Solve these four questions
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HELLO DEAR,
★ ( 1 ) ★
x = 1/(2 - √3)
x = (2 + √3)/(4 - 3(
x = (2 + √3)
now,
x² - 4x + 1
(2 + √3)² - 4(2 + √3) + 1
4 + 3 + 4√3 - 8 - 4√3 + 1
7 - 7 = 0
★ ( 2 ) ★
[ (3ⁿ + 3^(ⁿ - 1) ] / [ 3^(ⁿ + 1) - 3ⁿ ]
= [ (3ⁿ + 3ⁿ * 1/3 ] / [ 3ⁿ * 3 - 3ⁿ ]
= 3ⁿ [ 1 + 1/3 ] / 3ⁿ [ 3 - 1 ]
= (3 + 1)/3 / 2
= 4/3/2
= 2/3
★ ( 3 ) ★
plz open the link for your answer
https://brainly.in/question/3576525
★ ( 4 ) ★
x^a = y------( 1 )
y^ b = z------( 2 )
z^c = x--------( 3 )
from----( 1 ) , ----( 2 ) & -----( 3 )
x^a * y^b * z^c = xyz
on comparing both side,
we get,
a = 1 , b = 1 , c = 1
hence, abc = (1)(1)(1) = 1
I HOPE ITS HELP YOU DEAR,
THANKS
★ ( 1 ) ★
x = 1/(2 - √3)
x = (2 + √3)/(4 - 3(
x = (2 + √3)
now,
x² - 4x + 1
(2 + √3)² - 4(2 + √3) + 1
4 + 3 + 4√3 - 8 - 4√3 + 1
7 - 7 = 0
★ ( 2 ) ★
[ (3ⁿ + 3^(ⁿ - 1) ] / [ 3^(ⁿ + 1) - 3ⁿ ]
= [ (3ⁿ + 3ⁿ * 1/3 ] / [ 3ⁿ * 3 - 3ⁿ ]
= 3ⁿ [ 1 + 1/3 ] / 3ⁿ [ 3 - 1 ]
= (3 + 1)/3 / 2
= 4/3/2
= 2/3
★ ( 3 ) ★
plz open the link for your answer
https://brainly.in/question/3576525
★ ( 4 ) ★
x^a = y------( 1 )
y^ b = z------( 2 )
z^c = x--------( 3 )
from----( 1 ) , ----( 2 ) & -----( 3 )
x^a * y^b * z^c = xyz
on comparing both side,
we get,
a = 1 , b = 1 , c = 1
hence, abc = (1)(1)(1) = 1
I HOPE ITS HELP YOU DEAR,
THANKS
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