Question

Can you please solve this?!?​

Answer 1

$\bold{\huge{\blue{\underline{ Solution }}}}$

Given :-

Here, We have,

• $\sf{\dfrac{a}{b}}{\sf{=}}{\sf{\dfrac{b}{c}}}{\sf{=}}{\sf{\dfrac{c}{d}}}$

To Prove :-

• We have to prove that √(a - b - c)(b-c-d) = √ab-√bc-√cd

Let's Begin :-

$\sf{√(a-b-c)(b-c-d)=√ab-√bc-√cd}$

• Let $\sf{\dfrac{a}{b}}{\sf{=}}{\sf{\dfrac{b}{c}}}{\sf{=}}{\sf{\dfrac{c}{d}}}{\sf{ = k}}$

Therefore,

From above, We can conclude that,

a = bk, b = ck, c = dk

a = dk³ , b = dk² , c = dk

By solving RHS :-

$\sf{√(a - b - c)(b-c-d) }$

$\sf{√(dk³ - dk² - dk) (dk² - dk - d) }$

$\sf{√d(k³- k² - k) d(k² - k - 1)}$

$\sf{d√(k³- k² - k) d(k² - k - 1)}$

$\sf{d√k(k²- k - 1)(k² - k - 1)}$

$\sf{d(k²- k - 1)√k ...eq(1)}$

By solving LHS :-

$\sf{√ab - √bc - √cd}$

$\sf{√dk³(dk²) - √dk²(dk)-√dk(d)}$

$\sf{√d²k^5 - √ d²k³ - √d²k}$

$\sf{dk²√k - dk√k - d√k}$

$\sf{(dk² - dk - d)√k}$

$\sf{d(k²- k - 1)√k...eq(2)}$

From eq(1) and eq(2)

$\bold{\red{ LHS = RHS }}$

That is,

$\sf{√(a-b-c)(b-c-d)=√ab-√bc-√cd}$

Hence proved .

Answer 2

Yes, the RHS and the LHS are equal/

Calculations and Parameters:

Given:

$a/b= b/c = c/d$

We have to prove that √(a - b - c)(b-c-d) = √ab-√bc-√cd

Hence, after we expand the brackets and evaluate, we see that:

• a = bk, b = ck, c = dk
• a = dk³ , b = dk² , c = dk

After the evaluation of the RHS and LHS, we can see that

RHS= LHS .

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