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9 tháng 6 2017

\(\left\{\dfrac{-5< 0< -0,4}{x\in Z}\right\}\Rightarrow x\in\left\{-4;-3;-2;-1\right\}\)

31 tháng 8 2017

chiu thôi khocroikhocroikhocroileo nheo qua

23 tháng 10 2021

\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+...+\left|x+\dfrac{1}{110}\right|=11x\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{6}+x+\dfrac{1}{12}+...+x+\dfrac{1}{110}=11x\)

\(\Leftrightarrow10x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{10.11}\right)=11x\)

\(\Leftrightarrow x=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(\Leftrightarrow x=1-\dfrac{1}{11}=\dfrac{10}{11}\left(tm\right)\)

28 tháng 9 2021

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

29 tháng 10 2021

3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

23 tháng 6 2017

a )

\(4\dfrac{5}{9}:2\dfrac{5}{8}-7=-\dfrac{995}{189}=-5\) ( đã làm tròn )

\(\left(3\dfrac{1}{5}:3,2+4,5.1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\) \(=\dfrac{2}{5}=0\) ( đã làm tròn )

Ta có :

\(-5< x< 0\)

\(\Rightarrow x=\left\{-4;-3;-2;-1\right\}\)

b )

\(-\dfrac{17}{21}:\left(\dfrac{5}{4}-\dfrac{2}{5}\right)=-\dfrac{20}{21}=-1\) ( đã làm tròn )

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{7}{12}=1\) ( đã làm tròn )

Ta có : \(-1< x+0,6< 1\)

\(\Rightarrow x=\left\{-1;0\right\}\)

23 tháng 6 2017

a) \(4\dfrac{5}{9}:2\dfrac{5}{8}-7< x< \left(3\dfrac{1}{5}:3,2+4,5\cdot1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>4\dfrac{5}{9}:2\dfrac{5}{8}-7\\x< \left(3\dfrac{1}{5}:3,2+4,5\cdot1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{227}{36}\\x< -\dfrac{2}{5}\end{matrix}\right.\)

\(\Leftrightarrow-\dfrac{2}{5}>x>-\dfrac{227}{36}\)

b) \(-\dfrac{17}{21}:\left(\dfrac{5}{4}-\dfrac{2}{5}\right)< x+\dfrac{4}{7}< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{4}{7}>-\dfrac{17}{21}:\left(\dfrac{5}{4}-\dfrac{2}{5}\right)\\x+\dfrac{4}{7}< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{32}{21}\\x< \dfrac{1}{84}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{84}>x>-\dfrac{32}{21}\)

27 tháng 1 2018

Với \(\forall x\) ta có :

+) \(\left|x+\dfrac{1}{2}\right|\ge0\)

+) \(\left|x+\dfrac{1}{6}\right|\ge0\)

..........................

+) \(\left|x+\dfrac{1}{110}\right|\ge0\)

\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+.........+\left|x+\dfrac{1}{110}\right|\ge0\)

\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+........+\left|x+\dfrac{1}{110}\right|=11x\)

\(\Leftrightarrow11x\ge0\)

\(\Leftrightarrow x\ge0\)

Với \(x\ge0\) thì :

+) \(\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2}\)

+) \(\left|x+\dfrac{1}{6}\right|=x+\dfrac{1}{6}\)

.....................................

+) \(\left|x+\dfrac{1}{110}\right|=x+\dfrac{1}{110}\)

\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{6}+......+x+\dfrac{1}{110}=11x\)

\(\Leftrightarrow11x+\left(\dfrac{1}{2}+\dfrac{1}{6}+........+\dfrac{1}{110}\right)=11x\)

\(\Leftrightarrow0x=\dfrac{1}{2}+\dfrac{1}{6}+....+\dfrac{1}{110}\) (vô lí)

\(\Leftrightarrow x\in\varnothing\)

AH
Akai Haruma
Giáo viên
25 tháng 7 2021

Bài 1:

a.

$|x+\frac{7}{4}|=\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)

b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$

$|2x+1|=\frac{11}{15}$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)

c.

$3x(x+\frac{2}{3})=0$

\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)

d.

$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$

$\Leftrightarrow x=\frac{2}{5}$

AH
Akai Haruma
Giáo viên
25 tháng 7 2021

Nguyễn Quý Trung:

\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)

Bạn bớt 2 vế đi 1/3 thì \(x=\dfrac{2}{5}\)

20 tháng 8 2017

Bn tách ra đi,mỏi tay lắm luôn ik,đánh máy mà.

20 tháng 8 2017

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