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a)\(x-\dfrac{3}{4}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{3}{4}\)
\(x=-\dfrac{1}{4}\)
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
a) \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x=-\dfrac{5}{6}+\dfrac{7}{3}=\dfrac{3}{2}\\ =>x=\dfrac{3}{2}:\dfrac{-1}{4}=-6\)
b) \(\left|x-\dfrac{1}{6}\right|+-\dfrac{5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\\ < =>\left|x-\dfrac{1}{6}\right|=\left(\dfrac{4}{7}.\dfrac{14}{48}\right)-\left(-\dfrac{5}{12}\right)=\dfrac{1}{6}+\dfrac{5}{12}=\dfrac{7}{12}\\ \)
Xảy ra 2 trường hợp:
+) TH1: \(x-\dfrac{1}{6}=\dfrac{7}{12}\\ =>x=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{3}{4}->\left(a\right)\)
+) TH2" \(-\left(x-\dfrac{1}{6}\right)=\dfrac{7}{12}\\ < =>-x+\dfrac{1}{6}=\dfrac{7}{12}\\ < =>-x=\dfrac{7}{12}-\dfrac{1}{6}=\dfrac{5}{12}\\ =>x=-\dfrac{5}{12}->\left(b\right)\)
Từ (a) và (b) => \(x\in\left\{-\dfrac{5}{12};\dfrac{3}{4}\right\}\)
a, \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{-1}{4}x=\dfrac{-5}{6}+\dfrac{7}{3}\)
\(\Rightarrow\dfrac{-1}{4}x=\dfrac{3}{2}\Rightarrow x=-6\)
Vậy \(x=-6\)
b, \(\left|x-\dfrac{1}{6}\right|+\dfrac{-5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|-\dfrac{5}{12}=\dfrac{1}{6}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|=\dfrac{7}{12}\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{-7}{12}\\x-\dfrac{1}{6}=\dfrac{7}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{12}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-5}{12};\dfrac{3}{4}\right\}\)
Chúc bạn học tốt!!!
1. Tìm \(x\):
a) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
\(\dfrac{x}{5}=\dfrac{1}{5}\)
\(\Rightarrow x=1\)
b) \(\dfrac{-5}{6}-x=\dfrac{7}{12}-\dfrac{1}{3}.x\)
\(\dfrac{-5}{6}-\dfrac{7}{12}=x-\dfrac{1}{3}.x\)
\(x-\dfrac{1}{3}.x=\dfrac{-17}{12}\)
\(\dfrac{2}{3}.x=\dfrac{-17}{12}\)
\(x=\dfrac{-17}{12}:\dfrac{2}{3}\)
\(x=\dfrac{-17}{8}\)
c) \(2016^3.2016^x=2016^8\)
\(2016^x=2016^8:2016^3\)
\(2016^x=2016^{8-3}\)
\(2016^x=2016^5\)
\(\Rightarrow x=5\)
d) \(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=3\dfrac{1}{2}\)
\(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=\dfrac{7}{2}\)
\(\left(x+\dfrac{3}{4}\right)=\dfrac{7}{2}.\dfrac{5}{2}\)
\(x+\dfrac{3}{4}=\dfrac{35}{4}\)
\(x=\dfrac{35}{4}-\dfrac{3}{4}\)
\(x=\dfrac{32}{4}=8\)
e) \(\left(2,8.x-2^5\right):\dfrac{2}{3}=3^2\)
\(\left(2,8.x-2^5\right)=9.\dfrac{2}{3}\)
\(2,8.x-2^5=6\)
\(2,8.x=6+32\)
\(2,8.x=38\)
\(x=38:2,8\)
\(x=\dfrac{95}{7}\)
f) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{2}{5}\)
\(\dfrac{4}{7}.x=\dfrac{2}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}:\dfrac{4}{7}\)
\(x=\dfrac{28}{15}\)
g) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)
\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
\(\Rightarrow3x=-6\)
\(x=\left(-6\right):3\)
\(x=-2\)
2. Thực hiện phép tính:
a) \(\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}+1\dfrac{4}{5}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{3}+1\right)-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{1}{2}.\dfrac{5}{3}-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{9}{5}\)
\(=\dfrac{7}{18}+\dfrac{9}{5}\)
\(=\dfrac{197}{90}\)
b) \(\dfrac{7.5^2-7^2}{7.24+21}\)
\(=\dfrac{7.25-7.7}{7.24+7.3}\)
\(=\dfrac{7.\left(25-7\right)}{7.\left(24+3\right)}\)
\(=\dfrac{7.18}{7.27}\)
\(=\dfrac{2}{3}\)
c) \(\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{-4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)
b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)
c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)
d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)
lưu ý mk ko chép đầu bài
mình cần gấp lắm đến chiều mai là phải nộp rùi
giúp mình nha thanks cá bạn trước 
ko có tâm trạng mà cười nữa
Bài 1: Tính ( hợp lý nếu có thể )
\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)
\(=-1+1+\dfrac{2}{-5}\)
\(=0+\dfrac{2}{-5}\)
\(=\dfrac{2}{-5}\)
\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)
\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}\)
\(=\dfrac{2}{3}\)
\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)
\(=-1+1\)
\(=0\)
\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)
\(=\dfrac{1}{6}+\dfrac{-1}{6}\)
\(=0\)
Bài 2: Tìm x,biết:
a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{2}{3}\)
\(x=\dfrac{2}{15}\)
Vậy \(x=\dfrac{2}{15}\)
b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)
\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=\dfrac{3}{3}=1\)
Vậy \(x=1\)
c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!
\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)
\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)
\(x=\dfrac{1}{44}\)
Vậy \(x=\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\)
bài 1:
a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)
\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)
\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)
\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)
c) Cho mình hỏi x ở đâu vậy ???
d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)
\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)
\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)
f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)
\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)
\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)
\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)
\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)
\(x=\dfrac{-5}{7}\)
bài 2:
Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)
\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)
Vậy \(a=-15;b=81\)
1,
x =( -12 . ( -3) ) : 2
x = 18
2,
a, -7/9 . 6/11 + (-2/9) = -14/33 + (-2/9) = -64/99
b, -4/7 : 2 = -4/7 . 1/2 = -2/7
c, 115 - (24 - 5. 3) = 115 - ( 24 - 15) = 115 - 9 = 106
d,= -3/7. (5/9 + 4/9) + 17/7 = -3/7 . 1 +17/7 = -3/7 . 17/7 = -51/49
e, ??? mình cx k biết
c)
\(\left|\dfrac{2}{3}x-\dfrac{1}{2}\right|-1=\dfrac{5}{6}\)
\(\Leftrightarrow\left|\dfrac{4x-3}{6}\right|=1+\dfrac{5}{6}=\dfrac{11}{6}\)
\(\Leftrightarrow\left|4x-3\right|=11\Leftrightarrow\left[{}\begin{matrix}4x-3=11=>x=\dfrac{11+3}{4}=\dfrac{7}{2}\\4x-3=-11=>x=\dfrac{-8}{2}=-4\end{matrix}\right.\)
