Pertanyaan

Solve the following system for x dan y . { x 2 1 + y 2 1 = a 2 1 y 2 1 + x y 1 = b 2 1

Solve the following system for $x dan y$ .

${\frac{1}{x ^{2}} + \frac{1}{y ^{2}} = \frac{1}{a ^{2}} \frac{1}{y ^{2}} + \frac{1}{x y} = \frac{1}{b ^{2}}$

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D. Wahyu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

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Jawaban

solusi adalah .

solusi adalah $open parentheses square root of fraction numerator a squared over denominator 1 plus a squared end fraction end root space comma square root of fraction numerator b squared over denominator 1 plus b squared end fraction end root close parentheses space space dan space open parentheses negative square root of fraction numerator straight a squared over denominator 1 plus straight a squared end fraction end root space comma negative square root of fraction numerator straight b squared over denominator 1 plus straight b squared end fraction end root close parentheses space$ .

Pembahasan

Diketahui: Selanjutnya, persamaan (1) = persamaan (2) diperoleh Subtitusikan persamaan (5) ke persamaan (1),sehingga Selanjutnya, subtitusikan persamaan (5) ke persamaan (2) , sehingga Jadi, solusi adalah .

Diketahui:

$1 over x squared plus fraction numerator 1 over denominator x y end fraction equals 1 over a squared space fraction numerator y plus 1 over denominator x y end fraction equals 1 over a squared space left parenthesis Dikali space silang right parenthesis a squared left parenthesis y plus 1 right parenthesis equals x y... left parenthesis 1 right parenthesis 1 over y squared plus fraction numerator 1 over denominator x y end fraction equals 1 over b squared fraction numerator x y plus 1 over denominator x y end fraction equals 1 over b squared b squared space open parentheses x y plus 1 close parentheses equals x y space... left parenthesis 2 right parenthesis$

Selanjutnya, persamaan (1) = persamaan (2) diperoleh

Subtitusikan persamaan (5) ke persamaan (1),sehingga

$1 over x squared plus fraction numerator 1 over denominator left parenthesis negative 1 right parenthesis end fraction equals 1 over a squared 1 over x squared minus 1 equals 1 over a squared 1 over x squared equals 1 over a squared plus 1 1 over x squared equals fraction numerator 1 plus a squared over denominator a squared end fraction a squared equals x squared space left parenthesis 1 plus a squared right parenthesis x squared equals fraction numerator a squared over denominator left parenthesis 1 plus a squared right parenthesis end fraction x equals plus-or-minus square root of fraction numerator a squared over denominator 1 plus a squared end fraction end root x subscript 1 equals square root of fraction numerator a squared over denominator 1 plus a squared end fraction end root comma space space x subscript 2 equals negative square root of fraction numerator a squared over denominator 1 plus a squared end fraction end root$

Selanjutnya, subtitusikan persamaan (5) ke persamaan (2) , sehingga

$1 over y squared plus fraction numerator 1 over denominator left parenthesis negative 1 right parenthesis end fraction equals 1 over b squared 1 over y squared minus 1 equals 1 over b squared 1 over y squared equals 1 over b squared plus 1 1 over y squared equals fraction numerator 1 plus b squared over denominator b squared end fraction b squared equals y squared space left parenthesis 1 plus b squared right parenthesis y squared equals fraction numerator b squared over denominator left parenthesis 1 plus b squared right parenthesis end fraction y equals plus-or-minus square root of fraction numerator b squared over denominator 1 plus b squared end fraction end root y subscript 1 equals square root of fraction numerator b squared over denominator 1 plus b squared end fraction end root comma space y subscript 2 equals negative square root of fraction numerator b squared over denominator 1 plus b squared end fraction end root$

Jadi, solusi adalah $open parentheses square root of fraction numerator a squared over denominator 1 plus a squared end fraction end root space comma square root of fraction numerator b squared over denominator 1 plus b squared end fraction end root close parentheses space space dan space open parentheses negative square root of fraction numerator straight a squared over denominator 1 plus straight a squared end fraction end root space comma negative square root of fraction numerator straight b squared over denominator 1 plus straight b squared end fraction end root close parentheses space$ .

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