=>x/30-x/40=3/4

=>x/120=3/4

=>x=90

\(\dfrac{x}{30}=\dfrac{x}{40}+\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{4x}{120}=\dfrac{3x}{120}+\dfrac{90}{120}\)

\(\Leftrightarrow4x=3x+90\)

\(\Leftrightarrow4x-3x-90=0\)

\(\Leftrightarrow x-90=0\)

\(\Leftrightarrow x=90\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{90\right\}\)

7)    \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)

\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+36}{65}+1+\frac{x+40}{60}+1\)

\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)            (vì 1/75 + 1/70 - 1/65 - 1/60  \(\ne\)0)

\(\Leftrightarrow\)\(x=-100\)

Vậy.....

7)    \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)

\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)

\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)    (1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)

\(\Leftrightarrow\)\(x=-100\)

Vậy...

Lời giải:

PT $\Leftrightarrow \frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1$

$\Leftrightarrow \frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}$
$\Leftrightarrow (x+100)(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60})=0$

Dễ thấy $\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}<0$

$\Rightarrow x+100=0$

$\Leftrightarrow x=-100$ (tm)

=> (x + 1)(x + 5)(x + 2)(x + 4) - 40 = 0

=> (x² + 6x + 5)(x² + 6x + 8) - 40 = 0

Đặt x² + 6x + 5 = a (a > 0)

=> a.(a + 3) - 40 = 0

=> a² + 3a - 40 = 0

=> (a - 5)(a + 8) = 0

=> a = 5 (nhận) hoặc a = -8 (loại)

a = 5 => x² + 6x + 5 = 5 => x² + 6x = 0 => x(x + 6) = 0 => x = 0 hoặc x = -6

Vậy x = 0 , x = -6

x=0

x=0

x=4 hoặc -6

có cần chi tiết ko

<=>\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)-40=x\left(x+6\right)\left(x^2+6x+13\right)\)

=>x=0 và x=-6

=>\(x^2+6x+13=0\)

=> có biệt thức \(6^2-4\left(1.13\right)=-16\)

=>PT ko có nghiệm thực

=>x=-6 hoặc 0

9: \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)

=>x-99=0

hay x=99

7: \(\Leftrightarrow\left(\dfrac{x+25}{75}+1\right)+\left(\dfrac{x+30}{70}+1\right)=\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+40}{60}+1\right)\)

=>x+100=0

hay x=-100

8:

Sửa đề: \(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=-4\) 

\(\Leftrightarrow\left(\dfrac{99-x}{101}+1\right)+\left(\dfrac{97-x}{103}+1\right)+\left(\dfrac{95-x}{105}+1\right)+\left(\dfrac{93-x}{107}+1\right)=0\)

=>200-x=0

hay x=200