bai toan nay kho qua

x=-\(\frac{2737}{2115}\)

\(x+\frac{16}{49}+x+\frac{18}{47}=x+\frac{20}{45}-1\)

\(2x+\frac{16}{49}+\frac{18}{47}-x-\frac{20}{45}=-1\)

\(x+\frac{5494}{20727}=-1\)

\(x=-1-\frac{5494}{20727}=-\frac{26221}{20707}\)

(x+16)/49.... cơ Dung à 

\(\left(x+\frac{16}{49}\right)+\left(x+\frac{18}{47}\right)=\left(x+\frac{20}{45}\right)-1\)

\(\Rightarrow x+\frac{16}{49}+x+\frac{18}{47}=x+\frac{4}{9}-1\)

\(\Rightarrow x+\frac{16}{49}+x+\frac{18}{47}-x-\frac{4}{9}+1=0\)

\(\Rightarrow x+\frac{26221}{20727}=0\)

\(\Rightarrow x=-\frac{26221}{20727}\) (loại)

Vậy x thuộc rỗng

Thêm 2 vào pt có :

\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)                (1)

\(\Leftrightarrow\frac{x+16}{49}+1+\frac{x+18}{47}+1=\frac{x+20}{45}+1\)

\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\) (2)

\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)

Vì \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\ne0\)

\(\Leftrightarrow x+65=0\)

\(\Leftrightarrow x=-65\)

k mk nha!

thanks

thanks

pt \(\Leftrightarrow\left(\dfrac{x+16}{49}+1\right)+\left(\dfrac{x+18}{47}+1\right)=\left(\dfrac{x+20}{45}+1\right)\)

\(\Leftrightarrow\dfrac{x+65}{49}+\dfrac{x+65}{47}-\dfrac{x+65}{45}=0\)

\(\Leftrightarrow\left(x+65\right)\left(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\right)=0\)

\(\Leftrightarrow x+65=0\)

\(\Leftrightarrow x=-65\)

\(\dfrac{x+16}{49}+\dfrac{x+18}{47}=\dfrac{x+20}{45}+1\left(1\right)\)

Thêm (2) vào vế của p.trình (1) ta có :

\(\Leftrightarrow\dfrac{x+16}{49}+1+\dfrac{x+18}{47}+1=\dfrac{x+20}{45}+1\)

\(\Leftrightarrow\dfrac{x+65}{49}+\dfrac{x+65}{47}-\dfrac{x+65}{45}=0\) (2)

\(\Leftrightarrow\left(x+65\right)\left(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\right)=0\)

\(\Leftrightarrow x+65=0\) ( vì \(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\ne0\))

\(\Leftrightarrow x=-65\)

Lời giải:

PT $\Leftrightarrow \frac{x+16}{49}+1+\frac{x+18}{47}+1=\frac{x+20}{45}+1$

$\Leftrightarrow \frac{x+65}{49}+\frac{x+65}{47}=\frac{x+65}{45}$

$\Leftrightarrow (x+65)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0$

Thấy rằng $\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\neq 0$

Do đó $x+65=0\Rightarrow x=-65$

Bạn hỏi j dzậy!

giải phương trình á bạn

Mình mới học lớp 6 thui . kb voi mink nhé

bạn chỉ mình cách gõ phần như vậy đi mình mới chỉ bạn lời giải được

<=> (x+16/49 +2)+(x+18/47 +2)= (x+20/45 +2)-1

<=>x+65/49 + x+65/47 - x+65/45 +1=0

<=>(x+65)(1/49 + 1/47 - 1/45 + 1)=0

<=>x+65=0 (vì 1/49 + 1/47 - 1/45 + 1 khác 0)

<=>x= -65

Vậy S={-65}