Question
Download Solution PDFx3 + y3 = 22 மற்றும் x + y = 5 எனில் x4 + y4. இன் தோராயமான மதிப்பைக் கண்டறியவும்.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFநமக்குத் தெரியும்,
x3 + y3 = (x + y)(x2 + y2 – xy)
இப்போது நம்மிடம் இருக்கிறது x3 + y3 = 22 and x + y = 5
⇒ 22 = 5(x2 + y2 – xy)
⇒ 22 = 5[(x + y)2 − 3xy)]
⇒ 22 = 5[(5)2 − 3xy)]
⇒ xy = 103/15
இப்போது x3 + y3 = 22 ஐ x + y = 5 உடன் பெருக்கவும்
⇒ x4 + y4 + xy(x2 + y2) = 110
⇒ x4 + y4 = 110 – xy{(x2 + y2 − 2xy + 2xy)}
⇒ x4 + y4 = 110 – xy{(x + y)2 − 2xy}
xy = 103/15 and x + y = 5
⇒ x4 + y4 = 110 – 103/15{(5)2 − 2 × 103/15}
⇒ x4 + y4 = 110 – 6.87{(25 – 13.73}
⇒ x4 + y4 = 110 – 6.87 {(11.27)}
⇒ x4 + y4 = 110 – 77.42
⇒ x4 + y4 = 32.58
∴ Value of x4 + y4 is 33.
Shortcut Trick
x3 + y3 = (x + y)(x2 + y2 – xy)
⇒ 22 = 5(x2 + y2 – xy)
⇒ 22 = 5[(x + y)2 − 3xy)]
⇒ 22 = 5[(5)2 − 3xy)]
⇒ xy = 103/15
(x3 + y3) (x + y) = x4 + y4 + xy(x2 + y2)
(x3 + y3) (x + y)= (x4 + y4) + {xy[(x + y)2 – 2xy)]
⇒ 22 × 5 = x4 + y4 + 103/15[25 - 206/15]
⇒ x4 + y4 = 32.63 ≈ 33
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