3×2^x+2^x+3=88
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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
\(2^{x-2}-3\cdot2^x=-88\)
\(\Leftrightarrow\frac{2^x}{2^2}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\left(-\frac{11}{4}\right)=-88\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow x=5\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k->\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
Thay vào \(x^2+2y^2=88\)
\(=>\left(2k\right)^2+2.\left(3k\right)^2=88\)
\(=>4.k^2+18.k^2=88\)
\(=>k^2\left(4+18\right)=88\)
\(=>k^2=4\)
\(=>\left[{}\begin{matrix}k=-2\\k=2\end{matrix}\right.\)
Th1: \(k=-2\)
\(=>x=-4;y=-6\)
Th2:\(k=2\)
\(=>x=4;y=6\)
Vậy có 2 cặp \(\left(x;y\right)\) t/m: \(\left(-4;-6\right);\left(4;6\right)\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\) (k ∈ N*)
\(\Rightarrow x=2k;y=3k\) (*)
Thay (*) vào biểu thức \(x^2+2y^2=88\) , ta được:
\(\left(2k\right)^2+2\cdot\left(3k\right)^2=88\)
\(\Leftrightarrow4k^2+18k^2=88\)
\(\Leftrightarrow22k^2=88\)
\(\Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\) (tmđk)
\(+,\) Với \(k=2\) \(\Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\end{matrix}\right.\) (tm)
\(+,\) Với \(k=-2\)\(\Rightarrow\left\{{}\begin{matrix}x=-2\cdot2=-4\\y=-2\cdot3=-6\end{matrix}\right.\) (tm)
Ta có : 2x - 1 - 3.2x = -88
<=> -2.2x - 3.2x = -88
<=> 2x(-2 - 3) = -88
<=> 2x . -5 = -88
=> 2x = -88 : (-5)
=> 2x =
\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+...+\frac{2}{x\left(x+3\right)}=\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}\)
\(=\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Từ đó ta có:
\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(x+3=308\)
x = 305