plz solve this question need it urgent
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x = 1/(√2 +1)
rationalizing it,
x= 1×(√2 -1)/[(√2+1)×(√2-1)
= (√2-1)/(2-1) = (√2 -1)
now, expression is x³+2x² -x +1
putting the value,
(√2 -1)³ +2 (√2 -1)² - (√2 -1) +1
=2√2 -1 -3√2(√2 -1) + 2(2+1- 2√2) - √2+1 +1
=2√2 -1 -6+3√2 +6-4√2 - √2 +2
=1
so, the value will be 1
rationalizing it,
x= 1×(√2 -1)/[(√2+1)×(√2-1)
= (√2-1)/(2-1) = (√2 -1)
now, expression is x³+2x² -x +1
putting the value,
(√2 -1)³ +2 (√2 -1)² - (√2 -1) +1
=2√2 -1 -3√2(√2 -1) + 2(2+1- 2√2) - √2+1 +1
=2√2 -1 -6+3√2 +6-4√2 - √2 +2
=1
so, the value will be 1
qais:
see now, it's 1
yup
my pleasure
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Answered by
1
Firstly we have to rationalise the denominator
We have 1
√2+1
= 1 × √2+1
√2+1 √2-1
=(√2-1)
2 - 1
=(√2-1)
So x= (√2-1)
The given expression is x³+2x² -x +1
By implying value of x=√2-1,we get
(√2 -1)³ +2 (√2 -1)² - (√2 -1) +1
=2√2 -1 -3√2(√2 -1) + 2(2+1- 2√2) - √2+1 +1
=2√2 -1 -6+3√2 +6-4√2 - √2 +2
=1
We have 1
√2+1
= 1 × √2+1
√2+1 √2-1
=(√2-1)
2 - 1
=(√2-1)
So x= (√2-1)
The given expression is x³+2x² -x +1
By implying value of x=√2-1,we get
(√2 -1)³ +2 (√2 -1)² - (√2 -1) +1
=2√2 -1 -3√2(√2 -1) + 2(2+1- 2√2) - √2+1 +1
=2√2 -1 -6+3√2 +6-4√2 - √2 +2
=1
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