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a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(-\dfrac{5}{6}x=\dfrac{5}{12}\)
\(x=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)
\(3x-3.7=-\dfrac{19}{2}\)
\(3x=-5.8\)
\(x=-\dfrac{29}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
\(\dfrac{3}{4}x=\dfrac{5}{8}\)
\(x=\dfrac{5}{6}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
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: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
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) \(\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{-4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
c: \(\left|\dfrac{7}{5}x+\dfrac{2}{3}\right|=\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|\)
=>7/5x+2/3=4/3x-1/4 hoặc 7/5x+2/3=1/4-4/3x
=>1/15x=-11/12 hoặc 41/15x=-5/12
=>x=-55/4 hoặc x=-25/164
d: |7/8x+5/6|=|1/2x+5|
=>|42x+40|=|24x+240|
=>42x+40=24x+240 hoặc 42x+40=-24x-240
=>18x=200 hoặc 66x=-280
=>x=100/9 hoặc x=-140/33
a) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
x=\(\dfrac{1}{10}:-\dfrac{2}{3}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\).
b) \(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(\dfrac{2}{3}:x=-7-\dfrac{1}{3}\)
\(\dfrac{2}{3}:x=-\dfrac{22}{3}\)
\(x=\dfrac{2}{3}:-\dfrac{22}{3}\)
\(x=-\dfrac{1}{11}\)
Vậy \(x=-\dfrac{1}{11}\).
c) \(60\%x=\dfrac{1}{3}\cdot6\dfrac{1}{3}\)
\(60\%x=\dfrac{19}{9}\)
\(\dfrac{3}{5}x=\dfrac{19}{9}\)
\(x=\dfrac{19}{9}:\dfrac{3}{5}\)
\(x=\dfrac{95}{27}\)
Vậy \(x=\dfrac{95}{27}\).
d) \(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\)
\(\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\)
\(\dfrac{2}{3}-x=\dfrac{3}{20}\)
\(x=\dfrac{2}{3}-\dfrac{3}{20}\)
\(x=\dfrac{31}{60}\)
Vậy \(x=\dfrac{31}{60}\).
e) \(-2x-\dfrac{-3}{5}:\left(-0.5\right)^2=-1\dfrac{1}{4}\)
\(-2x-\dfrac{-12}{5}=-1\dfrac{1}{4}\)
\(-2x=-1\dfrac{1}{4}+\dfrac{-12}{5}\)
\(-2x=-\dfrac{73}{20}\)
\(x=-\dfrac{73}{20}:\left(-2\right)\)
\(x=\dfrac{73}{40}\)
Vậy \(x=\dfrac{73}{40}\).
tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!
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 \(a=-15;b=81\)
a: x=4/27-2/3=4/27-18/27=-14/27
b: =>3/4x-1/4x=1/6+7/3
=>1/2x=1/6+14/6=5/2
hay x=5
c: =>13/10x=7/2+5/2=6
=>x=13/10:6=13/60
d: (3x+2)(-2/5x-7)=0
=>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-35/2
a) x = 4/27 - 2/3
x = -14/27