Question

insert a fraction between 20/22, 25/26​

Answer 1

$\large\underline{\sf{Solution-}}$

Given rational numbers are

$\sf \: \dfrac{20}{22} \: and \: \dfrac{25}{26}$

can be further rewritten as

$\sf \: \dfrac{10}{11} \: and \: \dfrac{25}{26}$

We know, if a and b are two real numbers, then a rational number between a and b is $\frac{a+b}{2}$

So,

$\sf \: A \: rational \: number \: between \: \dfrac{10}{11} \: and \: \dfrac{25}{26} \: is \:$

$\sf \: = \: \frac{1}{2}\bigg(\dfrac{10}{11} \: + \: \dfrac{25}{26}\bigg) \:$

$\sf \: = \: \frac{1}{2}\bigg(\dfrac{260 + 275}{286}\bigg) \:$

$\sf \: = \: \frac{1}{2}\bigg(\dfrac{535}{286}\bigg) \:$

$\sf \: = \: \dfrac{535}{572} \:$

So,

$\boxed{ \sf{ \: A \: rational \: number \: between \: \dfrac{20}{22} \: and \: \dfrac{25}{26} \: is \: \frac{535}{572} \: }}$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

20/22 = 10/11

25/26 = 25/26

LCM of 11 and 26 = 286

(10/11)×(26/26) = 260/286

(25/26)×(11/11)= 275/286

fraction b/w 20/22 and 25/26 are :-

261/286 , 262/286..... 274/286