2,5-\(\dfrac{1}{7}.x=-2\)
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a, \(x\) : \(\dfrac{13}{3}\) = -2,5
\(x\) = -2,5 . \(\dfrac{13}{3}\)
\(x\) = \(\dfrac{65}{6}\)
b,\(\dfrac{3}{5}\)\(x\) = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)
\(\dfrac{3}{5}x\) = \(\dfrac{-3}{20}\)
\(x\) = \(\dfrac{-3}{20}\) : \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{-1}{4}\)
c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)
\(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)
\(\dfrac{12}{13}x=2\)
\(x=2:\dfrac{12}{13}\)
\(x=\dfrac{13}{6}\)
d.\(x:4\dfrac{1}{2}=-2,5\)
\(x:\dfrac{9}{2}=-\dfrac{5}{2}\)
\(x=-\dfrac{5}{2}\times\dfrac{9}{2}\)
\(x=-\dfrac{45}{4}\)
e.\(\left(\dfrac{x}{7}+1\right):\left(-4\right)=-\dfrac{1}{28}\)
\(\dfrac{x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{x}{7}=-\dfrac{6}{7}\)
\(x=-6\)
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
2:
a: =>x=1/2+1/6=3/6+1/6=4/6=2/3
b: =>x+3,5=7,8-6,3=1,5
=>x=1,5-3,5=-2
c: =>(35-x)/-105=1/5
=>35-x=-21
=>x=56
1
c hình như tính bình thường thôi:v
d
= 1,75 : 5 + 2,5 . (16 – 4 . 4,1)
= 1,75 : 5 + 2,5 . (16 – 16,4)
= 1,75 : 5 + 2,5 . (−0,4)
= 0,35 − 1
= −0,65.
2
a
\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3}{6}+\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b
\(\Rightarrow x+\dfrac{35}{10}=\dfrac{78}{10}-\dfrac{63}{5}:2\\ \Rightarrow x+\dfrac{35}{10}=\dfrac{78}{10}-\dfrac{63}{10}=\dfrac{15}{10}\\ \Rightarrow x=\dfrac{15}{10}-\dfrac{35}{10}=-2\)
c không rõ đề:v
c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)
\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)
d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)
hay \(x=\dfrac{2}{3}\)
\(75\%:2\dfrac{1}{5}+\left(0,5\right)^2.\left(-7\right)+2,5\left(7\dfrac{2}{3}+5\dfrac{2}{3}\right)\)
\(=\dfrac{3}{4}:\dfrac{11}{5}+\dfrac{1}{4}.\left(-7\right)+2,5\left(12+\dfrac{4}{3}\right)\)
\(=\dfrac{15}{44}-\dfrac{7}{4}+\dfrac{5}{2}.\dfrac{40}{3}\)
\(=-\dfrac{31}{22}+\dfrac{100}{3}\)
\(=\dfrac{2107}{66}\)
\(A=\dfrac{7}{3}+\dfrac{5}{7}+\dfrac{2}{3}-\dfrac{7}{12}+\dfrac{5}{2}=3+\dfrac{221}{84}=\dfrac{473}{84}\)
\(=\dfrac{\left(\dfrac{13}{84}\cdot\dfrac{7}{5}-\dfrac{5}{2}\cdot\dfrac{7}{180}\right):\dfrac{43}{18}+\dfrac{9}{2}\cdot\dfrac{1}{10}}{70.5-528:7.5}\)
\(=\dfrac{\left(\dfrac{13}{60}-\dfrac{7}{72}\right)\cdot\dfrac{18}{43}+\dfrac{9}{20}}{0.1}\)
\(=10\cdot\left[\dfrac{43}{360}\cdot\dfrac{18}{43}+\dfrac{9}{20}\right]\)
\(=10\cdot\left[\dfrac{1}{20}+\dfrac{9}{20}\right]=10\cdot\dfrac{10}{20}=5\)
d. \(\sqrt{9x^2+12x+4}=4\)
<=> \(\sqrt{\left(3x+2\right)^2}=4\)
<=> \(|3x+2|=4\)
<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow x=1\)
\(2,5-\dfrac{1}{7}x=-2\)
\(\Rightarrow\dfrac{5}{2}-\dfrac{1}{7}x=-2\)
\(\Rightarrow\dfrac{1}{7}x=\dfrac{5}{2}-\left(-2\right)\)
\(\Rightarrow\dfrac{1}{7}x=\dfrac{9}{2}\)
\(\Rightarrow x=\dfrac{63}{2}\)