Question

solve the following equations. (2x+1)²+(2x-1)²/(2x+1)²-(2x-1)²=17/8​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

x=2 and x=1/8

Step-by-step explanation:

$\frac{( {2x + 1})^{2} + ( {2x - 1})^{2} }{ ({2x + 1})^{2} - ({2x - 1})^{2} } = \frac{17}{8} \\ 8( ({2x + 1})^{2} + ({2x - 1})^{2} ) = 17(( {2x + 1})^{2} - ({2x - 1})^{2} ) \\ 8 ({2x + 1})^{2} + 8 ({2x - 1})^{2} = 17 ({2x + 1})^{2} - 17 ({2x - 1})^{2} \\ 8 ({(2x)}^{2} + 2(2x)(1) + {1}^{2} ) + 8( {(2x})^{2} - 2(2x)(1) + {1}^{2} ) = 17( ({2x})^{2} + 2(2x)(1) + {1}^{2} ) - 17( ({2x})^{2} - 2(2x)(1) + {1}^{2} ) \\ 8(4 {x}^{2} + 4x + 1) + 8( 4{x}^{2} - 4x + 1) = 17(4 {x}^{2} + 4x + 1) - 17(4 {x}^{2} - 4x + 1) \\ 32 {x}^{2} + 32x + 8 + 32 {x}^{2} - 32x + 8 = 68 {x}^{2} + 68x + 17 - 68 {x}^{2} + 68x - 17 \\ 32 {x}^{2} + 32 {x}^{2} + 32x - 32x + 8 + 8 = 68 {x}^{2} - 68 {x}^{2} + 68x + 68x + 17 - 17 \\ 64 {x}^{2} + 16 = 136x \\ 64 {x}^{2} - 136x + 16 = 0$

$64 {x}^{2} - 136x + 16 = 0 \\ 64 {x}^{2} -128x - 8x + 16 = 0 \\ 64x(x - 2) - 8(x - 2) = 0 \\ (64x - 8)(x - 2) = 0 \\ 64x - 8 = 0 \: \: \: x - 2 = 0 \\ 64x = 8 \: \: \: x = 2 \\ x = \frac{8}{64} \: \: \: x = 2 \\ x = \frac{1}{8} \: \: \: x = 2$