can somebody pls tell me the answer i will mrk brainliest!!
1.3z-1/2z+4=1 2.+1 /4− /8= 1
Answers
Answer:
1 afcourse hi hai na tu ofAnswer:
10
Step-by-step explanation:
As per the provided question, we have to solve for :
\begin{gathered} \longmapsto \bf { \dfrac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} + \dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}} } \\ \end{gathered}
⟼
3
2
+2
3
3
2
−2
3
+
3
2
−2
3
3
2
+2
3
Here, we have :
\begin{gathered} \odot \sf { Term \; 1 = \dfrac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} } \\ \end{gathered}
⊙Term1=
3
2
+2
3
3
2
−2
3
\begin{gathered} \odot \sf { Term \; 2 = \dfrac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}} } \\ \end{gathered}
⊙Term2=
3
2
−2
3
3
2
+2
3
Basically, we've to rationalize the denominator of both terms in order to find the required answer.
In order to rationalise the denominator, we multiply the rationalising factor of the denominator with both the numerator and the denominator.
Rationalising the denominator of term 1 :
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} }} \\ \end{gathered}
⟶
3
2
+2
3
3
2
−2
3
Here, the denominator is in the form of (a + b), so its rationalising factor will be (a – b). Thus, here the rationalising factor is (3√2 – 2√3).
Multiplying (3√2 – 2√3) with both the numerator and the denominator.
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} \times \dfrac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}} }} \\ \end{gathered}
⟶
3
2
+2
3
3
2
−2
3
×
3
2
−2
3
3
2
−2
3
Rearranging the terms.
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }{(3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3})} }} \\ \end{gathered}
⟶
(3
2
+2
3
)(3
2
−2
3
)
(3
2
−2
3
)(3
2
−2
3
)
This expression can be further written as,
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{(3\sqrt{2}-2\sqrt{3})^2 }{(3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3})} }} \\ \end{gathered}
⟶
(3
2
+2
3
)(3
2
−2
3
)
(3
2
−2
3
)
2
Now, by using the following identities,
⠀⠀⠀★ (a – b)² = a² + b² – 2ab
⠀⠀⠀★ (a + b)(a – b) = a² – b²
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{(3\sqrt{2})^2 + (2\sqrt{3})^2 - 2(3\sqrt{2} \times 2\sqrt{3}) }{(3\sqrt{2})^2-(2\sqrt{3})^2} }} \\ \end{gathered}
⟶
(3
2
)
2
−(2
3
)
2
(3
2
)
2
+(2
3
)
2
−2(3
2
×2
3
)
Simplifying further.
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{18 + 12 - 2(6\sqrt{6}) }{18-12} }} \\ \end{gathered}
⟶
18−12
18+12−2(6
6
)
Performing addition and multiplication in numerator and subtraction in denominator.
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{30- 12\sqrt{6} }{6} }} \\ \end{gathered}
⟶
6
30−12
6
Now, we can write is as ,
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{30}{6} - \dfrac{12\sqrt{6} }{6} }} \\ \end{gathered}
⟶
6
30
−
6
12
6
Performing division.
\begin{gathered} \longrightarrow \quad \underline{\boxed{\sf { 5 - 2\sqrt{6} }}} \\ \end{gathered}
⟶
5−2
6
Now, we have to rationalise the denominator of the second term.
Rationalising the denominator of term 2 :
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}} }} \\ \end{gathered}
⟶
3
2
−2
3
3
2
+2
3
Here, the denominator is in the form of (a – b), so its rationalising factor will be (a + b). Thus, here the rationalising factor is (3√2 + 2√3).
Multiplying (3√2 + 2√3) with both the numerator and the denominator.
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}} \times \dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} }} \\ \end{gathered}
⟶
3
2
−2
3
3
2
+2
3
×
3
2
+2
3
3
2
+2
3
Rearranging the terms.
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{(3\sqrt{2}+2\sqrt{3})(3\sqrt{2}+2\sqrt{3}) }{(3\sqrt{2}+2\sqrt{3})(3\sqrt{2}+2\sqrt{3})} }} \\ \end{gathered}
⟶
(3
2
+2
3
)(3
2
+2
3
)
(3
2
+2
3
)(3
2
+2
3
)
This expression can be further written as,
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{(3\sqrt{2}+2\sqrt{3})^2 }{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})} }} \\ \end{gathered}
⟶
(3
2
−2
3
)(3
2
+2
3
)
(3
2
+2
3
)
2
Now, by using the following identities,
⠀⠀⠀★ (a + b)² = a² + b² + 2ab
⠀⠀⠀★ (a + b)(a – b) = a² – b²
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{(3\sqrt{2})^2 + (2\sqrt{3})^2 + 2(3\sqrt{2} \times 2\sqrt{3}) }{(3\sqrt{2})^2-(2\sqrt{3})^2} }} \\ \end{gathered}
⟶
(3
2
)
2
−(2
3
)
2
(3
2
)
2
+(2
3
)
2
+2(3
2
×2
3
)
Simplifying further.
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{18 + 12 + 2(6\sqrt{6}) }{18-12} }} \\ \end{gathered}
⟶
18−12
18+12+2(6
6
)
Performing addition and multiplication in numerator and subtraction in denominator.
\begin{gathered} \longrightarrow \sf{\quad { \dfrac{30+ 12\sqrt{6} }{6} }} \\ \end{gathered}
⟶
6
30+12
6
i