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a: -3/100=-9/300; -2/3=-200/300
=>-3/100>-2/3
b: -3/5=-9/15
-2/3=-10/15
=>-3/5>-2/3
c: -5/4<-1<-3/8
d: -2/3=-8/12; -3/4=-9/12
=>-2/3>-3/4
e: -267/268>-1
-1>-1347/1343
=>-267/268>-1347/1343
a) Ta có : \(\dfrac{-1}{5}< 0< \dfrac{1}{1000}\)
\(\Rightarrow\dfrac{-1}{5}< \dfrac{1}{1000}\)
b) Ta có : \(\dfrac{267}{268}< 1< \dfrac{1347}{1343}\)
=> \(\dfrac{267}{-268}< -\dfrac{1347}{1343}\)
c) \(\dfrac{13}{38}>\dfrac{13}{39}=\dfrac{1}{3}=\dfrac{19}{87}>\dfrac{29}{88}\)
=> \(-\dfrac{13}{38}< \dfrac{29}{-88}\)
d) \(\dfrac{181818}{313131}=\dfrac{18}{31}\)
=> \(-\dfrac{18}{31}=-\dfrac{181818}{313131}\)
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
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.
$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$
$\Leftrightarrow x=\frac{2}{5}$
Nguyễn Quý Trung:
\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
Bạn bớt 2 vế đi 1/3 thì \(x=\dfrac{2}{5}\)
1, \(x\left(x+\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)
Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)
\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)
Vậy, ...
b, \(\left|x-\dfrac{1}{3}\right|\ge0\)
Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)
\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
Vậy, ...
1)
a)
\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2)
a)
\(\left|x+\dfrac{4}{6}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)
b)
\(\left|x-\dfrac{1}{3}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)
a) Bổ sung cho đầy đủ đề
b) (3x - 1)/4 = (2x - 5)/3
3(3x - 1) = 4(2x - 5)
9x - 3 = 8x - 20
9x - 8x = -20 + 3
x = -17
c) Điều kiện: x ≠ -1/3
3/(-2) = (x - 3)/(3x + 1)
3.(3x + 1) = -2(x - 3)
9x + 3 = -2x + 6
9x + 2x = 6 - 3
11x = 3
x = 3/11 (nhận)
Vậy x = 3/11
b, \(-x-2=\dfrac{5}{4}\Rightarrow-x=\dfrac{13}{4}\Rightarrow x=-\dfrac{13}{4}\)
c, \(\dfrac{4}{3}-\left(x-\dfrac{1}{5}\right)=\left|-\dfrac{3}{10}+\dfrac{1}{2}\right|-\dfrac{1}{6}\)
\(\Rightarrow\dfrac{4}{3}-x+\dfrac{1}{5}=\left|\dfrac{1}{5}\right|-\dfrac{1}{6}\)
\(\Rightarrow-x=\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{4}{3}-\dfrac{1}{5}\)
\(\Rightarrow-x=-\dfrac{3}{2}\Rightarrow x=\dfrac{3}{2}\)
d, \(\dfrac{1}{3}-\left(\dfrac{2}{3}-x+\dfrac{5}{4}\right)=\dfrac{7}{12}-\left(\dfrac{5}{2}-\dfrac{13}{6}\right)\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{2}{3}+x-\dfrac{5}{4}=\dfrac{7}{12}-\dfrac{5}{2}+\dfrac{13}{6}\)
\(\Rightarrow x=\dfrac{7}{12}-\dfrac{5}{2}+\dfrac{13}{6}-\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{11}{6}\)
Chúc bạn học tốt!!!
2. Tìm x
a) \(5-\left|x+\dfrac{1}{2}\right|=1\)
\(\left|x+\dfrac{1}{2}\right|=5-1\)
\(\left|x+\dfrac{1}{2}\right|=4\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=4\\x+\dfrac{1}{2}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{-9}{2}\end{matrix}\right.\)
Vậy x=7/2 hoặc -9/2
b) \(\dfrac{4}{3}+\left|2-\dfrac{1}{2}x\right|=7\)
\(\left|2-\dfrac{1}{2}x\right|=7-\dfrac{4}{3}\)
\(\left|2-\dfrac{1}{2}x\right|=\dfrac{17}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}2-\dfrac{1}{2}x=\dfrac{17}{3}\\2-\dfrac{1}{2}x=\dfrac{-17}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=\dfrac{-11}{3}\\\dfrac{1}{2}x=\dfrac{23}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-22}{3}\\x=\dfrac{46}{3}\end{matrix}\right.\)
Vậy x=-22/3 hoặc x=46/4
1. So sánh:
a. \(\dfrac{-1}{5}\) và \(\dfrac{1}{1000}\)
Ta có:
\(\dfrac{-1}{5}< 0\\ \dfrac{1}{1000}>0\\ \Rightarrow\dfrac{-1}{5}< \dfrac{1}{1000}\)
b.\(\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\)
Ta có:
\(\dfrac{267}{-268}>-1\\ \dfrac{-1347}{1343}< -1\\ \Rightarrow\dfrac{267}{-286}>\dfrac{-1347}{1343}\)
c.\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)
Ta có:
\(\dfrac{-13}{38}< \dfrac{-13}{39}=\dfrac{-1}{3}\\ \dfrac{-29}{88}>\dfrac{-29}{87}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{-13}{38}< \dfrac{-29}{88}\)