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\(a,\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\\dfrac{8}{5}+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{4}{5}\end{matrix}\right.\)
\(b,\dfrac{x-\dfrac{4}{7}}{x+\dfrac{1}{2}}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
\(c,\dfrac{2x-3}{x+\dfrac{7}{4}}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3< 0\\x+\dfrac{7}{4}>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3>0\\x+\dfrac{7}{4}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x >-\dfrac{7}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{7}{4}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-\dfrac{7}{4}< x< \dfrac{3}{2}\\x\in\varnothing\end{matrix}\right.\Leftrightarrow-\dfrac{7}{4}< x< \dfrac{3}{2}\)
Bài 2:
a: =>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
a: \(A=2\left(x+y\right)+3xy\left(x+y\right)+5x^2y^2\left(x+y\right)=0\)
b: \(B=3xy\left(x+y\right)+2x^2y\left(x+y\right)=0\)
a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x.y}{2.3}=\dfrac{54}{6}=9\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=81\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm9\end{matrix}\right.\)
b) \(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{5}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x^2-y^2}{5^2-3^2}=\dfrac{4}{16}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{25}{4}\\y^2=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{5}{2}\\y=\pm\dfrac{3}{2}\end{matrix}\right.\)
c: Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)
nên \(\dfrac{x}{10}=\dfrac{y}{15}\)
Ta có: \(\dfrac{y}{5}=\dfrac{z}{7}\)
nên \(\dfrac{y}{15}=\dfrac{z}{21}\)
mà \(\dfrac{x}{10}=\dfrac{y}{15}\)
nên \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{92}{46}=2\)
Do đó: x=20; y=30; z=42
a) \(\left|x+2\right|+\left|2y-1\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+2\right|=0\\\left|2y-1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+2=0\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=0,5\end{cases}}\)
Vậy (x; y) = (-2; 0,5)
b) \(\left|x-y\right|+\left|2x+3\right|=0\Leftrightarrow\hept{\begin{cases}\left|x-y\right|=0\\\left|2x+3\right|=0\end{cases}}\)
+) |2x + 3| = 0
2x + 3 = 0
2x = -3
x = -1,5
+) |x - y| = 0
x - y = 0
-1,5 - y = 0
y = -1,5
Vậy (x; y) = (-1,5; -1,5)
c, \(\left|2x+y\right|+\left|y+\left(1:4\right)\right|=0\)
\(\left|2x+y\right|+\left|y+\frac{1}{4}\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|2x+y\right|=0\\\left|y+\frac{1}{4}\right|=0\end{cases}}\)
\(\left|y+\frac{1}{4}\right|=0\Leftrightarrow y+\frac{1}{4}=0\Leftrightarrow y=-\frac{1}{4}\)
\(\left|2x+y\right|=0\Leftrightarrow2x+y=0\Leftrightarrow2x-\frac{1}{4}=0\Leftrightarrow2x=\frac{1}{4}\Leftrightarrow x=\frac{1}{8}\)
Vậy \(\left(x;y\right)=\left(\frac{1}{8};-\frac{1}{4}\right)\)
a) |x+2|+|2y-1|=0
=> |x+2| =0=> x+2=0 => x= -2
|2y-1|=0=> 2y-1=0 => 2y=1 => y = 1:2 =0,5
b) |x-y|+|2x+3|=0
=> |2x+3|=0 => 2x+3 => 2x= -3 => x = -3:2 = -1,5
|x-y|=0 => x-y =0 => y = -1,5 - 0 = -1,5