(x+1.(x-3) < 0
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KV
9 tháng 1
1) Do x ∈ Z và 0 < x < 3
⇒ x ∈ {1; 2}
2) Do x ∈ Z và 0 < x ≤ 3
⇒ x ∈ {1; 2; 3}
3) Do x ∈ Z và -1 < x ≤ 4
⇒ x ∈ {0; 1; 2; 3; 4}
TG
18 tháng 7 2021
đăng ít thôi bạn! Nếu bạn đăng lẻ ra thì bn sẽ nhận đc sự trợ giúp nhanh hơn !
18 tháng 7 2021
a) Ta có: \(\left|x-3\right|+\left|y-2x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x=2\cdot3=6\end{matrix}\right.\)
TA
1
10 tháng 8 2017
\(\left(x-3\right)^3+\left(x+3\right)^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)
\(\Leftrightarrow x^2\left(2x+54\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)
\(\Leftrightarrow6x^2=-2\)
\(\Leftrightarrow x^2=-3\) ( vô lí)
Vậy pt vô nghiệm
\(c,x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)
\(\Leftrightarrow-\left(2x+5\right)=0\)
\(\Leftrightarrow-2x-5=0\)
\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)
Học tốt nha you <3
TA
1
10 tháng 8 2017
\(\left(x-3\right)^3+\left(x+3\right)^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)
\(\Leftrightarrow x^2\left(2x+54\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)
\(\Leftrightarrow6x^2=-2\)
\(\Leftrightarrow x^2=-3\) ( vô lí)
Vậy pt vô nghiệm
\(c,x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)
\(\Leftrightarrow-\left(2x+5\right)=0\)
\(\Leftrightarrow-2x-5=0\)
\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)
Học tốt nha you <3
TA
1
10 tháng 8 2017
\(\left(x-3\right)^3+\left(x+3\right)^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)
\(\Leftrightarrow x^2\left(2x+54\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)
\(\Leftrightarrow6x^2=-2\)
\(\Leftrightarrow x^2=-3\) ( vô lí)
Vậy pt vô nghiệm
\(c,x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)
\(\Leftrightarrow-\left(2x+5\right)=0\)
\(\Leftrightarrow-2x-5=0\)
\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)
Học tốt nha you <3
7 tháng 5 2021
1) \(\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Vậy tập nghiệm \(S=\left\{-2;3\right\}\)
2) \(\left(2x+3\right)\left(-x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=7\end{matrix}\right.\)
Vậy...
3) \(\left(x-1\right)\left(x+5\right)\left(-3x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\\-3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy...
4) \(x\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
Vậy...
DT
25 tháng 6 2016
a)\(3x\left(x-1\right)+x-1=0\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\Leftrightarrow\hept{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)
\(S=\left\{1;\frac{1}{3}\right\}\)
b)\(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\Leftrightarrow\hept{\begin{cases}2-x=0\\x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=-3\end{cases}}}\)
\(S=\left\{2;-3\right\}\)
MS
3
NN
3 tháng 1 2018
\(a,\left(x+5\right)\left(x-4\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0.\\x-4=0.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5.\\x=4.\end{matrix}\right.\)
Vậy..........
\(b,\left(x-1\right)\left(x-3\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\x-3=0.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=3.\end{matrix}\right.\)
Vậy..........
\(c,\) Sửa đề:
\(\left(3-x\right)\left(x-3\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0.\\x-3=0.\end{matrix}\right.\)
\(\Leftrightarrow x=3.\)
Vậy..........
\(d,x\left(x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0.\\x+1=0.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0.\\x=-1.\end{matrix}\right.\)
Vậy..........