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  1. Top | #13

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  3. Top | #14

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    $Q10)(2x+7)(x^{2}-9)(2x-5)=91$

    $4x^{4}+4x^{3}-71x^{2}-36x+224=0$

    $4x^{4}+4x^{3}+x^{2}-72x^{2}-36x+224=0$

    $(2x^{2}+x)^{2}-36(2x^{2}+x)+224=0$

    $(2x^{2}+x-28)(2x^{2}+x-8)=0$

    $x=\frac{7}{2},x=-4,x=\frac{-1\mp \sqrt{65}}{4}$

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  5. Top | #15

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  7. Top | #16

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    \[\begin{array}{l}
    \left( {11} \right) & {5^{1 + x}} + {5^{1 - x}} = 26\,\, \\
    \left( {12} \right){2^{x + 1}} + {4^x} = 8 \\
    \left( {13} \right) & 5{x^{1/2n}} - 3{x^{1/n}} = 2\, \\
    \left( {14} \right){2^{{x^2}}}:{4^x} = 8:1 \\
    \left( {15} \right) & 9\left( {{x^2} + \frac{1}{{{x^2}}}} \right) - 27\left( {x + \frac{1}{x}} \right) + 8 = 0\, \\
    \left( {16} \right) & 2{\left( {x + \frac{1}{x}} \right)^2} - 7\left( {x + \frac{1}{x}} \right) + 6 = 0 \\
    \left( {17} \right) & \left( {{x^2} + \frac{1}{{{x^2}}}} \right) - 5\left( {x + \frac{1}{x}} \right) + 6 = 0 \\
    \left( {18} \right)\,\,{x^4} + {x^3} - 4{x^2} + x + 1 = 0 \\
    \left( {19} \right) & {\left( {x + \frac{1}{x}} \right)^2} - \frac{3}{2}\left( {x - \frac{1}{x}} \right) - 4 = 0 \\
    \left( {20} \right)\,\,\,x\left( {x + 2} \right)\left( {{x^2} - 1} \right) + 1 = 0 \\
    \end{array}\]







    ; 02-14-2017 01:34 AM

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  9. Top | #17

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  11. Top | #18

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    $Q12)2^{x+1}+4^{x}=8$

    $2.2^{x}+2^{2x}-8=0$

    $(2^{x}+4)(2^{x}-2)=0$

    $2^{x}=-4 (REF),2^{x}=2\Rightarrow x=1$

    $S=\left \{ 1\right \}$

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  13. Top | #19

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    $Q15)solve. in .R :9(x^{2}+\frac{1}{x^{2}})-27(x+\frac{1}{x})+8=0$

    $9(x^{2}+\frac{1}{x^{2}})^{2}-27(x+\frac{1}{x})-10=0$

    $\left [ 3(x+\frac{1}{x}) +1\right ]\left [ 3(x+\frac{1}{x}) -10\right ]=0$

    $\left [ 3(x+\frac{1}{x}) +1\right ](ref)$

    $3(x+\frac{1}{x}) -10=0\Leftrightarrow 3x^{2}-10x+3=0$

    $(x-3)(3x-1)=0\Rightarrow x=3,x=\frac{1}{3}$

    $S=\left \{ 3,\frac{1}{3} \right \}$
    ; 02-14-2017 08:59 AM

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  15. Top | #20

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  17. Top | #21

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  19. Top | #22

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  21. Top | #23

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    $Q20)x(x+2)(x^2-1)+1=0$

    $x^{4}+2x^{3}-x^{2}-2x+1=0$

    $x^{4}+2x^{3}+x^{2}-2x^{2}-2x+1=0$

    $(x^{2}+x)^{2}-2(x^{2}+x)+1=0$

    $((x^{2}+x)-1)^{2}=0$

    $x=\frac{-1\pm \sqrt{5}}{2}$

    $S=\left \{ \frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2} \right \}$
    ; 02-14-2017 10:33 PM

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  23. Top | #24

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