tìm x:
a)92x-1=\(\frac{1}{3}\)
b)\(3.8^x-2.2^{3x}-64=0\)
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\(3.8^x-2.2^{3x}-16=0\)
\(\Leftrightarrow3.\left(2^3\right)^x-2.2^{3x}=16\)
\(\Leftrightarrow3.2^{3x}-2^{3x}=16\)
\(\Leftrightarrow2^{3x}\left(3-1\right)=16\)
\(\Leftrightarrow2^{3x}.2=16\)
\(\Leftrightarrow2^{3x+1}=2^4\)
\(\Leftrightarrow3x+1=4\)
\(\Leftrightarrow x=1\)
Vậy ....
\(a,\Rightarrow\left(35x+3\right)\cdot19=152\\ \Rightarrow35x+3=8\\ \Rightarrow x=\dfrac{1}{7}\\ b,\Rightarrow3\left(x+7\right)=42\\ \Rightarrow x+7=14\Rightarrow x=7\\ c,\Rightarrow3\left(x+1\right)=48\\ \Rightarrow x+1=16\Rightarrow x=15\\ d,\Rightarrow120-5x+100\cdot2:5=4\cdot15\\ \Rightarrow120-5x+40=60\\ \Rightarrow5x=100\Rightarrow x=20\\ e,\Rightarrow4x-10=30\\ \Rightarrow4x=40\\ \Rightarrow x=10\\ g,\Rightarrow10x+10=70\\ \Rightarrow10x=60\\ \Rightarrow x=6\)
a.
\(\Leftrightarrow\left(3x-1\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow3x-1=-\dfrac{1}{2}\)
\(\Leftrightarrow3x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{6}\)
b.
\(\Leftrightarrow\left(2x-1\right)\left(x-4\right)-x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x-1-x\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\\\end{matrix}\right.\)
c.
\(\Leftrightarrow3x\left(5x-2\right)-2\left(5x-2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)
\(a,3.8^x-2.2^{3x}-16=0\)
\(\Leftrightarrow3.\left(2^3\right)^x-2.2^{3x}=16\)
\(\Leftrightarrow3.2^{3x}-2^{3x}=16\)
\(\Leftrightarrow2^{3x}\left(3-1\right)=16\)
\(\Leftrightarrow2^{3x+1}=2^4\)
\(\Leftrightarrow3x+1=4\)
\(\Leftrightarrow3x=3\)
\(\Leftrightarrow x=1\)
Vậy .....
a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)
\(\Leftrightarrow x^3-x^3-1=x\)
hay x=-1
c: Ta có: \(56x^4+7x=0\)
\(\Leftrightarrow7x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: Ta có: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
a) (32)2x-1 = 3-1 => 32(2x-1) = 3-1 => 2(2x -1) = -1 => 4x - 2 = -1 => 4x = 1 => x = 1/4. vậy,,,
b) 3.8x - 2. (23)x = 64
=> 3.8x - 2.8x = 64
=> 8x = 82 => x = 2
Vậy...