tìm phân số a/b
a)a/b-1/2=4/7
b)a/b-1/4-1/3=5/6
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a) \(\frac{3}{4}-\frac{1}{6}-\frac{a}{b}=\frac{1}{2}\)
\(\frac{7}{12}-\frac{a}{b}=\frac{1}{2}\)
\(\frac{a}{b}=\frac{7}{12}-\frac{1}{2}\)
\(\frac{a}{b}=\frac{1}{12}\)
b) \(\frac{a}{b}\times\frac{1}{4}\times\frac{2}{5}=\frac{1}{7}\)
\(\frac{a}{b}\times\frac{1}{10}=\frac{1}{7}\)
\(\frac{a}{b}=\frac{1}{7}:\frac{1}{10}\)
\(\frac{a}{b}=\frac{10}{7}\)
c) \(\frac{1}{3}:\frac{a}{b}=\frac{2}{3}:\frac{4}{3}\)
\(\frac{1}{3}:\frac{a}{b}=\frac{1}{2}\)
\(\frac{a}{b}=\frac{1}{3}:\frac{1}{2}\)
\(\frac{a}{b}=\frac{3}{2}\)
Bài 1:
a) \(2\)\(\dfrac{2}{3}\)\(=\dfrac{8}{3}\)
b) \(1\)\(\dfrac{5}{7}=\dfrac{12}{7}\)
a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)
\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)
Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)
Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)
- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;
a, \(\dfrac{a}{b}=\dfrac{4}{7}+\dfrac{1}{2}=\dfrac{8+7}{14}=\dfrac{15}{14}\)
b, \(\dfrac{a}{b}=\dfrac{5}{6}+\dfrac{1}{4}+\dfrac{1}{3}=\dfrac{10+3+4}{12}=\dfrac{17}{12}\)
\(a:\dfrac{a}{b}-\dfrac{1}{2}=\dfrac{4}{7}\\ \dfrac{a}{b}=\dfrac{4}{7}+\dfrac{1}{2}\\ \dfrac{a}{b}=\dfrac{8}{14}+\dfrac{7}{14}\\ \dfrac{a}{b}=\dfrac{15}{14}\\ b:\dfrac{a}{b}-\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{5}{6}\\ \dfrac{a}{b}-\left(\dfrac{1}{4}+\dfrac{1}{3}\right)=\dfrac{5}{6}\\ \dfrac{a}{b}-\dfrac{7}{12}=\dfrac{5}{6}\\ \dfrac{a}{b}=\dfrac{5}{6}+\dfrac{7}{12}\\ \dfrac{a}{b}=\dfrac{17}{12}\)