\(a,\frac{-2}{3}x=8\)<=> \(x=-12\)

\(b,\frac{1}{4}:x=-3+\frac{3}{4}\)<=>\(\frac{1}{4}:x=\frac{-9}{4}\)<=>\(x=-9\)

\(c,\orbr{\begin{cases}2x-\frac{1}{3}=0\\0.5x+0.25=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{-3}{4}\end{cases}}\)

a) \(\frac{-2}{3}x+4=12\)

\(\Rightarrow\frac{-2}{3}x=12-4\)

\(\Rightarrow\frac{-2}{3}x=8\)

\(\Rightarrow x=8:\frac{-2}{3}\)

\(\Rightarrow x=-12\)

Vậy x = -12

b) \(\frac{-3}{4}+\frac{1}{4}:x=-3\)

\(\Rightarrow\frac{1}{4}:x=-3-\left(\frac{3}{4}\right)\)

\(\Rightarrow\frac{1}{4}:x=\frac{-9}{4}\)

\(\Rightarrow x=\frac{1}{4}:\frac{-9}{4}\)

\(\Rightarrow x=\frac{-1}{9}\)

Vậy  \(x=\frac{-1}{9}\)

c) \(\left(2x-\frac{1}{3}\right)\left(0,5x+0,25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\0,5x+0,25=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\0,5x=-0,25\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-0,5\end{cases}}\)

Vậy  \(x=\frac{1}{6}\)hoặc \(x=-0,5\)

_Chúc bạn học tốt_

*Tính M(x) - N(x)

M(x)            = -x³ + 2,5x² - 0,5x - 1

N(x)            = -x³ + 2,5x² + 2x   - 6

------------------------------------

M(x) - N(x) =                    -2,5x + 5

=> M(x) - N(x) = A(x) = -2,5x + 5

Để đa thức A(x) có nghiệm => -2,5x + 5 = 0

=> -2,5x = -5

=> 2,5x = 5

=>  x = 2

Tính M(x) + N(x)  

M(x)          =   -x³ + 2,5x² - 0,5x - 1

N(x)          = -x³ + 2,5x² + 2x - 6 

---------------------------------------------

M(x) + N(x) = -2x³ + 5x² + 1,5x - 7

=> M(x) + N(x) = B(x) = -2x³ + 5x² + 1,5x - 7

Bậc của đa thức B(x) là 3

P/S : Cái dấu chấm đó là nhân hay phẩy?

1,3x+2/5x+7 =3x-1/5x+1

<=> 1,3x+2/5x-3x+1/5x = 1-7

<=> (1,3+2/5-3+1/5)x = -6

<=> -11/10x=-6

<=> x= -6 : (-11/10)

<=> x= 60/11

2.x+1/2x+1 = 0,5x+2/x +3

<=> 2x+1/2x-0,5x-2/1x = 3-1

<=> x(2+1/2-0,5-2 ) =2

<=>0x =2

<=> x=0 

Hinh nhu minh thay ban Kunzy Nguyen giai hoi sai

1,3x o dau ra ???????????????

 Hình như bạn Kunzy Nguyễn ghi nhầm 3x thành 1,3x nhỉ ?!!! 

1) áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\frac{3x+2-3x+1}{5x+7-5x-1}=\frac{3}{6}=\frac{1}{2}\)

suy ra :

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

=> 5x+1=6x-2

5x-6x=-2-1

-x=-3

x=3

2)áp dụng tính chất của dãy tỉ số bằng nhau ta có;

\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}=\frac{\left(x+1\right)-2.\left(0,5x+2\right)}{\left(2x+1\right)-2.\left(x+3\right)}=\frac{x+1-x-4}{2x+1-2x-6}=\frac{-3}{-5}=\frac{3}{5}\)

suy ra:

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

=>6x+3=5x+5

6x-5x=5-3

x=2

 chỉ có làm mới có ăn

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

\(\Leftrightarrow x^2+4x+3=x^2+4.5x+2\)

\(\Leftrightarrow0.5x+2=3\Leftrightarrow0.5x=1\Leftrightarrow x=2\)

\(\frac{x+1}{2x+1}\)\(=\)\(\frac{0,5x+2}{x+3}\)

  => (x+1).(x+3) = (2x+1).(0,5x+2)

  => x²+4x+3 = x²+4,5x+2

  => x²+4x-x²-4,5x = 2-3

  => -0,5x = -1

 =>  x      = -1:-0,5

 =>  x      = 2 

\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(0,5x+2\right)\left(2x+1\right)\)

\(\Leftrightarrow x^2+3x+x+3=x^2+0,5x+4x+2\)

\(\Leftrightarrow x^2+4x+3=x^2+4,5x+2\)

\(\Leftrightarrow4x+3=4,5x+2\)

\(\Leftrightarrow0,5x=-1\)

\(\Leftrightarrow x=-2\)

Tíck cho mìk vs nha Vân Nguyễn!

a,\(\left(0,5x-\dfrac{3}{7}\right):\dfrac{1}{2}=1\dfrac{1}{7}\)

\(0,5x-\dfrac{3}{7}=\dfrac{4}{7}\)

\(0,5x=1\)

\(x=2\)

b,Đề chưa rõ,bạn viết lại nhé!

c,\(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)

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

\(x=-\dfrac{13}{15}\)

\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Rightarrow\left(x+1\right)\left(x+3\right)=\left(0,5x+2\right)\left(2x+1\right)\)

\(\Rightarrow x^2+3x+x+3=x^2+0,5x+4x+2\)

\(\Rightarrow x^2+3x+x-x^2-0,5x-4x=2-3\)

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

\(\Rightarrow x=2\)

\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)

\(\Rightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)

\(\Rightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)

\(\Rightarrow x^2+3x+x+3-x^2-4x-0,5x-2=0\)

\(\Rightarrow1-0,5x=0\)

\(\Rightarrow\frac{1}{2}x=1\Leftrightarrow x=2\)