{x |xϵ\(ℕ^∗\),x=4k+2,k+10<25
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45,90,150 đều chia hết cho x
\(\Rightarrow x\inƯC\left(45;90;150\right)\\ 45=5.3^2;90=2.5.3^2;150=2.5^2.3\\ \RightarrowƯCLN\left(45;90;150\right)=5.3=15\\ \Rightarrow x\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)
Vì x< 10 => x=1 hoặc x=3 hoặc x=5
\(7⋮x-1\)
\(\Rightarrow x-1\inƯ\left(7\right)\)
Ta có:
\(Ư\left(7\right)=\left\{1;7\right\}\)
\(\Rightarrow x-1\in\left\{1;7\right\}\)
\(\Rightarrow x\in\left\{2;8\right\}\)
\(a,A=\left\{0;1;2;3;4\right\}\\ b,B=\left\{-16;-13;-10;-7;-4;-1;2;5;8\right\}\\ c,C=\left\{-9;-8;-7;...;7;8;9\right\}\\ d,x^2-3x+1=0\\ \Delta=9-4=5\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{5}}{2}\\x=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\\ \Leftrightarrow D=\left\{\dfrac{3-\sqrt{5}}{2};\dfrac{3+\sqrt{5}}{2}\right\}\)
\(e,2x^3-5x^2+2x=0\\ \Leftrightarrow x\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow E=\left\{0;2\right\}\\ f,F=\left\{0;3;6;9;12;15;18\right\}\)
a) ta có \(\frac{y1}{x1}=\frac{y2}{x2}=\frac{y1+y2}{x1+x2}=\frac{3k}{4k}=\frac{3}{4}\)
=>\(y=\frac{3}{4}.x;và:x=\frac{4}{3}y\)
b) y1 +x2 =5
x1+x2 = 16 ; x1 = 4/3 y1 => 4/3y1 + x2 =16 => 1/3y1 +(y1+x2) =16 => 1/3 y1 +5 =16 => 1/3 y1 =11 => y1 =33
=> x1 =4/3 y1 =4/3 .33 =44
k+10<25
=>k<15
x\(\in\)N*
=>x>0
=>4k+2>0
=>k>-0,5
=>-0,5<k<15
mà k nguyên
nên k\(\in\left\{0;1;2;3;...;14\right\}\)
=>\(x\in\left\{2;6;10;...;58\right\}\)