rationalize 2+√3√3-√2
Answers
Answer:
Consider √2 + √3 as a single term and √4 or 2 as another term
So
1 ÷ [ (√2+√3)-(2) ]
Now do what we normally do for 1÷ a-b
We multiply it by a+b in both numerator and denominator
Do the same here
Multiply (√2+√3)+2 on both numerator and denominator
{ 1×[(√2+√3)+2] } ÷ [(√2+√3)-2]×[(√2+√3)+2]
As we know that (a+b)×(a-b) = a²-b²
= (√2+√3+2) ÷ (√2+√3)²-2²
We know that (a+b)² = a² + b² +2ab so
= (√2+√3+2) ÷ (√2)²+(√3)²+2×√2×√3- 4
= (√2+√3+2) ÷ 2 + 3+ 2√6 -4
= (√2+√3+2) ÷ 5–4 + 2√6
= (√2 +√3 + 2) ÷ 1+2√6
Now rationalise this by multiplying 1–2√6 on both numerator and denominator
=[(√2+√3+2)× (1-2√6)] ÷ [(1+2√6)×(1–2√6)]
=[(√2+√3+2)×(1–2√6)]÷[1²-(2√6)²]
As (2√6)²=[2²×(√6)²]=(4×6)=24
=[(√2+√3+2)×(1–2√6)]÷[1–24]
=[(√2+√3+2)×(1–2√6)]÷(-23)
Now take - (minus) sign upwards
= — [(√2+√3+2)×(1–2√6)] ÷23
You can further simplyfy the numerator and do not forget the minus sign infrint of it
Step-by-step explanation:
hope it helps
Answer:
Consider √2 + √3 as a single term and √4 or 2 as another term
So
1 ÷ [ (√2+√3)-(2) ]
Now do what we normally do for 1÷ a-b
We multiply it by a+b in both numerator and denominator
Do the same here
Multiply (√2+√3)+2 on both numerator and denominator
{ 1×[(√2+√3)+2] } ÷ [(√2+√3)-2]×[(√2+√3)+2]
As we know that (a+b)×(a-b) = a²-b²
= (√2+√3+2) ÷ (√2+√3)²-2²
We know that (a+b)² = a² + b² +2ab so
= (√2+√3+2) ÷ (√2)²+(√3)²+2×√2×√3- 4
= (√2+√3+2) ÷ 2 + 3+ 2√6 -4
= (√2+√3+2) ÷ 5–4 + 2√6
= (√2 +√3 + 2) ÷ 1+2√6
Now rationalise this by multiplying 1–2√6 on both numerator and denominator
=[(√2+√3+2)× (1-2√6)] ÷ [(1+2√6)×(1–2√6)]
=[(√2+√3+2)×(1–2√6)]÷[1²-(2√6)²]
As (2√6)²=[2²×(√6)²]=(4×6)=24
=[(√2+√3+2)×(1–2√6)]÷[1–24]
=[(√2+√3+2)×(1–2√6)]÷(-23)
Now take - (minus) sign upwards
= — [(√2+√3+2)×(1–2√6)] ÷23
You can further simplyfy the numerator and do not forget the minus sign infrint of it
Step-by-step explanation:
hope it helps
Step-by-step explanation:
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