Question

please solve the que​

Answer 1

QUESTION:

Prove that:

$\sf \dfrac{{ 3}^{ - 3} \times {6}^{2} \times \sqrt{98 }}{ {5}^{2} \times \sqrt[3]{ \dfrac{1}{25}} \times {15}^{ - 4/ 3} \times {3}^{1 /3}} = 28 \sqrt{2}$

GIVEN AND TO PROVE:

$\sf \dfrac{{ 3}^{ - 3} \times {6}^{2} \times \sqrt{98 }}{ {5}^{2} \times \sqrt[3]{ \dfrac{1}{25}} \times {15}^{ - 4/ 3} \times {3}^{1 /3}} = 28 \sqrt{2}$

EXPLANATION:

$\sf Let \ { 3}^{ - 3} \times {6}^{2} \times \sqrt{98 } = x$

$\sf Let \ {5}^{2} \times \sqrt[3]{\dfrac{1}{25}} \times {15}^{ - 4/ 3} \times {3}^{1 /3}=y$

$\sf Then \ \dfrac{{ 3}^{ - 3} \times {6}^{2} \times \sqrt{98 }}{ {5}^{2} \times \sqrt[3]{\dfrac{1}{25}} \times {15}^{ - 4/ 3} \times {3}^{1 /3}} = \dfrac{x}{y}$

$\sf \leadsto x = \dfrac{1}{ 3^ 3} \times {6}^{2} \times \sqrt{2 \times 49 }$

$\sf \leadsto x = \dfrac{1}{ 9 \times 3} \times 36 \times7 \sqrt{2}$

$\sf \leadsto x = \dfrac{1}{ 3} \times 4\times7 \sqrt{2}$

$\sf \leadsto x = \dfrac{28}{ 3} \sqrt{2}$

$\sf \leadsto y = {5}^{2} \times \sqrt[3]{ \dfrac{1}{25}} \times \sqrt[3]{ \dfrac{1}{ {15}^{4} }} \times\sqrt[3]{ 3}$

$\sf \leadsto y = \sqrt[3]{ (25) {}^{3}} \times \sqrt[3]{ \dfrac{1}{25}} \times \sqrt[3]{ \dfrac{1}{ {15}^{4} }} \times\sqrt[3]{ 3}$

$\sf \leadsto y = \sqrt[3]{ \dfrac{25 \times 25 \times 25 \times 3}{25 \times 15 \times 15 \times 15 \times 15}}$

$\sf \leadsto y = \sqrt[3]{ \dfrac{25 \times 25}{5 \times 5 \times 3 \times 15 \times 15}}$

$\sf \leadsto y = \sqrt[3]{ \dfrac{25 }{ 3 \times 5 \times 3 \times 5 \times 3}}$

$\sf \leadsto y = \sqrt[3]{ \dfrac{1 }{ 3 \times 3 \times 3}}$

$\sf \leadsto y = \dfrac{1 }{ 3}$

$\sf \sf \leadsto \dfrac{x}{y} = \dfrac{\dfrac{28 \sqrt{2} }{ 3}} {\dfrac{1 }{ 3}}$

$\sf \sf \leadsto \dfrac{x}{y} = 28 \sqrt{2}$

$\sf \dfrac{{ 3}^{ - 3} \times {6}^{2} \times \sqrt{98 }}{ {5}^{2} \times \sqrt[3]{ \dfrac{1}{25}} \times {15}^{ - 4/ 3} \times {3}^{1 /3}} = 28 \sqrt{2}$

HENCE PROVED.

Answer 2

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