Ta có \(x1-\frac{1}{9}=x2-\frac{2}{8}=...=x9-\frac{9}{1}\)

\(=\frac{x1-1}{9}=\frac{x2-2}{8}=\frac{x3-3}{7}=...=\frac{x9-9}{1}\)

= \(\frac{x1-1+x2-2+x3-3+...+x9-9}{9+8+7+...+1}\)

\(=\frac{\left(x1+x2+x3+...+x9\right)-\left(1+2+3+...+9\right)}{9+8+7+....+1}\)

=\(\frac{90-45}{45}=\frac{45}{45}=1\)

=> \(\hept{\begin{cases}\begin{cases}x1=10\\x2=10\end{cases}\\.....\\x9=10\end{cases}}\)

Đáp án:ta có :

X1-1/9=X2-2/8=X3-3/7=......X9-9/1

Áp dụn t/c dãy tỉ số bằng nhau

⇒(X1 +X2+X3+........X9)-(1+2+3+...+9)/1=2+3+...+9

=90-45/45=1

⇒X1=X2=X3=X4=..=X9=10

1 là chữ số 0

2 là chữ số 5

a) tận cùng là chữ số 0

b) tan cùng là chữ số 5

\(9^{x-1}=9\)

\(9^x\div9^1=9\)

\(9^x=9\times9\)

\(9^x=9^2\)

\(\Rightarrow x=2\)

Ta có : 9^x-1=9

       <=> 9^x-1=9¹

        <=> x-1=1

         <=> x=1+1

          <=> x= 2

           Vậy x =2

\(\dfrac{1}{3}+\dfrac{5}{9}=\dfrac{6}{18}+\dfrac{10}{18}=\dfrac{16}{18}=\dfrac{8}{9}\)

\(\dfrac{1}{7}-\dfrac{1}{9}=\dfrac{9}{63}-\dfrac{7}{63}=\dfrac{2}{63}\)

\(3:\dfrac{5}{9}=3.\dfrac{9}{5}=\dfrac{27}{5}\)

\(3.\dfrac{5}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)

\(\dfrac{1}{9}.\dfrac{9}{3}=\dfrac{1}{3}\)

\(\dfrac{1}{3}:\dfrac{1}{7}=\dfrac{7}{3}\)

\(9+\dfrac{9}{3}=9+3=12\)

\(4-\dfrac{2}{4}=4-\dfrac{1}{2}=\dfrac{7}{2}\)

\(\dfrac{1}{3}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{9}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3+5}{9}\) \(=\) \(\dfrac{8}{9}\)

\(\dfrac{1}{7}\) \(-\) \(\dfrac{1}{9}\) \(=\) \(\dfrac{9}{63}\) \(-\) \(\dfrac{7}{63}\) \(=\) \(\dfrac{9-7}{63}\) \(=\) \(\dfrac{2}{63}\)

\(\dfrac{1}{9}\) \(\times\) \(\dfrac{9}{3}\) \(=\) \(\dfrac{1\times9}{9\times3}\) \(=\) \(\dfrac{1}{3}\)

\(\dfrac{1}{3}\) \(\div\) \(\dfrac{1}{7}\) \(=\) \(\dfrac{1}{3}\) \(\times\) \(\dfrac{7}{1}\) \(=\) \(\dfrac{1\times7}{3\times1}\) \(=\) \(\dfrac{7}{3}\)

\(3\) \(\div\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{1}\) \(\div\)  \(\dfrac{5}{9}\) \(=\dfrac{3}{1}\times\dfrac{9}{5}=\dfrac{3\times9}{1\times5}=\dfrac{27}{5}\)

\(3\times\dfrac{5}{9}=\dfrac{3}{1}\times\dfrac{5}{9}=\dfrac{3\times5}{1\times9}=\dfrac{5}{3}\)

\(9+\dfrac{9}{3}=\dfrac{9}{1}+\dfrac{9}{3}=\dfrac{27}{3}+\dfrac{9}{3}=\dfrac{27+9}{3}=\dfrac{36}{3}=12\)

\(4\) \(-\dfrac{2}{4}=\dfrac{4}{1}-\dfrac{2}{4}=\dfrac{16}{4}-\dfrac{2}{4}=\dfrac{14}{4}=\dfrac{7}{2}\)

Goi phân số cần tìm là x

  Theo bài :    \(\frac{7}{9}< x< \frac{8}{9}\)

                        \(\frac{70}{90}< x< \frac{80}{90}\)

     Mình chọn \(x=\frac{71}{90}\)

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Tùy bạn, chọn trong khoảng đấy số nào cũng được