2,35 : (y - 15,3) = \(\dfrac{1}{2}\)
y = ?
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y - 2,35 = 1,5 x 3,2
=> y - 2,35 = 4,8
=> y = 4,8 + 2,35
=> y = 7,15
\(y\cdot\frac{4}{9}=\frac{1}{5}\div\frac{3}{5}\)
\(\Rightarrow y\cdot\frac{4}{9}=\frac{1}{3}\)
\(\Rightarrow y=\frac{1}{3}\div\frac{4}{9}\)
\(\Rightarrow y=\frac{3}{4}\)
B)
B = \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{29.31}\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...\dfrac{1}{29.31}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{29}-\dfrac{1}{31}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{31}\right)\)
= \(\dfrac{1}{2}.\dfrac{30}{31}\)
= \(\dfrac{30}{62}\) = \(\dfrac{15}{31}\)
a: Sửa đề: \(\left[\left(6\dfrac{3}{7}-\dfrac{0,75x-2}{0,35}\right)\cdot2.8+1.75\right]\cdot0.05=2.35\)
\(\Leftrightarrow\left(\dfrac{45}{7}-\dfrac{15x-40}{7}\right)\cdot2.8+1,75=47\)
\(\Leftrightarrow\dfrac{45-15x+40}{7}\cdot2,8=45,25\)
\(\Leftrightarrow\dfrac{85-15x}{7}=\dfrac{905}{56}\)
=>8(85-15x)=905
=>680-120x=905
=>120x=-225
hay x=-15/8
b: \(\dfrac{3}{5}-\dfrac{2}{7}< \dfrac{2}{3}x+\dfrac{3}{4}< \dfrac{1}{2}+\dfrac{7}{9}\)
\(\Leftrightarrow\dfrac{11}{35}-\dfrac{3}{4}< \dfrac{2}{3}x< \dfrac{23}{18}-\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{61}{140}< \dfrac{2}{3}x< \dfrac{19}{36}\)
\(\Leftrightarrow-\dfrac{183}{280}< x< \dfrac{19}{24}\)
Y + 156,8% = 15,3 - 3,7
Y + 1,568 = 11,6
Y = 11,6 - 1,568
Y = 10,032
bạn ấy làm sai rồi
đáp số là 75/149
nếu mình làm đúng thì nhớ
nhé ! hihi!
=> (15,3+14,5)xy=15
29,8xy=15
=> y=15:29,8=\(\frac{15}{29,8}\)
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
\(2,35:\left(y-15,3\right)=\dfrac{1}{2}\)
\(y-15,3=2,35:\dfrac{1}{2}\)
\(y-15,3=4,7\)
\(y=4,7+15,3\)
\(y=20\)
\(2,35:\left(y-15,3\right)=\dfrac{1}{2}\)
\(y-15,3=2,35:\dfrac{1}{2}\)
\(y-15,3=4,7\)
\(y=4,7+15,3\)
\(y=20\)