Question

[Hanyou Inu]

d,\(\frac{5x-4}{3-2x}=\frac{7+4x}{x+2}\)

e,\(\frac{x-1}{99}+\frac{x-2}{98}=\frac{x-3}{97}+\frac{x-4}{96}\)

mong cac bn giup

Answer 1

[Ngọc Vĩ]

a/ \(\frac{5x-4}{3-2x}=\frac{7+4x}{x+2}\)     (ĐK: \(x\ne\frac{3}{2};x\ne-2\))

     \(\Rightarrow\left(x+2\right)\left(5x-4\right)=\left(7+4x\right)\left(3-2x\right)\)

     \(\Rightarrow5x^2-4x+10x-8=21-14x+12x-8x^2\)

     \(\Rightarrow13x^2+8x-29=0\)

      \(\Rightarrow13\left(x^2+\frac{8}{13}x-\frac{29}{13}\right)=0\)

     \(\Rightarrow13\left[x^2+2.\frac{4}{13}.x+\left(\frac{4}{13}\right)^2-\left(\frac{4}{13}\right)^2-\frac{29}{13}\right]=0\)

      \(\Rightarrow13\left[\left(x+\frac{4}{13}\right)^2-\frac{393}{169}\right]=0\)

       \(\Rightarrow13\left(x+\frac{4}{13}\right)^2-\frac{393}{13}=0\)

        \(\Rightarrow\left(x+\frac{4}{13}\right)^2=\frac{393}{169}\)

       \(\Rightarrow\orbr{\begin{cases}x+\frac{4}{13}=\sqrt{\frac{393}{169}}=\frac{\sqrt{393}}{13}\Rightarrow x=\frac{-4+\sqrt{393}}{13}\\x+\frac{4}{3}=-\sqrt{\frac{393}{169}}=-\frac{\sqrt{393}}{13}\Rightarrow x=\frac{-4-\sqrt{393}}{13}\end{cases}}\)

        Vậy biểu thức có 2 nghiệm \(x=\left\{\frac{-4+\sqrt{393}}{13};\frac{-4-\sqrt{393}}{13}\right\}\)

b/ \(\frac{x-1}{99}+\frac{x-2}{98}-\frac{x-3}{97}-\frac{x-4}{96}=0\)

   \(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1-\left(\frac{x-3}{97}-1\right)-\left(\frac{x-4}{96}-1\right)=0\)

   \(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}-\frac{x-100}{97}-\frac{x-100}{96}=0\)

   \(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

    => x - 100 = 0 => x = 100

                                                    Vậy x = 100