tìm x
\(2^{-x}+2^{-x-4}=17\)
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\(x+\frac{11}{17}+x+\frac{2}{17}+x+\frac{4}{17}=4x\)
\(\Rightarrow3x+1=4x\)
\(\Rightarrow x=1\)
Vì vế trái >=0 => vế phải >=0 => x>=0
Khi đó ta có: \(x+\frac{11}{17}+x+\frac{2}{17}+x+\frac{4}{17}=4x\)
=> \(3x+1=4x\)
=> \(x=1\)
\(x\left(x-5\right)\left(x+5\right)-\left(x-2\right)\left(x^2+2x+4\right)=-17\)
\(\Leftrightarrow x^3-25x-x^3+8=-17\)
\(\Leftrightarrow-25x=-25\)
hay x=1
\(\Rightarrow x^3-3x^2+3x-1+8-x^3=17-3x^2-6x\\ \Rightarrow9x=10\Rightarrow x=\dfrac{10}{9}\)
\(a,\left(2x-5\right)+17=6\\ \Rightarrow2x-5=-11\\ \Rightarrow2x=-6\\ \Rightarrow x=-3\\ b,10-2\left(4-3x\right)=-4\\ \Rightarrow2\left(4-3x\right)=14\\ \Rightarrow4-3x=7\\ \Rightarrow3x=-3\\ \Rightarrow x=-1\\ c,24:\left(3x-2\right)=-3\\ \Rightarrow3x-2=-8\\ \Rightarrow3x=-6\\ \Rightarrow x=-2\\ d,5-2x=-17+12\\ \Rightarrow5-2x=-5\\ \Rightarrow2x=10\\ \Rightarrow x=5\)
a: =>2x-5=-11
=>2x=-6
hay x=-3
b: =>2(4-3x)=14
=>4-3x=7
=>3x=-3
hay x=-1
c: =>3x-2=-8
=>3x=-6
hay x=-2
\(2^{-x}+2^{-x-4}=17\)
\(\Leftrightarrow2^{-x}\left(1+2^{-4}\right)=17\)
\(\Leftrightarrow2^{-x}.1,0625=17\)
\(\Leftrightarrow2^{-x}=17:1,0625\)
\(\Leftrightarrow2^{-x}=16\)
\(\Leftrightarrow2^{-x}=2^4\)
\(\Leftrightarrow x=-4\)
\(2^{-x}+2^{-x-4}=17=>2^{-x}\cdot1+2^{-4}\cdot\frac{1}{16}=17=>2^{-x}\cdot\left(1+\frac{1}{16}\right)\)\(=17=>2^{-x}=17:\frac{17}{16}=>2^{-x}=2^4=>-x=4=>x=-4\)