Câu này có câu trl r mà

`CCIV`

Lỗi kìa e.

Để \(\overline{1a6b}⋮2,3,5,9\) thì \(\overline{b}=0\)

Ta có:

\(\overline{1a60}=\overline{1}+\overline{a}+\overline{6}+\overline{0}=\overline{7a}\)

Mà \(a\) là số có `1` chữ số

\(a\in\left\{2\right\}\)

\(\Rightarrow\overline{a}=\overline{2};\overline{b}=\overline{0}\)

\(a,1-\left(x+\dfrac{2}{7}\right)=\dfrac{5}{7}\)

\(\Rightarrow x+\dfrac{2}{7}=1-\dfrac{5}{7}\)

\(\Rightarrow x+\dfrac{2}{7}=\dfrac{2}{7}\)

\(\Rightarrow x=\dfrac{2}{7}-\dfrac{2}{7}\)

\(\Rightarrow x=0\)

Vậy `x = 0`

\(b,\dfrac{4}{3}+\left(1,25-x\right)=2,25\)

\(\Rightarrow1,25-x=2,25-\dfrac{4}{3}\)

\(\Rightarrow1,25-x=\dfrac{9}{4}-\dfrac{4}{3}\)

\(\Rightarrow1,25-x=\dfrac{27}{12}-\dfrac{16}{12}\)

\(\Rightarrow1,25-x=\dfrac{11}{12}\)

\(\Rightarrow x=1,25-\dfrac{11}{12}\)

\(\Rightarrow x=\dfrac{15}{12}-\dfrac{11}{12}\)

\(\Rightarrow x=\dfrac{1}{3}\)

Vậy \(x=\dfrac{1}{3}\)

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`5/9 + 4/9 . 8/17 + 4/9 . 1/17 - 4/9 . 26/17`

` = 5/9 + 4/9 . (8/17 + 1/17 - 26/17)`

` = 5/9 + 4/9 . (-1)`

` = 5/9 + (-4/9)`

` = 1/9`

6,10,13,17

`209 ÷ (2x - 1) = 11`

`→ 2x - 1 = 209 ÷ 11`

`→ 2x - 1 = 19`

`→ 2x       = 19 + 1`

`→ 2x       = 20`

`→   x        = 20 ÷ 2`

`→   x        = 10`

Vậy `x = 10`

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