Bài 100613

Question

Giải các phương trình :

$\begin{array}{l} 1)\,{\left( {\frac{1}{4}} \right)^{x - 2}} = {2^{5 - x}} + 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,{\left( {\frac{1}{6}} \right)^{x - 3}} = {6^{5 - 2x}} - 12\\ 2)\,\frac{3}{{{2^{3 - x}}}} = {4^{x - 4}} - 7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,{5^{2x - 3}} = \frac{2}{{{5^{1 - x}}}} + 15 \end{array}$

score 0 · Answer 1

1)
TXĐ: R.
Phương trình đã cho tương đương với:

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

$\Leftrightarrow 16.(2^{-x})^2-32.2^{-x}-9=0$

$\Leftrightarrow (4.2^{-x}-9)(4.2^{-x}+1)=0$

$\Leftrightarrow \left[ \begin{array}{l} 2^{-x}=\frac{9}{4}\\ 2^{-x}=-\frac{1}{4}(L do 2^{-x}>0\forall x) \end{array} \right.$

$\Rightarrow x={\log _2}\frac{4}{9}.$

2)
TXĐ: R.
Phương trình đã cho tương đương với:

$3.2^{x-3}-4^{x-4}+7=0$

$\Leftrightarrow 3.\frac{2^x}{2^3}-\frac{4^x}{4^4}+7=0$

$\Leftrightarrow \frac{3}{2^3}.2^x-\frac{1}{4^4}.(2^x)^2+7=0$

$\Leftrightarrow (2^x+16)(2^x-112)=0$

$\Leftrightarrow 2^x=112 (do 2^x>0 \forall x)$

$\Rightarrow x = 4 + {\log _2}7.$

3)
TXĐ: R.
Phương trình đã cho tương đương với:

$6^{3-x}-6^{5-2x}+12=0$

$\Leftrightarrow 6^3.6^{-x}-6^5.(6^{-x})^2+12=0$

$\Leftrightarrow 7776.(6^{-x})^2-216.6^{-x}-12=0$

$\Leftrightarrow (18.6^{-x}-1)(36.6^{-x}+1)=0$

$\Leftrightarrow \left[ \begin{array}{l} 6^{-x}=\frac{1}{18}\\6^{-x}=-\frac{1}{36} \end{array} \right.$
Do

$6^{-x}>0\forall x\Rightarrow 6^{-x}=\frac{1}{18}$

$\Rightarrow x=2 - {\log _6}2.$

4)
TXĐ: R.
Phương trình đã cho tương đương với:

$5^{2x-3}=2.5^{x-1}+15$

$\Leftrightarrow \frac{5^{2x}}{5^3}=2.\frac{5^x}{5}+15$

$\Leftrightarrow \frac{1}{5^3}.(5^x)^2-\frac{2}{5}.5^x-15=0$

$\Leftrightarrow (5^x-75)(5^x+25)=0$
Do

$5^x>0\forall x\Rightarrow 5^x=75$

$\Rightarrow x = 2 + {\log _5}3.$

ĐS:

$\begin{array}{l} 1)\,x = {\log _2}\frac{4}{9}\,\,;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2)\,x = 4 + {\log _2}7\\ 3)\,x = 2 - {\log _6}2;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,x = 2 + {\log _5}3 \end{array}$

Bài 100613

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Bài 100613

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