Question

\(Cho\)\(x_1;\)\(x_2;\)\(x_3;\)\(...\)\(x_{99}\)\(bi\text{ết}\)

                                                       \(\frac{1-x_1}{99}=\frac{2-x_2}{98}=\frac{3-x_3}{97}=...=\frac{9-x_{99}}{1}\)

                                            \(v\text{à}\)\(x_1\)+\(x_2\)+\(x_3\)+ ... +\(x_{99}\)= \(4950\)

                                                  \(T\text{ìm}\)\(x_1\);\(x_2\);\(x_3\); ... ;\(x_{99}\)

                                          

Answer 1

Ta có ; \(\frac{1-x_1}{99}=\frac{2-x_2}{98}=\frac{3-x_3}{97}=...=\frac{99-x_{99}}{1}\)

\(\Leftrightarrow\frac{1-x_1}{99}+1=\frac{2-x_2}{98}+1=\frac{3-x_3}{97}+1=...=\frac{99-x_{99}}{1}+1\)

\(\Leftrightarrow\frac{100-x_1}{99}=\frac{100-x_2}{98}=\frac{100-x_3}{97}=...=\frac{100-x_{99}}{1}\)

Áp dụng t/c dãy tỉ số bằng nhau : \(\frac{100-x_1}{99}=\frac{100-x_2}{98}=\frac{100-x_3}{97}=...=\frac{100-x_{99}}{1}\)

\(=\frac{\left(100-x_1\right)+\left(100-x_2\right)+\left(100-x_3\right)=...=\left(100-x_{99}\right)}{1+2+3+...+98+99}\)

\(=\frac{100.99-\left(x_1+x_2+x_3+...+x_{99}\right)}{1+2+3+...+99}=\frac{100.99-4950}{\frac{99.100}{2}}=1\)

\(\Rightarrow x_i=100-\left(100-i\right)=i\)với \(i=1,2,3,...,99\)

Answer 2

\(\frac{1-x_1}{99}=\frac{2-x_2}{98}=\frac{3-x_3}{97}=...=\frac{99-x_{99}}{1}=\)\(\frac{\left(1+2+3+..+99\right)-\left(x_1+x_2+x_3+...+x_{99}\right)}{99+98+97+...+1}\)\(=\frac{4950-4950}{4950}=0\)

\(\Rightarrow1-x_1=2-x_2=3-x_3=...=99-x_{99}=0\)

\(\Rightarrow x_i=i-0\left(i=1,2,3,...,99\right)\)

Answer 3

sao may ngu the

Answer 4

cai loai oc cho nha may nua