[1+1/x+1] [1-1/x-1]=7/8 using factorisation method
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Answer:
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Step-by-step explanation:
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\begin{gathered}(1 + \frac{1}{x + 1} ) \times (1 - \frac{1}{x - 1} ) = \frac{7}{8} \\ \\ \frac{x + 1 + 1}{x + 1} \times \frac{x - 1 - 1}{x - 1} = \frac{7}{8} \\ \\ \frac{x + 2}{x + 1} \times \frac{x - 2}{x - 1} = \frac{7}{8} \\ \\ \frac{{x}^{2} - 2 {}^{2} }{ {x}^{2} - {1}^{2} } = \frac{7}{8} \\ \\ \frac{ {x}^{2} - 4}{ {x}^{2} - 1} = \frac{7}{8} \\ \\ = > 8( {x}^{2} - 4) = 7( {x}^{2} - 1) \\ \\ = > 8 {x}^{2} - 32 = {7}x^{2} - 7 \\ \\ = > 8 {x}^{2} - 7 {x}^{2} - 32 + 7 = 0 \\ \\ = > {x}^{2} - 25 = 0 \\ \\ = > {x}^{2} - {5}^{2} = 0 \\ \\ = > (x + 5)(x - 5) = 0 \\ \\ = > x = - 5 \: \: or \: \: x = 5\end{gathered}
(1+
x+1
1
)×(1−
x−1
1
)=
8
7
x+1
x+1+1
×
x−1
x−1−1
=
8
7
x+1
x+2
×
x−1
x−2
=
8
7
x
2
−1
2
x
2
−2
2
=
8
7
x
2
−1
x
2
−4
=
8
7
=>8(x
2
−4)=7(x
2
−1)
=>8x
2
−32=7x
2
−7
=>8x
2
−7x
2
−32+7=0
=>x
2
−25=0
=>x
2
−5
2
=0
=>(x+5)(x−5)=0
=>x=−5orx=5