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\(10A=\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)
\(10B=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
Vì \(10^{12}-1>10^{11}+1\)
nên \(-\dfrac{9}{10^{12}-1}>-\dfrac{9}{10^{11}+1}\)
hay A>B
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)
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, \(\dfrac{17}{7}-\dfrac{6}{5}x=\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{17}{7}-\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{-51}{28}\)
\(\Rightarrow x=\dfrac{-51}{28}:\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{-85}{56}\)
\(\dfrac{x-2}{3}=\dfrac{-1}{2y+1}\)
\(\Leftrightarrow\left(x-2\right)\left(2y+1\right)=-1.3\)
\(\left(x-2\right)\left(2y+1\right)=-3\)
\(\Leftrightarrow x-2;2y+1\inƯ\left(-3\right)\)
\(Ư\left(-3\right)=\left\{\pm1;\pm3\right\}\)
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!!!
\(\left(x-2\right)\left(x-4\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2< 0\\x-4>0\end{matrix}\right.=>4< x< 2\left(1\right)\\\left\{{}\begin{matrix}x-2>0\\x-4< 0\end{matrix}\right.=>2< x< 4\left(2\right)}\end{matrix}\right.\)(1 ) vô lý=> loại
=> (x-2).(x-4)<0 <=> 2<x<4
b. ta có\(x^2+1>0\forall x\)
=>(x2 -1).(x2+1)<0 <=> (x2 -1)<0 <=> x2<1
<=> -1<x<1
câu c bạn làm tương tự
\(\left(\dfrac{2}{3}+x\right)\left(\dfrac{3}{5}-2x\right)=10\)
\(\Leftrightarrow\dfrac{2}{3}+x;\dfrac{3}{5}-2x\inƯ\left(10\right)\)
\(Ư\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
\(\Leftrightarrow\dfrac{2}{3}+x=1\Rightarrow x=\dfrac{1}{3}\)
\(\dfrac{3}{5}-2x=10\Rightarrow2x=\dfrac{-47}{5}\Rightarrow x=\dfrac{-47}{10}\)
\(\Leftrightarrow\dfrac{2}{3}+x=-1\Rightarrow x=\dfrac{-5}{3}\)
\(\dfrac{3}{5}-2x=-10\Rightarrow2x=\dfrac{53}{5}\Rightarrow x=\dfrac{53}{10}\)
\(\Leftrightarrow\dfrac{2}{3}+x=2\Rightarrow x=\dfrac{4}{3}\)
\(\dfrac{3}{5}-2x=5\Rightarrow2x=\dfrac{-22}{5}\Rightarrow x=\dfrac{-11}{5}\)
\(\Leftrightarrow\dfrac{2}{3}+x=-2\Rightarrow x=\dfrac{-8}{3}\)
\(\dfrac{3}{5}-2x=-5\Rightarrow2x=\dfrac{28}{5}\Rightarrow x=\dfrac{14}{5}\)
\(\Leftrightarrow\dfrac{2}{3}+x=5\Rightarrow x=\dfrac{13}{3}\)
\(\dfrac{3}{5}-2x=2\Rightarrow2x=\dfrac{-7}{5}\Rightarrow x=\dfrac{-7}{10}\)
\(\Leftrightarrow\dfrac{2}{3}+x=-5\Rightarrow x=\dfrac{-17}{3}\)
\(\dfrac{3}{5}-2x=-2\Rightarrow2x=\dfrac{13}{5}\Rightarrow x=\dfrac{13}{10}\)
\(\Leftrightarrow\dfrac{2}{3}+x=10\Rightarrow x=\dfrac{28}{3}\)
\(\dfrac{3}{5}-2x=1\Rightarrow2x=\dfrac{-2}{5}\Rightarrow x=\dfrac{-1}{5}\)
\(\Leftrightarrow\dfrac{2}{3}+x=-10\Rightarrow x=\dfrac{-32}{3}\)
\(\dfrac{3}{5}-2x=-1\Rightarrow2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\)
Mk hơi băn khoăn chỗ này . Tại sao mà trong 1 trường hợp mà x lại khác nhau ?
\(6\dfrac{2}{9}.x+3\dfrac{10}{27}=22\dfrac{1}{7}\)
\(\dfrac{56}{9}.x+\dfrac{91}{27}=\dfrac{155}{7}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{155}{7}-\dfrac{91}{27}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{4185}{189}-\dfrac{637}{189}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{3548}{189}\)
\(x\) \(=\dfrac{3548}{189}:\dfrac{56}{9}\)
\(x\) \(=\dfrac{3548}{189}.\dfrac{9}{56}\)
\(x\) \(=\dfrac{887}{294}\)
Vậy \(x\\\) \(=\dfrac{887}{294}\)
Chúc bạn học tốt