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\(c,20=2^2\cdot5\\ 45=3^2\cdot5\\ ƯCLN\left(20,45\right)=5\\ \RightarrowƯC\left(20,45\right)=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\\ C=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(d,\left(6x^2-7x+1\right)\left(x^3-x\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6x-1\right)x\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{6}\\x=1\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow D=\left\{-1;0;\dfrac{1}{6};1\right\}\)
Sửa: \(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{6}\\x=1\\x=-1\end{matrix}\right.\Leftrightarrow D=\left\{-1;0;\dfrac{1}{6};1\right\}\)
a: \(\left(\dfrac{1}{2}\right)^n>=\dfrac{1}{32}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^n>=\left(\dfrac{1}{2}\right)^5\)
=>n<=5
=>M={1;1/2;1/4;1/8;1/16;1/32}
b: \(x^2+x+3=0\)
\(\text{Δ}=1^2-4\cdot1\cdot3=1-12=-11< 0\)
=>Phương trình vô nghiệm
=>\(C=\varnothing\)
a: \(\left(2x^2-5x+3\right)\left(x^2-4x+3\right)=0\)
=>(2x-3)(x-1)(x-3)(x-1)=0
=>x=1; x=3;x=3/2
=>A={1;3;3/2}
b: \(\left\{{}\begin{matrix}x+3< 2x+4\\5x-3< 4x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x< 1\\x< 2\end{matrix}\right.\Leftrightarrow-1< x< 2\)
mà x là số tự nhiên
nên B={0;1}
\(a,C=\left\{0;5;10;15;20;25;30\right\}\\ b,x^2+3x-4=0\\ \Leftrightarrow x^2-x+4x-4=0\\ \Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+4\right)=0\\ \Leftrightarrow x-1=0.hoặc.x+4=0\\ \Leftrightarrow x=1.hoặc.x=-4\\ Vậy:D=\left\{-4;1\right\}\)
\(x^4-6x^2+8=0\\ \Leftrightarrow x^4-2x^2-4x^2+8=0\\ \Leftrightarrow x^2\left(x-2\right)-4\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)^2\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(F=\left\{-2;2\right\}\)
Ta có: \(x^4-6x^2+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)+2x^2\left(x-2\right)-2x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
Vậy \(F=\left\{2;-2;\sqrt{2};-\sqrt{2}\right\}\)