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\(\frac{x}{32}=\frac{2}{x}\)
NHÂN CHÉO CHO NHAU TA CÓ:
\(x.x=32.2\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x^2=8^2\)
\(\Leftrightarrow x=8\)
BẠN KIA LÀM SAI RỒI X.X MÀ BẰNG 2X CÓ PHẢI X+X ĐÂU
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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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\(\frac{2^{4-x}}{16^5}=32^6\)
\(\Rightarrow\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
\(\Rightarrow\frac{2^{4-x}}{2^{20}}=2^{30}\)
\(\Rightarrow2^{4-x}=2^{30}.2^{20}\)
\(\Rightarrow2^{4-x}=2^{50}\)
\(\Rightarrow4-x=50\)
\(\Rightarrow x=-46\)
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\(\frac{2^x}{4}.2^x=32\)
<=>\(\frac{2^x}{4}.2^x=2^5\)
<=>\(2^x:4=2^5:2^x\)
<=>\(2^x:2^2=2^5:2^x\)
<=>\(2^{x-2}=2^{5-x}\)
<=>\(x-2=5-x\)
<=>\(x+x=5+2\)
<=>\(2x=7\)
<=>\(x=3,5\)
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\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\Leftrightarrow\frac{2^x}{2}+5.\frac{2^x}{2^2}=\frac{7}{32}\Leftrightarrow2^x\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\Leftrightarrow2^x=\frac{1}{8}=2^{-3}\)
<=> x=-3
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Nếu bạn đang thì và câu đó với điều kiện x<0 thì bạn cứ điền là -8
100 %
\(\frac{x}{2}=\frac{32}{x}=>x.x=32.2=>x^2=64=8^2=\left(-8\right)^2=>\orbr{\begin{cases}x=8\\x=-8\end{cases}}\)
x=8
k giùm mk nha mấy bn.