Math, asked by manjushahanwate9678, 2 months ago

if 15a²+4b²÷15a²-4b² =47÷7 then find the value of b³-2a³÷b³+2a³​

Answers

Answered by navneetsohal80
0

Answer:

7a+3b

7a−3b

=

6

1

; if a>0

7a+3b

7a−3b

=6 ; if a<0

Step-by-step explanation:

Given:

\rm \dfrac{15a^2+4b^2}{15a^2-4b^2}=\dfrac {47}{7}

15a

2

−4b

2

15a

2

+4b

2

=

7

47

The Componendo and Dividendo theorem states that if there is a fraction such as,

\rm \dfrac {N^m}{D^m}=\dfrac pq

D

m

N

m

=

q

p

,

then,

\rm \dfrac{N^m+D^m}{N^m-D^m}=\dfrac{p+q}{p-q}

N

m

−D

m

N

m

+D

m

=

p−q

p+q

.

Applying this rule to the given equation by putting,

\begin{gathered}\rm N^m= 15a^2+4b^2\\D^m=15a^2-4b^2\\p=47\\q=7\end{gathered}

N

m

=15a

2

+4b

2

D

m

=15a

2

−4b

2

p=47

q=7

we get,

\begin{gathered}\rm \dfrac{(15a^2+4b^2)+(15a^2-4b^2)}{(15a^2+4b^2)-(15a^2-4b^2)}=\dfrac{47+7}{47-7}\\\dfrac{15a^2+15a^2+4b^2-4b^2}{15a^2-15a^2+4b^2-(-4b^2)}=\dfrac{54}{40}\\\dfrac{30a^2}{8b^2} =\dfrac{27}{20}\\\dfrac{a^2}{b^2} = \dfrac {8}{30} \times \dfrac{27}{20} = \dfrac {9}{25}\\\Rightarrow \dfrac ab=\pm \sqrt{\dfrac{9}{25}} = \pm \dfrac 35\end{gathered}

(15a

2

+4b

2

)−(15a

2

−4b

2

)

(15a

2

+4b

2

)+(15a

2

−4b

2

)

=

47−7

47+7

15a

2

−15a

2

+4b

2

−(−4b

2

)

15a

2

+15a

2

+4b

2

−4b

2

=

40

54

8b

2

30a

2

=

20

27

b

2

a

2

=

30

8

×

20

27

=

25

9

b

a

25

9

5

3

Therefore,

\rm a = \pm \dfrac 35 ba=±

5

3

b

Taking positive value of a.

\begin{gathered}\rm a=\dfrac 35 b\\\\\dfrac{7a-3b}{7a+3b}=\dfrac{7\left( \dfrac 35 b\right) -3b}{7\left( \dfrac 35 b\right) +3b}\\=\dfrac{\dfrac {21b}{5}-3b}{\dfrac {21b}{5} +3b}\\=\dfrac{\dfrac{21b-15b}{5}}{\dfrac{21b+15b}{5}}\\=\dfrac{6b}{36b}\\=\dfrac 16\end{gathered}

a=

5

3

b

7a+3b

7a−3b

=

7(

5

3

b)+3b

7(

5

3

b)−3b

=

5

21b

+3b

5

21b

−3b

=

5

21b+15b

5

21b−15b

=

36b

6b

=

6

1

Taking negative value of a.

\begin{gathered}\rm a=-\dfrac 35 b\\\\\dfrac{7a-3b}{7a+3b}=\dfrac{7\left( -\dfrac 35 b\right) -3b}{7\left( -\dfrac 35 b\right) +3b}\\=\dfrac{-\dfrac {21b}{5}-3b}{-\dfrac {21b}{5} +3b}\\=\dfrac{\dfrac{-21b-15b}{5}}{\dfrac{-21b+15b}{5}}\\=\dfrac{-36b}{-6b}\\=6.\end{gathered}

a=−

5

3

b

7a+3b

7a−3b

=

7(−

5

3

b)+3b

7(−

5

3

b)−3b

=

5

21b

+3b

5

21b

−3b

=

5

−21b+15b

5

−21b−15b

=

−6b

−36b

=6.

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