Question

[Mai gia bảo]

Answer 1

[『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』]

`@` `\text {Ans}`

`\downarrow`

a)

\(2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\)

`\Rightarrow`\(\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)

`\Rightarrow`\(\dfrac{10}{4}-a=\dfrac{3}{2}\)

`\Rightarrow`\(a=\dfrac{10}{4}-\dfrac{3}{2}\)

`\Rightarrow`\(a=\dfrac{5}{2}-\dfrac{3}{2}\)

`\Rightarrow`\(a=\dfrac{2}{2}=1\)

Vậy, `a=1`

b)

\(3\dfrac{1}{4}-a-1\dfrac{3}{4}=\dfrac{7}{9}\)

`\Rightarrow`\(a+\dfrac{3}{2}=\dfrac{7}{9}\)

`\Rightarrow`\(a=\dfrac{7}{9}-\dfrac{3}{2}\)

`\Rightarrow`\(a=-\dfrac{13}{18}\)

Vậy, \(a=-\dfrac{13}{18}\)

c)

\(2\dfrac{5}{6}-1\dfrac{1}{2}-a=\dfrac{1}{6}\)

`\Rightarrow`\(\dfrac{4}{3}-a=\dfrac{1}{6}\)

`\Rightarrow`\(a=\dfrac{4}{3}-\dfrac{1}{6}\)

`\Rightarrow`\(a=\dfrac{7}{6}\)

Vậy, \(a=\dfrac{7}{6}\)