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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) \(2x-\frac{2}{3}-7x=\frac{3}{2}-1\\ 2x-7x-\frac{2}{3}=\frac{1}{2}\\ -5x=\frac{1}{2}+\frac{2}{3}\\ -5x=\frac{7}{6}\\ x=\frac{7}{6}:\left(-5\right)\\ x=\frac{-7}{30}\)Vậy \(x=\frac{-7}{30}\)
b) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\\ \frac{3}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{1}{4}\\ \frac{7}{6}x=\frac{3}{20}\\ x=\frac{3}{20}:\frac{7}{6}\\ x=\frac{9}{70}\)Vậy \(x=\frac{9}{70}\)
c) \(\frac{2}{3}-\frac{5}{3}x=\frac{7}{10}x+\frac{5}{6}\\ \frac{2}{3}-\frac{5}{6}=\frac{7}{10}x+\frac{5}{3}x\\ \frac{-1}{6}=\frac{71}{30}x\\ x=\frac{-1}{6}:\frac{71}{30}\\ x=\frac{-5}{71}\)Vậy \(x=\frac{-5}{71}\)
d) \(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\\ 2x+\frac{1}{2}x=\frac{5}{6}+\frac{1}{4}\\ \frac{5}{2}x=\frac{13}{12}\\ x=\frac{13}{12}:\frac{5}{2}\\ x=\frac{13}{30}\)Vậy \(x=\frac{13}{30}\)
e) \(3x-\frac{5}{3}=x-\frac{1}{4}\\ 3x-x=\frac{5}{3}-\frac{1}{4}\\ 2x=\frac{17}{12}\\ x=\frac{17}{12}:2\\ x=\frac{17}{24}\)Vậy \(x=\frac{17}{24}\)
Èo, chăm thế? Chăm hơn cả mik cơ, gần 11 h rồi onl thì thấy bài được bạn HISI làm hết rồi :((
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, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4
a) \(\frac{5}{6}=\frac{x-1}{x}\)
\(5x=6x-6\)
\(6x-5x=6\)
\(x=6\)
các câu còn lại lm tương tự
hok tốt!!
b) \(\frac{1}{2}=\frac{x+1}{3x}\)
\(\Rightarrow1.3x=2.\left(x+1\right)\)
\(3x=2x+2\)
\(3x-2x=2\)
\(x=2\)
Vậy x=2
các câu khác bạn làm tương tự
Bạn ơi ; tách từng bài ra cho dễ làm :
1.7C-C= 7^2016-7
C = ( 7^2016-7 ) :6
\(C=7+7^2+7^3+.....+7^{2016}\)
\(\Rightarrow7C=7^2+7^3+7^4+...+7^{2017}\)
\(\Rightarrow7C-C=\left(7^2+7^3+.....+7^{2017}\right)-\left(7+7^2+7^3+....+7^{2016}\right)\)
\(\Rightarrow6C=2^{2017}-7\)
\(\Rightarrow C=\frac{2^{2017}-7}{6}\)
a) Ta có:
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
b)Ta có:
\(\frac{5}{8x-2}=\frac{-4}{7-x}\)
\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)
\(\Rightarrow35-5x=-32x+8\)
\(\Rightarrow-5x+32x=8-35\)
\(\Rightarrow27x=-27\)
\(\Rightarrow x=-1\)
c) Ta có:
\(\frac{4}{3}=\frac{2x-1}{x}\)
\(\Rightarrow4x=3\left(2x-1\right)\)
\(\Rightarrow4x=6x-3\)
\(\Rightarrow3=6x-4x=2x\)
\(\Rightarrow x=\frac{3}{2}\)
d)Ta có:
\(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)
\(\Rightarrow8x-4=9x+3\)
\(\Rightarrow8x-9x=3+4\)
\(\Rightarrow-x=7\Rightarrow x=-7\)
e)Ta có:
\(\frac{4}{x+2}=\frac{7}{3x+1}\)
\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)
\(\Rightarrow12x+4=7x+14\)
\(\Rightarrow12x-7x=14-4\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
