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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
\(2^{x-2}-3\cdot2^x=-88\)
\(\Leftrightarrow\frac{2^x}{2^2}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\left(-\frac{11}{4}\right)=-88\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow x=5\)
a) \(-\left|2+1\right|-2x=1\)
\(\Leftrightarrow-\left|3\right|-2x=1\)
\(\Leftrightarrow-3-2x=1\)
\(\Leftrightarrow-2x=1-3\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=\dfrac{-2}{-2}=1\)
Vậy x=1
b) \(\left|7x-1\right|-\left|5+6\right|=0\)
\(\Leftrightarrow\left|7x-1\right|-\left|11\right|=0\)
\(\Leftrightarrow\left|7x-1\right|-11=0\)
\(\Leftrightarrow\left|7x-1\right|=11\)
\(\Leftrightarrow7x-1=\pm11\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x-1=11\\7x-1=-11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=11\\7x=-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{12}{7}\\x=\dfrac{-10}{7}\end{matrix}\right.\)
Vậy \(x=\dfrac{12}{7}\) hoặc \(x=\dfrac{-10}{7}\)
\(a) - \left | 2 + 1 \right | - 2x = 1\)
\(- 3 - 2x = 1\)
\(2x = - 3 - 1\)
2x = - 4
x = - 2
Vậy ......
Đặt 20212020=x
=>\(A=\dfrac{3\left(x+1\right)\left(x+3\right)-5x-2\cdot\left(x+1\right)^2-5}{\left(x+1\right)}\)
\(=\dfrac{3\left(x^2+4x+3\right)-5x-2x^2-4x-2-5}{\left(x+1\right)}\)
\(=\dfrac{3x^2+12x+9-2x^2-9x-7}{x+1}=\dfrac{x^2+3x+2}{x+1}=x+2\)
=20212022