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\(2^{x-2}-3.2^x=-88\)
\(\Rightarrow2^x.\frac{1}{4}-3.2^x=-88\)
\(\Rightarrow2^x.\left(\frac{1}{4}-3\right)=-88\)
\(\Rightarrow2^x.\frac{-11}{4}=-88\)
\(\Rightarrow2^x=-88:\frac{-11}{4}\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
2x-2 - 3.2x = -88
2x . 2-2 - 3.2x = -88
2x . (2-2 - 3) = -88
2x . (1/4 - 3) = -88
2x . (-11/4) = -88
2x = -88 : (-11/4)
2x = (-88).(4/-11)
2x = 32
=> x = 5
Ta có:
\(2^{x-2^{ }}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x:2^2-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\cdot\frac{1}{4}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\left(\frac{1}{4}-3\right)=-88\)
\(\Leftrightarrow2^x\cdot\left(-\frac{11}{4}\right)=-88\)
\(\Leftrightarrow2^x=-88:\left(-\frac{11}{4}\right)\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2x-2 - 3.2x = -88
=> 2x-2 - 3.22.2x-2 = -88
=> 2x-2 - 3.4.2x-2 = -88
=> 2x-2 - 12.2x-2 = -88
=> 2x-2.(1 - 12) = -88
=> 2x-2.(-11) = -88
=> 2x-2 = -88 : (-11)
=> 2x-2 = 8 = 23
=> x - 2 = 3
=> x = 3 + 2 = 5
Vậy x = 5
Ta có: \(2^{x-2}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\cdot\dfrac{1}{4}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\cdot\dfrac{-11}{4}=-88\)
\(\Leftrightarrow2^x=32\)
hay x=5
\(2^{x-2}-3\cdot2^x=-88\)
\(\Leftrightarrow\frac{2^x}{2^2}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\left(-\frac{11}{4}\right)=-88\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow x=5\)
\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(x+2\right)\left(x+5\right)}\right)=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+..+\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{2}-\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{20}\)
\(\Rightarrow x+5=20\)
\(\Rightarrow x=20-5\)
\(\Rightarrow x=15\)
\(\Leftrightarrow\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)
\(\Leftrightarrow\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{202}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Leftrightarrow x+3=308\)
\(\Leftrightarrow x=305\)
Vậy x=305
Ta có : 2x - 1 - 3.2x = -88
<=> -2.2x - 3.2x = -88
<=> 2x(-2 - 3) = -88
<=> 2x . -5 = -88
=> 2x = -88 : (-5)
=> 2x =
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