1. √(1/3)-√(1/27)
2. 1/√5+√(1/25)
3. √(3 )+2+1/(√3-2 )
Can this be solved WITHOUT RATIONALISING THE DENOMINATOR????
1.
√(1/3)-√(1/27)
√(1/3) -(√(1/3))/3
(√3)/3 - (√3)/9
(2/9)(√3)
2.
1/√5 + √(1/25)
(√5)/5 + 1/5
(1/5) + (1/5)(√5)
3.
√3 + 2 + 1/(√3 - 2)
√3 + 2 +(√3 + 2)/[(√3 - 2)(√3 + 2)]
√3 + 2 + (√3 + 2)/-1
√3 + 2 - √3 - 2
0
EDIT:
Yes it can
√3 + 2 + 1/[√3 - 2]
[(√3 + 2)(√3 - 2) + 1]/[√3 - 2]
[3 + 2√3 - 2√3 - 4 + 1]/[√3 - 2]
0/[√3 - 2]
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Answers & Comments
Verified answer
1.
√(1/3)-√(1/27)
√(1/3) -(√(1/3))/3
(√3)/3 - (√3)/9
(2/9)(√3)
2.
1/√5 + √(1/25)
(√5)/5 + 1/5
(1/5) + (1/5)(√5)
3.
√3 + 2 + 1/(√3 - 2)
√3 + 2 +(√3 + 2)/[(√3 - 2)(√3 + 2)]
√3 + 2 + (√3 + 2)/-1
√3 + 2 - √3 - 2
0
EDIT:
Yes it can
√3 + 2 + 1/[√3 - 2]
[(√3 + 2)(√3 - 2) + 1]/[√3 - 2]
[3 + 2√3 - 2√3 - 4 + 1]/[√3 - 2]
0/[√3 - 2]
0