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a: A={0;1;2;3}
b: B={-16;-13;-10;-7;-4;-1;2;5;8}
c: C={-9;-8;-7;...;7;8;9}
d: \(D=\varnothing\)
a: \(A=\left\{1;-5\right\}\)
b: \(B=\left\{1;2\right\}\)
c: \(C=\left\{0;1;4;9;16;25;36;49\right\}\)
d: \(D=\left\{1;2;3;6\right\}\)
e: E={8}
a: A={0;1;2;3;4;5}
b: B={2;3;4;5}
c: C={-3;-2;-1;0;1;2;3}
d: D={3;-3}
e: E={1;-1;-5}
\(\left(2x+1\right)\left(x^2+x-1\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x^2+x-1=0\\2x^2-3x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\) (pt \(x^2+x-1=0\) ko có nghiệm hữu tỉ nên ko cần quan tâm)
\(A=\left\{-\dfrac{1}{2};\dfrac{1}{2};1\right\}\)
a) \(2x^3-3x^2-5x=0\)
\(x\left(x+1\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=-1\left(TM\right)\\x=\dfrac{5}{2}\left(L\right)\end{matrix}\right.\)
\(A=\left\{-1\right\}\)
b) \(x< \left|3\right|\)\(\Leftrightarrow-3< x< 3\)
\(B=\left\{-2;-1;1;2\right\}\)
c) \(C=\left\{-3;3;6;9\right\}\)
a) \(A=\left\{x\in Z|2x^3-3x^2-5x=0\right\}\)
\(2x^3-3x^2-5x=0\)
\(\Leftrightarrow x\left(2x^2-3x-5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow A=\left\{0;-1\right\}\)
b) \(B=\left\{-2;-1;0;1;2\right\}\)
c) \(C=\left\{-3;3;6;9\right\}\)
\(a,C=\left\{0;5;10;15;20;25;30\right\}\\ b,x^2+3x-4=0\\ \Leftrightarrow x^2-x+4x-4=0\\ \Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+4\right)=0\\ \Leftrightarrow x-1=0.hoặc.x+4=0\\ \Leftrightarrow x=1.hoặc.x=-4\\ Vậy:D=\left\{-4;1\right\}\)
\(A=\left\{2;9;22\right\}\)
\(B=\left\{-2;1\right\}\)