Question

find the solution common to both inequalities
$\frac{ ({x - 1})^{3} {( {x}^{2} + 3x + 2 })^{5} |x + 4| }{ {( {x}^{2} + 4x + 4 })^{2} } < 0 \: and \: 1 < |x - 3| < 5$
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Answer 1

Answer:

common solution=(-1,1)✔✔

Step-by-step explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

$\frac{ ({x - 1})^{3} {( {x}^{2} + 3x + 2 })^{5} |x + 4| }{ {( {x}^{2} + 4x + 4 })^{2} } < 0 \\ = ) \frac{ {(x - 1})^{3}( {x + 2})^{2} ({x + 1})^{2} |x + 4| }{ ({ ({x + 2})^{2} })^{7} } \\ NOTE:\: \\THE\:FACTOR \:HAVING \\ODD \:POWER \:CAN\: BE \:TAKEN\\AS \:A\:POWER\:1\:AND\:THE\\FACTORS\:HAVING\\EVEN\:POWER\\CAN\:BE\:EXCLUDED\:FROM\:THE\:QUESTION\\HAVING\:MODULUS\\ALSO\:CAN\:BE\:EXCLUDED\\BECAUSE\:THEY\:DO\:NOT\:PARTICULATE\\IN-CHANGING\:THE\:SIGN\:INEQUALITY\\AGAIN,\\(x - 1)(x + 2)(x + 1) < 0 \: \: \: \: \: ( |x + 4| not = 0) \\ x( - \infty , - 2)u( - 1,1) \: and \: 1 < |x - 3| < 5 \\ = )1 < x - 3 < 5 \: or \: - 5 < x - 3 < - 1 \\ = )4 < x < 8 \: or \: - 2 < x < 2 \\ hence \: common \: solution \: is \: ( - 1,1)$