Melody08idn
Answered

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please help with this question and give detailed explanation!! thank you!!

Please Help With This Question And Give Detailed Explanation Thank You class=

Sagot :

Mhanifaidn

Answer:

  • [tex]-\sqrt{2}[/tex]

Step-by-step explanation:

Make the following operations:

  • a² + b² = 6ab
  • a² + 2ab + b² = 8ab
  • (a + b)² = 8ab
  • a + b = [tex]2\sqrt{2ab}[/tex]

and

  • a² + b² = 6ab
  • a² - 2ab + b² = 4ab
  • (a - b)² = 4ab
  • a - b = [tex]-2\sqrt{ab}[/tex], since a < b

The required value is:

  • [tex]\cfrac{a+b}{a-b} =\cfrac{2\sqrt{2ab} }{-2\sqrt{ab} } =-\sqrt{2}[/tex]
Semsee45idn

Answer:

[tex]\dfrac{a+b}{a-b}=-\sqrt{2}[/tex]

Step-by-step explanation:

Given:

[tex]a^2+b^2=6ab[/tex]

[tex]0 < a < b[/tex]

Add 2ab to both sides of the given equation:

[tex]\implies a^2+b^2+2ab=6ab+2ab[/tex]

[tex]\implies a^2+2ab+b^2=8ab[/tex]

Factor the left side:

[tex]\implies (a+b)^2=8ab[/tex]

Subtract 2ab from both sides of the given equation:

[tex]\implies a^2+b^2-2ab=6ab-2ab[/tex]

[tex]\implies a^2-2ab+b^2=4ab[/tex]

Factor the left side:

[tex]\implies (a-b)^2=4ab[/tex]

Therefore:

[tex]\implies \dfrac{(a+b)^2}{(a-b)^2}=\dfrac{8ab}{4ab}[/tex]

[tex]\implies \dfrac{(a+b)^2}{(a-b)^2}=2[/tex]

[tex]\textsf{Apply exponent rule} \quad \dfrac{a^c}{b^c}=\left(\dfrac{a}{b}\right)^c:[/tex]

[tex]\implies \left(\dfrac{a+b}{a-b}\right)^2=2[/tex]

Square root both sides:

[tex]\implies \sqrt{\left(\dfrac{a+b}{a-b}\right)^2}=\sqrt{2}[/tex]

[tex]\implies \dfrac{a+b}{a-b}=\pm\sqrt{2}[/tex]

As 0 < a < b then:

  • a + b > 0
  • a - b < 0

Therefore:

[tex]\implies \dfrac{a+b}{a-b}=\dfrac{+}{-}=-[/tex]

So:

[tex]\implies \dfrac{a+b}{a-b}=-\sqrt{2}[/tex]

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