=-6\end{cases}}\)

Vậy: ...

Mình làm bài 1, bài 2 bạn tự làm nhé!

Bài 1: 

a) \(2\left(4x-8\right)-7\left(3+x\right)=\left|4\right|\left(3-2\right)\)

\(\Leftrightarrow2\left(4x-8\right)-7\left(3+x\right)=4.1\)

\(\Leftrightarrow2\left(4x-8\right)-7\left(3+x\right)=4\)

\(\Leftrightarrow8x-16-21-7x=4\)

\(\Leftrightarrow x-37=4\)

\(\Leftrightarrow x=4+37\)

\(\Leftrightarrow x=41\)

Vậy \(x=41\)

b) \(8\left(x-\left|-7\right|\right)-6\left(x-2\right)=\left|-8\right|.6-50\)

\(\Leftrightarrow8\left(x-7\right)-6\left(x-2\right)=\left|-8\right|.6-50\)

\(\Leftrightarrow8\left(x-7\right)-6\left(x-2\right)=8.6-50\)

\(\Leftrightarrow8\left(x-7\right)-6\left(x-2\right)=48-50\)

\(\Leftrightarrow8\left(x-7\right)-6\left(x-2\right)=-2\)

\(\Leftrightarrow8x-56-6x+12=-2\)

\(\Leftrightarrow2x-44=-2\)

\(\Leftrightarrow2x=-2+44\)

\(\Leftrightarrow2x=42\)

\(\Leftrightarrow x=42:2\)

\(\Leftrightarrow x=21\)

Vậy \(x=21\)