Tìm x:
b) 1/3.x+2/5.(x-1)=0

\(<=> \dfrac{1}{3} .x +\dfrac{2}{5}x - \dfrac{2}{5} =0\)

\(<=> \dfrac{11}{15}x = \dfrac{2}{5}\)

\(<=> x= \dfrac{6}{11}\)

Vậy \( x= \dfrac{6}{11}\)
c) (2x-3).(6-2x)=0
\(<=> \begin{cases} 2x-3=0 \\ 6-2x=0 \end{cases}\) \(<=> \begin{cases} 2x=3 \\ -2x=-6 \end{cases}\) \(<=>\begin{cases} x=\dfrac{3}{2} \\ x=3 \end{cases}\)

Vậy \(x=( \dfrac{3}{2} ; 3)\)
d) -2/3-1/3.(2x-5)= 3/2

\(<=> 2x-5= \dfrac{5}{2}\)

\(<=> 2x= \dfrac{15}{2}\)

\(<=> x= \dfrac{15}{4}\)

Vậy \(x= \dfrac{15}{4}\)
f) 1/3.x-1/2=4 và 1/2 (Hỗn số ý '^')

\(<=> \dfrac{1}{3} x -\dfrac{1}{2} = \dfrac{9}{2}\)

\(<=> \dfrac{1}{3}x =5\)

\(<=> x= 15\)

Vậy \(x= 15\)

Cảm ơn cậu nhiều nhé!

a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)

b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)

\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)

c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)

\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)

e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)

f) \(\left|12-x\right|-7=5\)

th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)

th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)

i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)

k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)

a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)

b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)

Làm nốt nha

e) \(\left(x-3\right)\left(x^2+1\right)=0\)

\(\Rightarrow\left(x-3\right)=0\)  ( \(x^2+1>0\forall x\))

\(\Rightarrow x=3\)

đ) \(4.8^2=2^x\)

\(2^2.\left(2^3\right)^2=2^x\)

\(2^2.2^6=2^x\)

\(2^8=2^x\)

\(\Rightarrow x=8\)

d) \(\left|x+3\right|=8\)

\(\Rightarrow\orbr{\begin{cases}x+3=8\\x+3=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-11\end{cases}}\)

mấy câu trên dễ rồi tự làm em nhé 

a, 2x + 5 = 7

=> 2x = 2

=> x = 1

vậy____

\(a,234-\left(x-56\right)=789\)

\(\Leftrightarrow x-56=234-789\)

\(\Leftrightarrow x-56=-555\)

\(\Leftrightarrow x=\left(-555\right)+56=-499\)

Vậy x = -499

b) \(\frac{x+3}{-5}=\frac{x-15}{4}\)

\(\Leftrightarrow4\left(x+3\right)=-5\left(x-15\right)\)

\(\Leftrightarrow4x+12=-5x+75\)

\(\Leftrightarrow4x+12-\left(-5x\right)=75\)

\(\Leftrightarrow4x-\left(-5x\right)+12=75\)

\(\Leftrightarrow4x+5x=63\)

\(\Leftrightarrow9x=63\)

\(\Leftrightarrow x=7\)

Vậy x = 7

c) \(8\left(x-1\right)-7=2\left(x+2\right)+5\)

\(\Leftrightarrow8x-8-7=2x+4+5\)

\(\Leftrightarrow8x-8-7-2x+4=5\)

\(\Leftrightarrow8x-2x-8-7+4=5\)

\(\Leftrightarrow8x-2x=5-4+7+8\)

\(\Leftrightarrow4x=16\)

\(\Leftrightarrow x=4\)

Vậy x = 4

d) Đặt \(D=\frac{2x+3}{x-1}=\frac{2x-2+5}{x-1}=\frac{2\left(x-1\right)+5}{x-1}=2+\frac{5}{x-1}\)

=> \(5⋮x-1\)

=> \(x-1\inƯ\left(5\right)\)

=> \(x-1\in\left\{\pm1;\pm5\right\}\)

=> \(x\in\left\{2;0;6;-4\right\}\)

a) x-12:(-2)=4

=> x-12=-8

=> x=4
b ) 6-|x| = 5

=> /x/=1

=> x=1;-1
c ) 7⋮ ( x-3)

=> (x-3) thuộc Ư(7)

=> x-3=1 => x=4

=> x-3=-1 => x=2

=> x-3=7 => x= 10

=> x-3=-7 => x=-4

d ) 3⋮ ( 2x+1 )

=> (2x+1) thuộc Ư(3)

=> (2X+1)= 1 => x= 0

=> (2x+1)=-1 => x= -1

=> 2x+1= 3 => x= 1

=> 2x+1=-3 => x= -2

a) \(x-12:\left(-2\right)=4\Rightarrow x-\left(-6\right)=4\Rightarrow x=\left(-6\right)+4=-2\)

b) \(6-\left|x\right|=5\Rightarrow\left|x\right|=6-5=1\Rightarrow x=\left\{\pm1\right\}\)

c)\(7⋮x-3\Rightarrow x-3\inƯ\left(7\right)\)

\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(x=\left\{4;2;10;-4\right\}\)

d) \(3⋮2x+1\Rightarrow2x+1\inƯ\left(3\right)\)

\(Ư\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(x\in\left\{0;-1;1;-2\right\}\)

tưởng dễ mà khó

_Bài dễ :vvv

Nhìu thế

a,\(\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)

\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x};Đkxđ:x\ne1\)

\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}\left(\frac{-9}{20}\right)=\frac{5}{2-2x}\)

\(\Rightarrow\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2-2x}\)

\(\Rightarrow\frac{1}{x-1}-\frac{5}{2-2x}=\frac{-3}{10}\)

\(\Rightarrow\frac{1}{x-1}-\frac{5}{-2\left(x-1\right)}=\frac{-3}{10}\)

\(\Rightarrow\frac{1}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{7}{2\left(x-1\right)}=\frac{-3}{10}\)

\(\Rightarrow70=-6\left(x-1\right)\)

\(\Rightarrow6x=6-70\)

\(\Rightarrow6x=-64\)

\(\Rightarrow x=\frac{-32}{3}x\ne1\)