Question

Giải phương trình :

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

Answer 1

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\)nên \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}< 0\)

Suy ra x + 10 = 0

Vậy x = -10

Answer 2

Pt ban đầu tương đương :

\(\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà : \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x+10=0\)

\(\Leftrightarrow x=-10\) ( thỏa mãn )

Vậy pt đã cho có tập nghiệm \(S=\left\{-10\right\}\)

Answer 3

a.$\frac{x+1}{9}\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}$

=>$\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1$

<=> $\frac{x+1+9}{9}+\frac{x+2+8}{8}=\frac{x+3+7}{7}+\frac{x+4+6}{6}$

<=>$\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}$

<=> $\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0$

<=> $\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0$

<=> x+10=0

<=> x=-10

Vậy tập nghiệm của phương trình trên là S=$\left\{-10\right\}$

Answer 4

Tks ctk_07

       Lê Danh Vinh 

~~~

Answer 5

\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow x=-10\)

Vậy ...

Answer 6

Từ đẳng thức, ta suy ra:

\(1+\frac{x+1}{9}+1+\frac{x+2}{8}=1+\frac{x+3}{7}+1+\frac{x+4}{6}\)

\(\frac{10+x}{9}+\frac{10+x}{8}=\frac{10+x}{7}+\frac{10+x}{6}\)

\(\left(10+x\right)\left(\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)

Vì \(\left(\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)>0\)nên 10+x=0

<=>x=-10

Vậy phương trình có nghiệm duy nhất x=-10

Answer 7

Bài này của lớp 6 mà bạn ? :D

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(< =>\frac{x+1+9}{9}+\frac{x+2+8}{8}-\frac{x+3+7}{7}-\frac{x+4+6}{6}=0\)

\(< =>\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(< =>\left(x+10\right).\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Do \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(=>x+10=0\)

\(< =>x=-10\)

Vậy nghiệm của phương trình trên là {-10}

Answer 8

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(x+10\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(x+10=0\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{8}\ne0\right)\)

x=-10

Answer 9

x=10 bạn nha