Question

please solve it asap...with clear explanation
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Answer 1

Ello user

Refer to the attachment for the solution.

Hope this helps.

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{a^{2}+b^{2}+c^{2}}{a+c-d}\:\:is\:a\:Integer}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Rational \: number = \frac{ a\sqrt{2} + b}{b \sqrt{2} + c } \\ \\ \tt: \implies a,b,c \: are \: positive \: integer \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Which \: option \: is \: always \: integer = ?$

• According to given question :

$\tt: \implies \frac{a \sqrt{2} + b}{b \sqrt{2} + c} \\ \\ \bold{As \: we \: know \:that} \\ \tt: \implies \frac{a \sqrt{2} + b}{b \sqrt{2} + c} \times \frac{b \sqrt{2} - c }{b \sqrt{2} - c } \\ \\ \tt: \implies \frac{2ab - \sqrt{2}ac + {b}^{2} \sqrt{2} - bc }{2 {b}^{2} - {c}^{2} } \\ \\ \tt: \implies \frac{2ab - bc + \sqrt{2}( {b}^{2} - ac) }{ {2b}^{2} - {c}^{2} } \\ \\ \tt: \implies {b}^{2} - ac =0 \\ \\ \tt: \implies {b}^{2} = ac \\ \\ \tt \because Numbers \: are \: a,ar,a{r}^{2} \\ \\ {\bold{For \: option \: A}} \\ \tt: \implies \frac{2 {a}^{2} + {b}^{2} }{2 {b}^{2} + c} \\ \\ \tt: \implies \frac{2 {a}^{2} + ac}{2ac + {c}^{2} }$

$\tt: \implies \frac{2a(a + c)}{2c(c + a)} \\ \\ \tt: \implies \frac{a}{c} \\ \\ \tt: \implies \frac{1}{ {r}^{2} } \\ \\ \tt\green{ \therefore \frac{1}{ {r}^{2} } \: may \: or \: may \: not \: be \: integer} \\ \\ \bold{For \: option \:B} \\ \tt: \implies \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{a + b - c} \\ \\ \tt: \implies \frac{ {a}^{2} + {a}^{2} {r}^{2} - {a}^{2} {r}^{4} }{a + ar - a {r}^{2} } \\ \\ \tt: \implies a(\frac{1 + {r}^{2} + {r}^{4} }{1 + r - {r}^{2} } ) \\ \\ \green{\tt\therefore It \: is \: also \: may \: or \: may \: not \: be \: a \: integer}$

$\bold{For \: option \: C} \\ \tt: \implies \frac{ {a}^{2} + 2 {b}^{2} }{ {b}^{2} + 2 {c}^{2} } \\ \\ \tt: \implies \frac{ {a}^{2} + 2 {a}^{2} {r}^{2} }{ {a}^{2} {r}^{2} + {2a}^{2} {r}^{4} } \\ \\ \tt: \implies \frac{ {a}^{2} (2 {r}^{2} + 1) }{ {a}^{2} {r}^{2}( {2r}^{2} + 1) } \\ \\ \tt: \implies \frac{1}{ {r}^{2} } \\ \\ \tt\green{ \therefore \frac{1}{ {r}^{2} } \: may \: or \: may \: not \: be \: integer} \\ \\ \bold{For \: option \: D} \\ \tt : \implies \frac{ {a}^{2} + {b}^{2} + {c}^{2} }{a + c - d} \\ \\ \tt : \implies \frac{ {a}^{2}(1 + {r}^{2} + {r}^{4} )}{a( {r}^{2} - r + 1)} \\ \\ \tt : \implies a( {r}^{2} + r + 1) \\ \\ \tt : \implies a + b + c \\ \\ \tt\green{ \therefore a + b + c \: is \: a \: integer}$