34x-1÷3x=243
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\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
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a) \(12x^3+8x^2-3x-2=4x^2\left(3x+2\right)-\left(3x+2\right)\)
\(=\left(3x+2\right)\left(4x^2-1\right)=\left(3x+2\right)\left(2x-1\right)\left(2x+1\right)\)
b) \(18x^3+27x^2-2x-3=9x^2\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(9x^2-1\right)=\left(2x+3\right)\left(3x-1\right)\left(3x+1\right)\)
c) \(8x^3+4x^2-34x+15=4x^2\left(2x-3\right)+8x\left(2x-3\right)-5\left(2x-3\right)\)
\(=\left(2x-3\right)\left(4x^2+8x-5\right)=\left(2x-3\right)\left(2x-1\right)\left(2x+5\right)\)
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a)2x.2x+1=128
\(4x^2+1=128\)
\(\Rightarrow4x^2=127\)
\(\Rightarrow x^2=127\div4\)
\(\Rightarrow x^2=3.175\)
mà \(x\in N\) hay\(x^2\in N\)
\(\Rightarrow\)Không có x
b) 3x+1.3x=243
\(\Rightarrow\)3x+3x=243
\(\Rightarrow\)6x=243
\(\Rightarrow\)x=40,5
Mà \(x\in N\)
\(\Rightarrow\)Không có x
34x - 1 : 3x = 243
=> 34x - 1 - x = 35
=> 33x - 1 = 35
=> 3x - 1 = 5
=> 3x = 6
=> x = 2
Vậy x = 2
\(3^{4x-1}:3^x=243\)
\(\Rightarrow3^{4x-1}:3^x=3^5\)
\(\Rightarrow4x-1=5\)
\(\Rightarrow4x=5+1\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\frac{3}{2}\)