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3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
\(1,3.\dfrac{15}{39}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
\(=\dfrac{13}{10}.\dfrac{15}{39}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(=\dfrac{1}{2}-\dfrac{2}{3}=-\dfrac{1}{6}\)
\(10A=\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)
\(10B=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
Vì \(10^{12}-1>10^{11}+1\)
nên \(-\dfrac{9}{10^{12}-1}>-\dfrac{9}{10^{11}+1}\)
hay A>B
\(6\dfrac{2}{9}.x+3\dfrac{10}{27}=22\dfrac{1}{7}\)
\(\dfrac{56}{9}.x+\dfrac{91}{27}=\dfrac{155}{7}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{155}{7}-\dfrac{91}{27}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{4185}{189}-\dfrac{637}{189}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{3548}{189}\)
\(x\) \(=\dfrac{3548}{189}:\dfrac{56}{9}\)
\(x\) \(=\dfrac{3548}{189}.\dfrac{9}{56}\)
\(x\) \(=\dfrac{887}{294}\)
Vậy \(x\\\) \(=\dfrac{887}{294}\)
Chúc bạn học tốt
\(\dfrac{x-2}{3}=\dfrac{-1}{2y+1}\)
\(\Leftrightarrow\left(x-2\right)\left(2y+1\right)=-1.3\)
\(\left(x-2\right)\left(2y+1\right)=-3\)
\(\Leftrightarrow x-2;2y+1\inƯ\left(-3\right)\)
\(Ư\left(-3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng sau:
| x-2 | 2y+1 | x | y |
| 1 | -3 | 3 | -2 |
| -1 | 3 | 1 | 1 |
| 3 | -1 | 5 | -1 |
| -3 | 1 | -1 | 0 |
Ta có:
\(\dfrac{x-2}{3}=\dfrac{-1}{2y+1}\)
\(\Rightarrow\left(x-2\right).\left(2y+1\right)=-3\)
\(\Rightarrow\left(x-2\right).\left(2y+1\right)\inƯ\left(-3\right)\)
\(\Rightarrow x-2;2y+1\in\left\{-3;-1;1;3\right\}\)
Ta có bảng sau:
| \(x-2\) | -3 | -1 | 1 | 3 |
| \(2y+1\) | 1 | 3 | -3 | -1 |
| x | -1 | 1 | 3 | 5 |
| y | 0 | 1 | -2 | -1 |
| Chọn or loại | Chọn | Chọn | Chọn | Chọn |
Vậy \(\left(x;y\right)\in\left\{\left(-1;0\right);\left(1;1\right);\left(3;-2\right);\left(5;-1\right)\right\}\)
Chúc bạn học tốt!!!
a, \(x+\dfrac{2}{3}=0,2\)
\(\Rightarrow x+\dfrac{2}{3}=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{-7}{10}\)
b, \(\dfrac{17}{7}-\dfrac{6}{5}x=\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{17}{7}-\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{-51}{28}\)
\(\Rightarrow x=\dfrac{-51}{28}:\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{-85}{56}\)
\(a)\) \(x+\dfrac{2}{3}=0,2\)
\(\Rightarrow x+\dfrac{2}{3}=\dfrac{1}{5}\)
\(\Rightarrow x=-\dfrac{7}{15}\)
Vậy ...
\(b)\) \(\dfrac{17}{7}-\dfrac{6}{5}x=\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{-51}{28}\)
\(\Rightarrow x=-\dfrac{85}{56}\)
Vậy ...