Mn bt câu nào thì giúp mình với

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Mn bt câu nào thì giúp mình với
mn-bt-cau-nao-thi-giup-minh-voi

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Huyền Thanh 5 years 2021-05-16T07:59:34+00:00 1 Answers 21 views 0

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  1. yenoanh
    0
    2021-05-16T08:00:53+00:00
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     \(\begin{array}{l}
    1)\\
    a)\quad 3^{9-2x} = 27\\
    \Leftrightarrow 3^{9-2x} = 3^3\\
    \Leftrightarrow 9 – 2x = 3\\
    \Leftrightarrow 2x = 6\\
    \Leftrightarrow x = 3\\
    \text{Vậy}\ S = \{3\}\\
    b)\quad 8^{x^2 + x } = 64\\
    \Leftrightarrow 8^{x^2 + x} = 8^2\\
    \Leftrightarrow x^2 + x = 2\\
    \Leftrightarrow x^2 + x – 2 =0\\
    \Leftrightarrow \left[\begin{array}{l}x = 1\\x = -2\end{array}\right.\\
    \text{Vậy}\ S = \{-2;1\}\\
    c)\quad 25^{x-2} = \left(\dfrac15\right)^x\\
    \Leftrightarrow 5^{2x – 4} = 5^{-x}\\
    \Leftrightarrow 2x – 4 = -x\\
    \Leftrightarrow 3x = 4\\
    \Leftrightarrow x = \dfrac43\\
    \text{Vậy}\ S = \left\{\dfrac43\right\}\\
    d)\quad 16^{x^2 – 4x + 1} = 1\\
    \Leftrightarrow x^2 – 4x + 1 =0\\
    \Leftrightarrow x = 2\pm \sqrt3\\
    \text{Vậy}\ S = \{2 \pm \sqrt3\}\\
    2)\\
    a)\quad 3^{x-4} >9\\
    \Leftrightarrow 3^{x-4} > 3^2\\
    \Leftrightarrow x – 4 > 2\\
    \Leftrightarrow x > 6\\
    \text{Vậy}\ S = (6;+\infty)\\
    b)\quad 25^{x-3} > \dfrac15\\
    \Leftrightarrow 5^{2x-6} > 5^{-1}\\
    \Leftrightarrow 2x – 6 > – 1\\
    \Leftrightarrow 2x > 5\\
    \Leftrightarrow x > \dfrac52\\
    \text{Vậy}\ S = \left(\dfrac52;+\infty\right)\\
    3)\\
    a)\quad \log_4(3-6x) = 2\qquad \left(ĐK: x < \dfrac12\right)\\
    \Leftrightarrow 3-6x = 4^2\\
    \Leftrightarrow 6x = -13\\
    \Leftrightarrow x = -\dfrac{13}{6}\quad (nhận)\\
    \text{Vậy}\ S = \left\{-\dfrac{13}{6}\right\}\\
    b)\quad \log_{\sqrt2}(x^2 +6x) = 2\qquad (ĐK: x >0\ \lor\ x < -4)\\
    \Leftrightarrow x^2 + 6x = \left(\sqrt2\right)^2\\
    \Leftrightarrow x^2 + 6x – 2 =0\\
    \Leftrightarrow x = -3 \pm \sqrt{11}\quad (nhận)\\
    \text{Vậy}\ S = \{-3 \pm \sqrt{11}\}\\
    4)\\
    a)\quad A = \log_49 + \log_\sqrt227 – \log_\tfrac123\\
    \to A = \log_{2^2}3^2 + \log_{2^{\tfrac12}}3^3 – \log_{2^{-1}}3\\
    \to A = \log_23 + 6\log_23 + \log_23\\
    \to A = 8\log_23\\
    b)\quad B = \log_427.\log_98\\
    \to B = \log_{2^2}3^3.\log_3^2.2^3\\
    \to B = \dfrac32\log_23\cdot\dfrac32\log_32\\
    \to B = \dfrac94\\
    5)\\
    \quad \log_\tfrac14x + \log_8x =- 6\qquad (ĐK: x >0)\\
    \Leftrightarrow \log_{2^{-2}}x + \log_{2^3}x = -6\\
    \Leftrightarrow -\dfrac12\log_2x + \dfrac13\log_2x = -6\\
    \Leftrightarrow -\dfrac16\log_2x = -6\\
    \Leftrightarrow \log_2x = 36\\
    \Leftrightarrow x = 2^{36}\quad (nhận)\\
    \text{Vậy}\ S = \left\{2^{36}\right\}
    \end{array}\)

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