9th marathi setu abhyas kram chachani 2 Question bank
Answers
Answer:
Since b is a mean proportion between a and c.
Hence by geometric mean property:
\begin{gathered} {b}^{2} = ac \: \: ...(1) \\ LHS = \frac{ {a}^{4} + {a}^{2} {b}^{2} + {b}^{4} }{{b}^{4} + {b}^{2} {c}^{2} + {c}^{4} } \: \\ = \frac{ {a}^{4} + {a}^{2} \times ac + {(ac)}^{2} }{{(ac)}^{2} + ac \times {c}^{2} + {c}^{4} } \: \\ = \frac{ {a}^{4} + {a}^{3} c \: + {a}^{2} {c}^{2} }{{a}^{2} {c}^{2} + a {c}^{3} + {c}^{4} } \\ = \frac{ {a}^{2} ({a}^{2} + {a} c \: + {c}^{2} )}{ {c}^{2} ({a}^{2} + a {c} + {c}^{2} )} = \frac{ {a}^{2} }{ {c}^{2} } \\ = \: RHS \\ < /p > < p > THUS \: PROVED\end{gathered}
b
2
=ac...(1)
LHS=
b
4
+b
2
c
2
+c
4
a
4
+a
2
b
2
+b
4
=
(ac)
2
+ac×c
2
+c
4
a
4
+a
2
×ac+(ac)
2
=
a
2
c
2
+ac
3
+c
4
a
4
+a
3
c+a
2
c
2
=
c
2
(a
2
+ac+c
2
)
a
2
(a
2
+ac+c
2
)
=
c
2
a
2
=RHS
</p><p>THUSPROVED