If lambda=(n_(1))/(n_(2))=(m_(1))/(m_(2)) prove that lambda=(n_(1)+-m_(1))/(n_2+-m_2) I know this is true if I add numbers to it 1/2=2/4 *(1+2)/(2+4)=3/6=1/2 (1−2)/(2−4)=(−1)/(−2)=1/2

Paxton Hoffman 2022-07-21 Answered
Prove equivalent fractions with ± sign
If
λ = n 1 n 2 = m 1 m 2
prove that
λ = n 1 ± m 1 n 2 ± m 2
I know this is true if I add numbers to it
1 2 = 2 4
1 + 2 2 + 4 = 3 6 = 1 2
1 2 2 4 = 1 2 = 1 2
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Answers (2)

eishale2n
Answered 2022-07-22 Author has 15 answers
Compute the difference: From n 1 n 2 = m 1 m 2 , we know tha tthe numerator of
n 1 n 2 m 1 m 2 = n 1 m 2 n 2 m 1 n 2 m 2
is zero. Hence
n 1 ± m 1 n 2 ± m 2 n 1 n 2 = ( n 1 ± m 1 ) n 2 n 1 ( n 2 ± m 2 ) ( n 2 ± m 2 ) m 1 = n 1 n 2 ± m 1 n 2 n 1 n 2 n 1 m 2 ( n 2 ± m 2 ) m 1 = ± ( n 1 m 2 n 2 m 1 ) ( n 2 ± m 2 ) m 1 = 0
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Baladdaa9
Answered 2022-07-23 Author has 2 answers
λ = n 1 n 2 = m 1 m 2 n 1 = λ . n 2 m 1 = λ . m 2
so
n 1 + m 1 n 2 + m 2 = ( λ . n 2 ) + ( λ . m 2 ) n 2 + m 2 = λ ( n 2 ) + ( m 2 ) n 2 + m 2 = λ .1 = λ
also for
n 1 m 1 n 2 m 2 = ( λ . n 2 ) ( λ . m 2 ) n 2 m 2 = λ ( n 2 ) ( m 2 ) n 2 m 2 = λ .1 = λ
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Prove that ( 2 a + b + c ) 2 2 a 2 + ( b + c ) 2 + ( 2 b + c + a ) 2 2 b 2 + ( c + a ) 2 + ( 2 c + a + b ) 2 2 c 2 + ( a + b ) 2 8
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