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a) \(\left(x-2,5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2,5=3\\x-2,5=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5,5\\x=-0,5\end{matrix}\right.\)
b) \(\left(1-x\right)^3=-8=\left(-2\right)^3\)
\(1-x=-2\)
\(x=3\)
c) \(2^4-x=16\)
\(16-x=16\)
\(x=0\)
\(a,\left(x-2,5\right)^2=9\)
\(+,TH1:x-2,5=3\)
\(\Rightarrow x=3+2,5\\\Rightarrow x=5,5\)
\(+,TH2:x-2,5=-3\)
\(\Rightarrow x=-3+2,5\\\Rightarrow x=-0,5\)
\(b,(1-x)^3=-8\)
\(\Rightarrow\left(1-x\right)^3=\left(-2\right)^3\)
\(\Rightarrow1-x=-2\)
\(\Rightarrow x=1-\left(-2\right)\)
\(\Rightarrow x=3\)
\(c,2^4-x=16\)
\(\Rightarrow16-x=16\)
\(\Rightarrow x=0\)
#\(Toru\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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a) \(2,5:2,7=x:\frac{3}{5}\)
\(\Rightarrow\frac{5}{2}:\frac{27}{10}=x.\frac{5}{3}\)
\(\Rightarrow\frac{5}{2}.\frac{10}{27}=\frac{5x}{3}\)
\(\Rightarrow\frac{25}{27}=\frac{5x}{3}\)
\(\Rightarrow5x=\frac{25.3}{27}=\frac{75}{27}=\frac{25}{9}\)
\(\Rightarrow x=\frac{25}{9}:5=\frac{25}{9}.\frac{1}{5}=\frac{5}{9}\)
b) \(\frac{19}{5}:\frac{608}{15}=0,25:x\)
\(\Rightarrow\frac{19}{5}.\frac{15}{608}=\frac{1}{4}:x\)
\(\Rightarrow\frac{3}{32}=\frac{1}{4}:x\)
\(\Rightarrow x=\frac{1}{4}:\frac{3}{32}=\frac{1}{4}.\frac{32}{3}=\frac{8}{3}\)
c/ \(\frac{8}{3}:x=\frac{16}{9}:0,2\)
\(\Rightarrow\frac{8}{3}:x=\frac{16}{9}:\frac{1}{5}=\frac{16}{9}.5=\frac{80}{9}\)
\(\Rightarrow x=\frac{8}{3}:\frac{80}{9}=\frac{8}{3}.\frac{9}{80}=\frac{3}{10}\)
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Giải:
a) Áp dụng tính chất của dãy tỉ số bằng nhau, có:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x+y}{5+3}=\dfrac{16}{8}=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=6\end{matrix}\right.\)
Vậy ...
b) \(2x=-3y\Leftrightarrow\dfrac{x}{-3}=\dfrac{y}{2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, có:
\(\dfrac{x}{-3}=\dfrac{y}{2}=\dfrac{y-x}{-3-2}=\dfrac{15}{-5}=-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=-6\end{matrix}\right.\)
Vậy ...
a) \(1,5x=2,5y\Leftrightarrow\dfrac{x}{2,5}=\dfrac{y}{1,5}\Leftrightarrow\dfrac{2x}{5}=\dfrac{2y}{3}\Leftrightarrow\dfrac{2x}{5}=\dfrac{4y}{6}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, có:
\(\dfrac{2x}{5}=\dfrac{4y}{6}=\dfrac{2x+4y}{5+6}=\dfrac{-11}{11}=-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy ...
\(\left(3x-2,5\right)^2=16\)
\(\Leftrightarrow\left(3x-2,5\right)^2=4^2\)
\(\Leftrightarrow\left|3x-2,5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2,5=4\\3x-2,5=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
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