Thus, The Answer of 1) is This are 1st and 2nd image and the solution of 2) is the 3rd image...
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[tex]\large\underline{\sf{Solution-1}}[/tex]
Given expression is
[tex]\sf \: \dfrac{2 \sqrt{3} - 3 \sqrt{2} }{3 \sqrt{2} + 2 \sqrt{3} } \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: = \: \dfrac{2 \sqrt{3} - 3 \sqrt{2} }{3 \sqrt{2} + 2 \sqrt{3} } \times \dfrac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \\ [/tex]
[tex]\sf \: = \: \dfrac{(2 \sqrt{3} - 3 \sqrt{2} )(3 \sqrt{2} - 2 \sqrt{3})}{(3 \sqrt{2})^{2} - (2 \sqrt{3} )^{2} } \\ [/tex]
[tex]\left[ \because \:\sf \: (x + y)(x - y) = {x}^{2} - {y}^{2} \right] \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \sqrt{3}(3 \sqrt{2} - 2 \sqrt{3}) - 3 \sqrt{2} (3 \sqrt{2} - 2 \sqrt{3})}{18 - 12 } \\ [/tex]
[tex]\sf \: = \: \dfrac{6 \sqrt{6} - 12 - 18 + 6 \sqrt{6} }{6} \\ [/tex]
[tex]\sf \: = \: \dfrac{12\sqrt{6} - 30}{6} \\ [/tex]
[tex]\sf \: = \: \dfrac{6(2\sqrt{6} - 5)}{6} \\ [/tex]
[tex]\sf \: = \: 2\sqrt{6} - 5 \\ [/tex]
Hence,
[tex]\implies\sf \: \boxed{\bf \: \dfrac{2 \sqrt{3} - 3 \sqrt{2} }{3 \sqrt{2} + 2 \sqrt{3} } = 2 \sqrt{6} - 5 \: }\\ [/tex]
[tex]\large\underline{\sf{Solution-2}}[/tex]
Given rational numbers are 1/3 and 1/2.
We know, If a and b are two rational numbers, then one rational number between a and b using mean method is
[tex]\boxed{ \sf \frac{a + b}{2} \: } \\ \\ [/tex]
So,
[tex]\sf \: First\:rational\:number \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left(\dfrac{1}{3} + \dfrac{1}{2}\right) \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left(\dfrac{2 + 3}{6} \right) \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left(\dfrac{5}{6} \right) \\ [/tex]
[tex]\sf \: = \: \dfrac{5}{12} \\ [/tex]
[tex]\sf\implies \sf \: First\:rational\:number = \: \dfrac{5}{12} \\ [/tex]
Now,
[tex]\sf \: Second\:rational\:number \: between \: \dfrac{1}{3} \: and \: \dfrac{5}{12} \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left(\dfrac{1}{3} + \dfrac{5}{12}\right) \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left( \dfrac{4 + 5}{12}\right) \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left( \dfrac{9}{12}\right) \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left( \dfrac{3}{4}\right) \\ [/tex]
[tex]\sf \: = \: \dfrac{3}{8} \\ [/tex]
[tex]\sf\implies \sf \:Second\:rational\:number= \: \dfrac{3}{8} \\ [/tex]
[tex]\sf\implies Two\:rational\:number \: between \: \dfrac{1}{3} \: and \: \dfrac{1}{2} \: are \dfrac{3}{8} \:and \: \dfrac{5}{12} \\ [/tex]
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Answers & Comments
Thus, The Answer of 1) is This are 1st and 2nd image and the solution of 2) is the 3rd image...
THANKS A LOT!
[tex]\large\underline{\sf{Solution-1}}[/tex]
Given expression is
[tex]\sf \: \dfrac{2 \sqrt{3} - 3 \sqrt{2} }{3 \sqrt{2} + 2 \sqrt{3} } \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: = \: \dfrac{2 \sqrt{3} - 3 \sqrt{2} }{3 \sqrt{2} + 2 \sqrt{3} } \times \dfrac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \\ [/tex]
[tex]\sf \: = \: \dfrac{(2 \sqrt{3} - 3 \sqrt{2} )(3 \sqrt{2} - 2 \sqrt{3})}{(3 \sqrt{2})^{2} - (2 \sqrt{3} )^{2} } \\ [/tex]
[tex]\left[ \because \:\sf \: (x + y)(x - y) = {x}^{2} - {y}^{2} \right] \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \sqrt{3}(3 \sqrt{2} - 2 \sqrt{3}) - 3 \sqrt{2} (3 \sqrt{2} - 2 \sqrt{3})}{18 - 12 } \\ [/tex]
[tex]\sf \: = \: \dfrac{6 \sqrt{6} - 12 - 18 + 6 \sqrt{6} }{6} \\ [/tex]
[tex]\sf \: = \: \dfrac{12\sqrt{6} - 30}{6} \\ [/tex]
[tex]\sf \: = \: \dfrac{6(2\sqrt{6} - 5)}{6} \\ [/tex]
[tex]\sf \: = \: 2\sqrt{6} - 5 \\ [/tex]
Hence,
[tex]\implies\sf \: \boxed{\bf \: \dfrac{2 \sqrt{3} - 3 \sqrt{2} }{3 \sqrt{2} + 2 \sqrt{3} } = 2 \sqrt{6} - 5 \: }\\ [/tex]
[tex]\large\underline{\sf{Solution-2}}[/tex]
Given rational numbers are 1/3 and 1/2.
We know, If a and b are two rational numbers, then one rational number between a and b using mean method is
[tex]\boxed{ \sf \frac{a + b}{2} \: } \\ \\ [/tex]
So,
[tex]\sf \: First\:rational\:number \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left(\dfrac{1}{3} + \dfrac{1}{2}\right) \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left(\dfrac{2 + 3}{6} \right) \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left(\dfrac{5}{6} \right) \\ [/tex]
[tex]\sf \: = \: \dfrac{5}{12} \\ [/tex]
So,
[tex]\sf\implies \sf \: First\:rational\:number = \: \dfrac{5}{12} \\ [/tex]
Now,
[tex]\sf \: Second\:rational\:number \: between \: \dfrac{1}{3} \: and \: \dfrac{5}{12} \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left(\dfrac{1}{3} + \dfrac{5}{12}\right) \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left( \dfrac{4 + 5}{12}\right) \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left( \dfrac{9}{12}\right) \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} \left( \dfrac{3}{4}\right) \\ [/tex]
[tex]\sf \: = \: \dfrac{3}{8} \\ [/tex]
So,
[tex]\sf\implies \sf \:Second\:rational\:number= \: \dfrac{3}{8} \\ [/tex]
Hence,
[tex]\sf\implies Two\:rational\:number \: between \: \dfrac{1}{3} \: and \: \dfrac{1}{2} \: are \dfrac{3}{8} \:and \: \dfrac{5}{12} \\ [/tex]