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

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

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

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

  => (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 

Bạn ghi lại đề đi bạn

\(\frac{2+x}{5}-0.5x=\frac{1-2x}{4}+0,\:25\)

a ) \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)

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

\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=3x\left(5x+7\right)-\left(5x+7\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)

\(\Leftrightarrow15x^2+13x+2=15x^2+16x-7\)

\(\Leftrightarrow13x+2=16x-7\)

\(\Leftrightarrow13x-16x=-7-2\)

\(\Leftrightarrow-3x=-9\)

\(\Rightarrow x=3\)

b ) tương tự

       \(\frac{1}{3}x-0,5x=0,75\)

\(\Leftrightarrow-\frac{1}{6}x=0,75\)

\(\Leftrightarrow\)\(x=-\frac{9}{2}\)

1/3x -0.5x = 0.75

(1/3 - 0.5)x = 0.75

-1/6x        =0.75

x              =0.75 : -1/6

x             =-9/2

tk mình nha .thank bạn nhiều:))

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 ???????????????

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

\(1,\)\(\frac{x+2}{x+3}+\frac{x-1}{x+1}=\frac{2}{x^2+4x+3}+1\)

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

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

\(\Rightarrow x^2-3x-6=0\)

.....

\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)

\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)

.....

\(a,\frac{-2}{3}x=8\)<=> \(x=-12\)

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

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

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_