2(1/2-2x)^2=8
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\(a^n=1\Rightarrow a^n=a^0\Rightarrow\left\{{}\begin{matrix}n=0\\a\in N\end{matrix}\right.\)
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\(\left(3x-\dfrac{1}{3}\right)^2=16=4^2\)
\(\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{3}=4\\3x-\dfrac{1}{3}=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=4+\dfrac{1}{3}\\3x=-4+\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{13}{3}\\3x=-\dfrac{11}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{9}\\x=-\dfrac{11}{9}\end{matrix}\right.\)
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B = \(\dfrac{x^2-2x+1}{x+1}\)
Với \(x\in\)Z, để B là số nguyên thì \(x^2-2x+1\)⋮ \(x+1\)
Theo Bezout ta có: F(\(x\)) = \(x^2\) - 2\(x\) + 1 ⋮ \(x+1\) ⇔ F(\(-1\)) ⋮ \(x+1\)
⇒ (-1)2 - 2.(-1) + 1 ⋮ \(x\) + 1 ⇔ 4 ⋮ \(x\) + 1
⇔ \(x\) + 1 \(\in\) Ư(4) = { -4; -2; -1; 1; 2; 4}
\(\Leftrightarrow\) \(x\) \(\in\) { -5; -3; -2; 0; 1; 3}
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Lời giải:
$7^{3x-2}-3.7^3=7^3.4$
$7^{3x-2}=3.7^3+7^3.4=7^3(3+4)=7^3.7=7^4$
$\Rightarrow 3x-2=4$
$\Rightarrow 3x=6$
$\Rightarrow x=2$
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\(\left(3x-\dfrac{1}{3}\right)^2=16=4^2=\left(-4\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{3}=4\\3x-\dfrac{1}{3}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{13}{3}\\3x=-\dfrac{11}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{9}\\x=-\dfrac{11}{9}\end{matrix}\right.\)
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`(2x-1/3)^2=4`
\(< =>\left[{}\begin{matrix}2x-\dfrac{1}{3}=2\\2x-\dfrac{1}{3}=-2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=2+\dfrac{1}{3}\\2x=-2+\dfrac{1}{3}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=-\dfrac{5}{3}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{3}:2\\x=-\dfrac{5}{3}:2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{3}\cdot\dfrac{1}{2}\\x=-\dfrac{5}{3}\cdot\dfrac{1}{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\)
\(\left(2x-\dfrac{1}{3}\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=2\\2x-\dfrac{1}{3}=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=2+\dfrac{1}{3}\\2x=-2+\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=-\dfrac{5}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
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\(\dfrac{4}{7}\times X=\dfrac{1}{2}\\ X=\dfrac{1}{2}:\dfrac{4}{7}=\dfrac{1}{2}\times\dfrac{7}{4}=\dfrac{7}{8}\)
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a/ x+65,27=72,6
x=72,6-65,27
x=7,33
b/x-43,502=21,73
x=21,73+43,502
x=65,232
c/x.6,3=187
x=187:6,3
x=29,682....
d/ 1602:x=7,2
x=1602:7,2
x=222,5
\(X+65,27=72,6\\ X=72,6-65,27=7,33\\ ---\\ X-43,502=21,73\\ X=21,73+43,502=65,232\\ ---\\ X.6,3=187\\ X=\dfrac{187}{6,3}=\dfrac{1870}{63}\\ ---\\ 1602:X=7,2\\ X=1602:7,2=222,5\)
\(2\left(\dfrac{1}{2}-2x\right)^2=8\)
\(\Rightarrow\left(\dfrac{1}{2}-2x\right)^2=4=2^2\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-2x=2\\\dfrac{1}{2}-2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-2\\2x=\dfrac{1}{2}+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-\dfrac{3}{2}\\2x=\dfrac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{5}{4}\end{matrix}\right.\)