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1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
\(=0.75-\dfrac{7}{3}-0.75+9\cdot\left(-\dfrac{1}{9}\right)=-\dfrac{7}{3}-1=-\dfrac{10}{3}\)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
=\(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
=\(\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
=\(\left(\dfrac{5}{10}+\dfrac{8}{10}\right)+0\)
=\(\dfrac{13}{10}\)
\(-\dfrac{7}{25}.\dfrac{11}{13}+\left(-\dfrac{7}{25}\right).\dfrac{2}{13}-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.1-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}-\dfrac{18}{25}\)
=\(-\dfrac{25}{25}\) = \(-1\)
4:
a: =4/15-2,9+11/15=1-2,9=-1,9
b: \(=-36,75+3,7-63,25+6,3=10-100=-90\)
c: \(=6,5+3,5-\dfrac{10}{17}-\dfrac{7}{17}=10-1=9\)
d: \(=\dfrac{13}{25}\left(-39,1-60,9\right)=\dfrac{13}{25}\left(-100\right)=-52\)
e: =-5/12-7/12-3,7-6,3=-1-10=-11
f: =2,8(-6/13-7/13)-7,2=-2,8-7,2=-10
a) \(-\frac{7}{2}:\left(3-x\right)-0,75=\frac{1}{4}\)
\(\Leftrightarrow\frac{-7}{2\left(3-x\right)}=\frac{1}{4}+0,75=0,25+0,75=1\)
\(\Leftrightarrow2\left(3-x\right)=-7\)
\(\Leftrightarrow3-x=-\frac{7}{2}\)
\(\Leftrightarrow x=3+\frac{7}{2}\)
\(\Leftrightarrow x=\frac{13}{2}\)
b) \(\left|2x-\frac{4}{3}\right|-1\frac{1}{3}=-\frac{8}{9}\)
\(\Leftrightarrow\left|2x-\frac{4}{3}\right|=\frac{4}{3}-\frac{8}{9}=\frac{4}{9}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{4}{3}=\frac{4}{9}\\2x-\frac{4}{3}=-\frac{4}{9}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{4}{3}+\frac{4}{9}=\frac{16}{9}\\2x=\frac{4}{3}-\frac{4}{9}=\frac{8}{9}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{9}\\x=\frac{4}{9}\end{cases}}\)
\(\left(\dfrac{-2}{3}\right).0,75+\dfrac{1}{\dfrac{2}{3}}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)\)
\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}:\left(\dfrac{-8}{18}\right)+\left(\dfrac{-9}{18}\right)\)
\(=\dfrac{-1}{2}+\dfrac{5}{3}.\dfrac{-18}{8}+\dfrac{-9}{18}\)
\(=\dfrac{-1}{2}+\dfrac{-15}{4}+\dfrac{-9}{18}\)
\(=\dfrac{-72}{144}+\dfrac{-540}{144}+\dfrac{-72}{144}\)
\(=\dfrac{-684}{144}\)
\(=\dfrac{-19}{4}.\)