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(2x+15) ÷ (x-2)
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a) đặt [41-(2x-5)] là a <=> 1440 : a = 48
=> a = 30 <=> [41-(2x-5)] = 30 => 41-(2x-5) =30
<=> -(2x-5) = -11 <=> 2x-5 = 11 <=> 2x = 16 <=> x=2
b) tương tự câu a
gợi ý đáp án : x = 210
\(A=\left\{25;26;27;28;29;30\right\}\\ B=\left\{11;12;13;14\right\}\\ C=\left\{1;2;5;10\right\}\\ D=\left\{14;16;18;20\right\}\)
\(A=\left\{25;26;27;28;29;30\right\}\)
\(B=\left\{11;12;13;14\right\}\)
\(C=\left\{1;2;5;10\right\}\)
\(D=\left\{14;16;18;20\right\}\)
a
\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)
b
\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)
c
\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)
a: =>(2x+15)(x^2+4)=0
=>2x+15=0
=>2x=-15
=>x=-15/2
b; =>(x-2)(5x-3)=0
=>x=2 hoặc x=3/5
c: =>(x+3)(2-x)=0
=>x=2 hoặc x=-3
\(x=\left\{x\in N;2x\left(2x-6\right)\left(3x-15\right)\right\}=0\)
\(\Leftrightarrow2x=0\Rightarrow x=0\)
\(2x-6=0\Rightarrow2x=6\Rightarrow x=3\)
\(3x-15=0\Rightarrow3x=15\Rightarrow x=5\)
\(\Rightarrow x=\left\{0;3;5\right\}\)
a: A có 5 phần tử
b: B có (2024-0):2+1=1013(số)
c: C có (101-1):5+1=21(số)
d: D={0;1;2;3;4}
=>D có 5 phần tử
e: E={0;2;...;998}
E có (998-0):2+1=500(số)
\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)
Vậy ...
2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
Vậy ...
3) PT \(\Rightarrow3x-4-2x+5=3\)
\(\Rightarrow x=2\)
Vậy ...
4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy ...
3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)
\(\Leftrightarrow3x-4-2x+5=3\)
\(\Leftrightarrow x+1=3\)
hay x=2
\(\dfrac{2\left(x-2\right)+19}{x-2}=2+\dfrac{19}{x-2}\Rightarrow x-2\inƯ\left(19\right)=\left\{\pm1;\pm19\right\}\)
Vi x là stn nên x = 3 ; 20
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