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c) Ta có: \(\left|x-\dfrac{2}{3}\right|+2.25=\dfrac{3}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{3}\right|=\dfrac{3}{4}-\dfrac{9}{4}=\dfrac{-3}{2}\)(vô lý)
Vậy: \(x\in\varnothing\)
a) Ta có: \(x+\dfrac{-3}{7}=\dfrac{4}{7}\)
\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{4}{7}\)
hay x=1
Vậy: x=1
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)
\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)
\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)
a) \(\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}\)
\(=\left(-\dfrac{3}{5}+\dfrac{-4}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}\)
\(=0+\dfrac{7}{21}\)
\(=\dfrac{7}{21}\)
b) `-3/17 + ( 2/3 + 3/17)`
` = -3/17 + 2/3 + 3/17`
` = 2/3 + ( -3/17 +3/17)`
` = 2/3 + 0`
` = 2/3`
c)
` -5/21 + ( -16/21 +1)`
\(=\dfrac{-5}{21}+\dfrac{-16}{21}+1\)
\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)
\(=-1+1=0\)
Bài 2:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
a: Sửa đề: \(\left[\left(6\dfrac{3}{7}-\dfrac{0,75x-2}{0,35}\right)\cdot2.8+1.75\right]\cdot0.05=2.35\)
\(\Leftrightarrow\left(\dfrac{45}{7}-\dfrac{15x-40}{7}\right)\cdot2.8+1,75=47\)
\(\Leftrightarrow\dfrac{45-15x+40}{7}\cdot2,8=45,25\)
\(\Leftrightarrow\dfrac{85-15x}{7}=\dfrac{905}{56}\)
=>8(85-15x)=905
=>680-120x=905
=>120x=-225
hay x=-15/8
b: \(\dfrac{3}{5}-\dfrac{2}{7}< \dfrac{2}{3}x+\dfrac{3}{4}< \dfrac{1}{2}+\dfrac{7}{9}\)
\(\Leftrightarrow\dfrac{11}{35}-\dfrac{3}{4}< \dfrac{2}{3}x< \dfrac{23}{18}-\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{61}{140}< \dfrac{2}{3}x< \dfrac{19}{36}\)
\(\Leftrightarrow-\dfrac{183}{280}< x< \dfrac{19}{24}\)
a) Ta có: \(2\dfrac{3}{3}\cdot4\cdot\left(-0.4\right)+1\dfrac{3}{5}\cdot1.75+\left(-7.2\right):\dfrac{9}{11}\)
\(=-4.8+\dfrac{8}{5}\cdot\dfrac{7}{4}-\dfrac{36}{5}\cdot\dfrac{11}{9}\)
\(=\dfrac{-24}{5}+\dfrac{14}{5}-\dfrac{44}{5}\)
\(=\dfrac{-54}{5}\)
b) Ta có: \(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}\cdot\left(-5\right)^0\)
\(=\left(\dfrac{2}{48}-\dfrac{15}{48}\right)\cdot\dfrac{8}{-3}+1\cdot1\)
\(=\dfrac{-13}{48}\cdot\dfrac{-8}{3}+1\)
\(=\dfrac{13}{18}+\dfrac{18}{18}=\dfrac{31}{18}\)
=1,75.(−1621)−(133+2,25):15860
=−4,3−7912:7930
=−43−52
à mk nhầm, cái này mới đúng:
=1,75.(−1621)−(133+2,25):15860G=1,75.(−1621)−(133+2,25):15860
=−4,3−7912:7930G=−4,3−7912:7930
=−43−52G=−43−52
=−236