Giải phương trình [ (x+2)/98 + (x+4)/96] = [(x+6)/94 + (x+8)/92]

Giải Toán 8 Bài tập ôn cuối năm

Bài 9 trang 130-131 SGK Toán lớp 8 tập 2:

$\frac{x+2}{98}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\frac{x+4}{96}=\frac{x+6}{94}\text{\hspace{0.17em}\hspace{0.17em}}+\frac{x+8}{92}\text{\hspace{0.17em}}$

Lời giải:

$\frac{x+2}{98}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\frac{x+4}{96}=\frac{x+6}{94}\text{\hspace{0.17em}\hspace{0.17em}}+\frac{x+8}{92}\iff \frac{x+2}{98}\text{\hspace{0.17em}\hspace{0.17em}}+1+\text{\hspace{0.17em}}\frac{x+4}{96}+\text{\hspace{0.17em}}1=\frac{x+6}{94}\text{\hspace{0.17em}\hspace{0.17em}}+1+\frac{x+\text{}8}{92}\text{\hspace{0.17em}\hspace{0.17em}}+1\iff \frac{x+100}{98}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\frac{x+100}{96}=\frac{x+100}{94}\text{\hspace{0.17em}\hspace{0.17em}}+\frac{x+\text{}100}{92}\iff \frac{x+100}{98}\text{\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}}\frac{x+100}{94}+\frac{x+100}{96}\text{\hspace{0.17em}\hspace{0.17em}}-\frac{x+\text{}100}{92}=0\iff (x+100).\left(\frac{1}{98}-\frac{1}{94}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{96}-\frac{1}{92}\right)=0\iff \begin{array}{c}x+100=0\\ \frac{1}{98}-\frac{1}{94}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{96}-\frac{1}{92}=0\end{array}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}$

Nếu x + 1000 = 0 thì x = -100

Ta có:

$\frac{1}{98}<\frac{1}{94};\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{96}<\frac{1}{92}\Rightarrow \text{\hspace{0.17em}}\frac{1}{98}-\frac{1}{94}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{96}-\frac{1}{92}<0$

Vậy phương trình có nghiệm x = -100.