Expand using suitable identity (2 − 3) 2
Answers
Answer:
1
Step-by-step explanation:
by using identity (a-b)²=a²+b²-2ab
(2)²+(3)²-2(2)(3)
4+9-12
13-12
=1
Answer:
ok varnika can I call you V
Step-by-step explanation:
\large\underline{\sf{Solution-}}
Solution−
Given expression is
\begin{gathered}\sf \: \left( \dfrac{2x}{3} - \dfrac{3}{2x} \right)^{2} \\ \\ \end{gathered}
(
3
2x
−
2x
3
)
2
Expanding using Binomial theorem, we have
\begin{gathered}\sf \: = \: ^2C_{0} {\bigg[\dfrac{2x}{3} \bigg]}^{0} { \bigg[ - \dfrac{3}{2x} \bigg]}^{2} + \: ^2C_{1} {\bigg[\dfrac{2x}{3} \bigg]}^{1} { \bigg[ - \dfrac{3}{2x} \bigg]}^{2 - 1} + \: ^2C_{2} {\bigg[\dfrac{2x}{3} \bigg]}^{2} {\bigg[ - \dfrac{3}{2x} \bigg]}^{2 - 2} \\ \\ \end{gathered}
=
2
C
0
[
3
2x
]
0
[−
2x
3
]
2
+
2
C
1
[
3
2x
]
1
[−
2x
3
]
2−1
+
2
C
2
[
3
2x
]
2
[−
2x
3
]
2−2
\begin{gathered}\sf \: = \: 1 \times 1 \times \frac{9}{ {4x}^{2} } + \: 2 {\bigg[\dfrac{2x}{3} \bigg]}^{1} { \bigg[ - \dfrac{3}{2x} \bigg]}^{1} + \: 1 {\bigg[\dfrac{2x}{3} \bigg]}^{2} {\bigg[ - \dfrac{3}{2x} \bigg]}^{0} \\ \\ \end{gathered}
=1×1×
4x
2
9
+2[
3
2x
]
1
[−
2x
3
]
1
+1[
3
2x
]
2
[−
2x
3
]
0
\begin{gathered}\sf \: = \: \dfrac{9}{ {4x}^{2} } - 2 + \: \dfrac{ {4x}^{2} }{9} \\ \\ \end{gathered}
=
4x
2
9
−2+
9
4x
2
Hence,
\begin{gathered}\bf\implies \: \left( \frac{2x}{3} - \frac{3}{2x} \right)^{2} = \: \dfrac{9}{ {4x}^{2} } - 2 + \: \dfrac{ {4x}^{2} }{9} \\ \\ \end{gathered}
⟹(
3
2x
−
2x
3
)
2
=
4x
2
9
−2+
9
4x
2
\rule{190pt}{2pt}
Formulae Used :-
\begin{gathered}\boxed{ \sf{ \: {(x + y)}^{n} = \: ^nC_{0} {x}^{0} {y}^{n} + \: ^nC_{1} {x}^{1} {y}^{n - 1} + \: ^nC_{2} {x}^{2} {y}^{n - 2} + - - - + \: ^nC_{n} {x}^{n} {y}^{0} \: }} \\ \\ \end{gathered}
(x+y)
n
=
n
C
0
x
0
y
n
+
n
C
1
x
1
y
n−1
+
n
C
2
x
2
y
n−2
+−−−+
n
C
n
x
n
y
0
\begin{gathered}\boxed{ \sf{ \:^nC_{0} \: = \: ^nC_{n} \: = \: 1 \: }} \\ \\ \end{gathered}
n
C
0
=
n
C
n
=1
\begin{gathered}\boxed{ \sf{ \:^nC_{1} \: = \: ^nC_{n - 1} \: = \: \frac{n(n - 1)}{2} \: }} \\ \\ \end{gathered}
n
C
1
=
n
C
n−1
=
2
n(n−1)
\rule{190pt}{2pt}
{{ \mathfrak{Additional\:Information}}}AdditionalInformation
General term in the expansion of (x + y)^n(x+y)
n
is
\begin{gathered}\boxed{ \sf{ \: t_{r + 1} \: = \: ^nC_{r} \: {x}^{n - r} \: {y}^{r} \: }} \\ \\ \end{gathered}
t
r+1
=
n
C
r
x
n−r
y
r
General term in the expansion of (x + y)^n(x+y)
n
from the end is
\begin{gathered}\boxed{ \sf{ \: t_{r + 1} \: = \: ^nC_{r} \: {x}^{n} \: {y}^{n - r} \: }} \\ \\ \end{gathered}
t
r+1
=
n
C
r
x
n
y
n−r