(x+1/2)*(2/3-2x)=0
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TN
2
25 tháng 7 2021
Đề bài: \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{2}y+4\right|=0\)
PT \(\Leftrightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{2}y+4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-8\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\dfrac{1}{6};-8\right)\)
25 tháng 7 2021
Ta có: \(\left|3x-\dfrac{1}{2}\right|\ge0\forall x\)
\(\left|\dfrac{1}{2}y+4\right|\ge0\forall y\)
Do đó: \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{2}y+4\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{2}y+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-8\end{matrix}\right.\)
LQ
2
11 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\\dfrac{8}{5}+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{4}{5}\end{matrix}\right.\)
\(b,\dfrac{x-\dfrac{4}{7}}{x+\dfrac{1}{2}}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
\(c,\dfrac{2x-3}{x+\dfrac{7}{4}}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3< 0\\x+\dfrac{7}{4}>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3>0\\x+\dfrac{7}{4}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x >-\dfrac{7}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{7}{4}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-\dfrac{7}{4}< x< \dfrac{3}{2}\\x\in\varnothing\end{matrix}\right.\Leftrightarrow-\dfrac{7}{4}< x< \dfrac{3}{2}\)
PM
1
10 tháng 1
2(1-2x)-5=3(x+2)
=>\(2-4x-5=3x+6\)
=>\(-4x-3=3x+6\)
=>\(-7x=9\)
=>\(x=-\dfrac{9}{7}\)
S
2
5 tháng 3 2022
\(\Leftrightarrow2x\left(x+5\right)-3\left(x-2\right)=7x+1\)
\(\Leftrightarrow2x^2+10x-3x+6-7x-1=0\)
\(\Leftrightarrow2x^2+5=0\)(vô lý)
5 tháng 3 2022
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne2\\x\ne-5\end{matrix}\right.\)
\(\dfrac{2x}{x-2}-\dfrac{3}{x+5}=\dfrac{7x+1}{x^2+3x-10}\\ \Leftrightarrow\dfrac{2x\left(x+5\right)}{\left(x+5\right)\left(x-2\right)}-\dfrac{3\left(x-2\right)}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x^2-2x+5x-10}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x\left(x-2\right)+5\left(x-2\right)}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}-\dfrac{7x+1}{\left(x+5\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+10x-3x+6-7x-1}{\left(x+5\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{2x^2+5}{\left(x+5\right)\left(x-2\right)}=0\\ \Rightarrow2x^2+5=0\left(vô.lí\right)\)
Vậy pt vô nghiệm
HH
4
3 tháng 9 2021
\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)
\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)
NT
3 tháng 9 2021
a. \(\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\)
\(\Leftrightarrow2\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
TC
0
KN
13 tháng 11 2019
\(9\left(x+3\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(3x+9\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(3x+9+2x-1\right)\left(3x+9-2x+1\right)=0\)
\(\Leftrightarrow\left(5x+8\right)\left(x+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x+8=0\\x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-8}{5}\\x=-10\end{cases}}\)