Chỉ mình
$0,5V_1+3V_2=2,5(V_1+V_2)$ |
đang như này làm sao biến đổi về được thành \(\dfrac{V1}{V2}=\dfrac{2,5-0,5}{3-2,5}\) vậy ạ các bạn
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Ý bạn là sao nhỉ?
Theo mình hiểu thì bạn muốn biến 72 thành căn đúng không? Vậy thì bạn chỉ cần biểu diễn $72=\sqrt{72^2}=\sqrt{5184}$ thôi.
Bạn chỉ cần phân tích nó ra thành thừa số nguyên tố là xong
\(75\%:2\dfrac{1}{5}+\left(0,5\right)^2.\left(-7\right)+2,5\left(7\dfrac{2}{3}+5\dfrac{2}{3}\right)\)
\(=\dfrac{3}{4}:\dfrac{11}{5}+\dfrac{1}{4}.\left(-7\right)+2,5\left(12+\dfrac{4}{3}\right)\)
\(=\dfrac{15}{44}-\dfrac{7}{4}+\dfrac{5}{2}.\dfrac{40}{3}\)
\(=-\dfrac{31}{22}+\dfrac{100}{3}\)
\(=\dfrac{2107}{66}\)
\(\dfrac{0.375-0.3+\dfrac{3}{11}+\dfrac{3}{12}}{-0.625+0.5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1.5+1-0.75}{2.5+\dfrac{5}{3}-1.25}\)
=\(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
=\(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)
=\(\dfrac{3}{-5}+\dfrac{3}{5}\)
=\(0\)
a) Ta có: \(\left|2.5-x\right|=1.3\)
\(\Leftrightarrow\left|x-2.5\right|=1.3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2.5=1.3\\x-2.5=-1.3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.8\\x=-1.3+2.5=1.2\end{matrix}\right.\)
Vậy: \(x\in\left\{3.8;1.2\right\}\)
b) Ta có: \(1.6-\left|x-0.2\right|=0\)
\(\Leftrightarrow\left|x-0.2\right|=1.6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0.2=1.6\\x-0.2=-1.6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.8\\x=-1.4\end{matrix}\right.\)
Vậy: \(x\in\left\{1.8;-1.4\right\}\)
c) Ta có: \(13^x=169\)
\(\Leftrightarrow13^x=13^2\)
\(\Leftrightarrow x=2\)
Vậy: x=2
d) Ta có: \(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\)
\(\Leftrightarrow\dfrac{2}{x}=\dfrac{x}{\dfrac{8}{25}}\)
\(\Leftrightarrow x^2=\dfrac{16}{25}\)
hay \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)
Vậy: \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)
\(|2,5-x|=1,3\)
\(\Rightarrow2,5-x=1,3\) hoặc -(2,5-x)=1,3
=>x=2,5-1,3 x-2,5=1,3
=>x=1,2 x=1,3+2,5=3,8
Vậy \(x\in\left\{1,2;3,8\right\}\)
\(1,6-|x-0,2|=0\Rightarrow|x-0,2|=1,6\)
=> x-0,2=1,6 hoặc -(x-0,2)=1,6
=>x=1,6+0,2 0,2-x=1,6
=>x=3,8 x=0,2-1,6=-1,4
Vậy \(x\in\left\{3,8;-1,4\right\}\)
\(13^x=169\Rightarrow13^x=13^2\Rightarrow x=2\)
\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\Rightarrow-2\times\dfrac{8}{25}=-x\times x\Rightarrow\dfrac{-16}{25}=-x^2\Rightarrow\dfrac{16}{25}=x^2\Rightarrow x^2=\left(\dfrac{4}{5}\right)^2=\left(-\dfrac{4}{5}\right)^2\)
\(\Rightarrow x=\dfrac{4}{5}\) hoặc \(x=-\dfrac{4}{5}\)
\(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\\ A=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\\ A=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}=\dfrac{3\left(0,125-0,1+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(0,125-0,1+\dfrac{5}{11}+\dfrac{5}{12}\right)}+\dfrac{\dfrac{3}{5}\left(2,5+\dfrac{5}{3}-1,25\right)}{2,5+\dfrac{5}{3}-1,25}=-\dfrac{3}{5}+\dfrac{3}{5}=0\)
\(C=\dfrac{1\dfrac{1}{2}+0,6-\dfrac{3}{7}}{2,5+1-\dfrac{35}{49}}=\dfrac{\dfrac{3}{2}+\dfrac{3}{5}-\dfrac{3}{7}}{\dfrac{5}{2}+\dfrac{5}{5}-\dfrac{5}{7}}=\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{7}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{7}\right)}=\dfrac{3}{5}\)
0,5V1+3V2=2,5(V1+V2)
=> 2V1=0,5V2 (có dạng a.d=b.c)
=> \(\dfrac{V_1}{V_2}=\dfrac{0,5}{2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow a.d=b.c\) đó bạn camcon