Heyy
Please help me to solve this problem.... by brief explaining no shortcut method is allowed!
Answers
Step-by-step explanation:
11) For this one, we want numbers between 5/7 and 9/11, so
5/7 < x < 9/11
Squaring, we need
5² / 7² < x² < 9² / 11²
Might be safer if we work with numbers over a common denominator, which where would be 7² 11² = 77². So just rewriting the numbers, we need
55² / 77² = 3025 / 77² < x² < 3969 / 77² = 63² / 77²
Now as long as we choose x² in this range so that it's not a perfect square, we will then have a suitable irrational value of x by just taking the square root.
Three non-square numbers between 3025 and 3969 are 3026, 3027 and 3028, so they will do the job.
Taking 3026 / 77² , 3027 / 77² , and 3028 / 77² as our three values for x², we then have three irrational numbers between 5/7 and 9/11 as
√3026 / 77 , √3027 / 77 , and √3028 / 77.
12) 8^(1/3) = ∛8 = 2 and 27^(1/3) = ∛27 = 3
So
[ 5 { 8^(1/3) + 27^(1/3) }³ ]^(1/4)
= [ 5 { 2 + 3 }³ ]^(1/4)
= [ 5 × 5³ ]^(1/4)
= [ 5⁴ ]^(1/4)
= 5
13) ( a/4 - b/2 + 1 )²
= (a/4) ( a/4 - b/2 + 1 ) - b/2 ( a/4 - b/2 + 1 ) + ( a/4 - b/2 + 1 )
= (a/4)² - ab/8 + a/4 - ab/8 + (b/2)² - b/2 + a/4 - b/2 + 1
= a²/16 + b²/4 + 1 - ab/4 + a/2 - b