f)Ta có:
\(\frac{-3}{x+1}=\frac{4}{2-2x}\)
\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)
\(\Rightarrow-6+6x=4x+4\)
\(\Rightarrow6x-4x=4+6\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
\(a.\dfrac{1}{2}-3x=-\dfrac{2}{5}\\ 3x=\dfrac{1}{2}+\dfrac{2}{5}\\ 3x=\dfrac{9}{10}\\ x=\dfrac{9}{10}:3\\ x=\dfrac{3}{10}\\ b.-x+\dfrac{1}{2}=-\dfrac{5}{6}\\ x=\dfrac{1}{2}+\dfrac{5}{6}\\ x=\dfrac{4}{3}\\ c.x+\dfrac{3}{5}=\left(-\dfrac{2}{5}\right)^2\\ x+\dfrac{3}{5}=\dfrac{4}{25}\\ x=\dfrac{4}{25}-\dfrac{3}{5}\\ x=-\dfrac{11}{25}\\ d.\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\\ \dfrac{1}{7}:x=\dfrac{3}{14}-\dfrac{3}{7}=-\dfrac{3}{14}\\ x=\dfrac{1}{7}:-\dfrac{3}{14}=-\dfrac{2}{3}\\ e.-\dfrac{1}{3}\left(\dfrac{1}{7}-x\right)=\dfrac{1}{21}\\ \dfrac{1}{7}-x=\dfrac{1}{21}:-\dfrac{1}{3}=-\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{2}{7}\\ h.\dfrac{1}{4}-3x+\dfrac{3}{2}=-0,75\\ \dfrac{1}{4}-3x+\dfrac{3}{2}=-\dfrac{3}{4}\\ 3x=\dfrac{1}{4}+\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{5}{2}\\ x=\dfrac{5}{2}:3\\ x=\dfrac{5}{6}\\ i.\dfrac{2}{7}-\left(\dfrac{2}{3}+2x\right)=\dfrac{5}{7}\\ \dfrac{2}{3}+2x=\dfrac{2}{7}-\dfrac{5}{7}=-\dfrac{3}{7}\\ 2x=-\dfrac{3}{7}-\dfrac{2}{3}=-\dfrac{23}{21}\\ x=\dfrac{-23}{21}:2=-\dfrac{23}{42}\)
a: \(\dfrac{1}{2}-3x=-\dfrac{2}{5}\)
=>\(3x=\dfrac{1}{2}+\dfrac{2}{5}=\dfrac{5}{10}+\dfrac{4}{10}=\dfrac{9}{10}\)
=>\(x=\dfrac{9}{10}:3=\dfrac{9}{30}=\dfrac{3}{10}\)
b: \(-x+\dfrac{1}{2}=-\dfrac{5}{6}\)
=>\(-x=-\dfrac{5}{6}-\dfrac{1}{2}=-\dfrac{5}{6}-\dfrac{3}{6}=-\dfrac{8}{6}=-\dfrac{4}{3}\)
=>\(x=\dfrac{4}{3}\)
c: \(x+\dfrac{3}{5}=\left(-\dfrac{2}{5}\right)^2\)
=>\(x+\dfrac{3}{5}=\dfrac{4}{25}\)
=>\(x=\dfrac{4}{25}-\dfrac{3}{5}=\dfrac{4}{25}-\dfrac{15}{25}=-\dfrac{11}{25}\)
d: \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)
=>\(\dfrac{1}{7}:x=\dfrac{3}{14}-\dfrac{3}{7}=-\dfrac{3}{14}\)
=>\(x=-\dfrac{1}{7}:\dfrac{3}{14}=-\dfrac{1}{7}\cdot\dfrac{14}{3}=-\dfrac{2}{3}\)
e: \(-\dfrac{1}{3}\left(\dfrac{1}{7}-x\right)=\dfrac{1}{21}\)
=>\(\dfrac{1}{7}-x=\dfrac{1}{21}:\dfrac{-1}{3}=\dfrac{-1}{21}\cdot3=-\dfrac{1}{7}\)
=>\(x=\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{2}{7}\)
h: \(\dfrac{1}{4}-3x+\dfrac{3}{2}=-0,75\)
=>\(-3x+\dfrac{5}{4}=-\dfrac{3}{4}\)
=>\(-3x=-\dfrac{3}{4}-\dfrac{5}{4}=-\dfrac{8}{4}=-2\)
=>\(x=\dfrac{-2}{-3}=\dfrac{2}{3}\)
i: \(\dfrac{2}{7}-\left(\dfrac{2}{3}+2x\right)=\dfrac{5}{7}\)
=>\(2x+\dfrac{2}{3}=\dfrac{2}{7}-\dfrac{5}{7}=-\dfrac{3}{7}\)
=>\(2x=-\dfrac{3}{7}-\dfrac{2}{3}=-\dfrac{9}{21}-\dfrac{14}{21}=-\dfrac{23}{21}\)
=>\(x=-\dfrac{23}{21}:2=-\dfrac{23}{42}\)