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Q
2 tháng 4 2017
a) \(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\)
\(\dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\)
\(\left(\dfrac{1}{2}-\dfrac{2}{3}\right)x=\dfrac{7}{12}\)
\(\dfrac{-1}{6}x=\dfrac{7}{12}\)
\(x=\dfrac{7}{12}:\dfrac{-1}{6}\)
\(x=\dfrac{-7}{2}\)
b) \(x:4\dfrac{1}{3}=-2,5\)
\(x:\dfrac{13}{3}=\dfrac{-5}{2}\)
\(x=\dfrac{13}{3}.\dfrac{-5}{2}\)
\(x=\dfrac{-65}{6}\)
c) \(5,5x=\dfrac{13}{15}\)
\(\dfrac{11}{2}x=\dfrac{13}{15}\)
\(x=\dfrac{13}{15}:\dfrac{11}{2}\)
\(x=\dfrac{26}{165}\)
d) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)
\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
\(\Rightarrow3x=-6\\ x=\left(-6\right):3\\ x=-2\)
C
2 tháng 4 2017
câu d mình nhầm bạn nha
d)\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\dfrac{3x}{7}+1=\dfrac{-1}{28}.\left(-4\right)\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
Ta có : 3x =-6
x = -6:3
x=-2
28 tháng 4 2017
a) \(\dfrac{x}{2}+\dfrac{y}{3}=\dfrac{x+y}{2+3}\)
\(\dfrac{x}{2}=\dfrac{x+y}{2+3}-\dfrac{y}{3}\)
\(\dfrac{x}{2}=\dfrac{x+y}{5}-\dfrac{y}{3}\)
\(\dfrac{x}{2}=\dfrac{3\left(x+y\right)}{15}-\dfrac{5y}{15}\)
\(\dfrac{x}{2}=\dfrac{3x-2y}{15}\)
\(\Rightarrow15x=2\left(3x-2y\right)\)
\(15x=6x-4y\)
\(15x-6x=4y\)
\(9x=4y\)
(CÒN LẠI MÌNH KHÔNG BIẾT LÀM)
b) \(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{4}{y}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{20}{5y}\)
\(\dfrac{x}{3}=\dfrac{1+4}{y+1}\)
\(\Rightarrow x\left(y+1\right)=15\)
(CÒN NHIÊU TỰ LÀM NHÉ)
MV
2 tháng 5 2017
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
27 tháng 6 2019
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
29 tháng 3 2017
Tìm x:
a, x+30%=-1,3
<=> x+0,3=-1,3
<=> x=-1,3-0,3
<=> x=-1,6
Vậy x=-1,6
b, 0,5x-\(\dfrac{2}{3}x\) =\(\dfrac{7}{12}\)
<=>\(\dfrac{-1}{6}x\)=\(\dfrac{7}{12}\)
<=> x=\(\dfrac{-7}{2}\)
Vậy x=\(\dfrac{-7}{2}\)
TA
1
29 tháng 6 2017
\(\left|x-3,5\right|=7,5\)
\(\Leftrightarrow x-3,5=7,5\Rightarrow x=11\)
\(\Leftrightarrow x-3,5=-7,5\Rightarrow x=-4\)
\(\left|x+\dfrac{4}{5}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow x+\dfrac{4}{5}=\dfrac{1}{2}\Rightarrow x=\dfrac{-3}{10}\)
\(\Leftrightarrow x+\dfrac{4}{5}=-\dfrac{1}{2}\Rightarrow x=\dfrac{-13}{10}\)
\(3,6-\left|x-0,4\right|=0\)
\(\Leftrightarrow\left|x-0,4\right|=3,6\)
\(\Leftrightarrow x-0,4=3,6\Rightarrow x=4\)
\(\Leftrightarrow x-0,4=-3,6\Rightarrow x=-3,2\)
16 tháng 3 2017
Bài 4: 26 \(\dfrac{1}{4}\)= 26, 25 . Quãng đương là 26,25 . 2,4 = 63 km.
Thời gian đi từ B đến A là : 63 : 30 = 2,1 h
7 tháng 8 2017
Bài 1 :
a) \(x+2\dfrac{3}{4}=5\dfrac{2}{3}\)
\(x+\dfrac{11}{4}=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}-\dfrac{11}{4}\)
\(x=\dfrac{35}{12}\)
Vậy .........................
b) \(x.3\dfrac{1}{2}=4\dfrac{3}{4}\)
\(x.\dfrac{7}{2}=\dfrac{19}{4}\)
\(x=\dfrac{19}{4}:\dfrac{7}{2}\)
\(x=\dfrac{19}{14}\)
Vậy .................
c) \(x:3\dfrac{1}{2}=4\dfrac{3}{4}\)
\(x:\dfrac{7}{2}=\dfrac{19}{4}\)
\(x=\dfrac{19}{4}.\dfrac{7}{2}\)
\(x=\dfrac{133}{8}\)
Vậy ...................
e) \(x-\dfrac{3}{4}=6.\dfrac{3}{8}\)
\(x-\dfrac{3}{4}=\dfrac{9}{4}\)
\(x=\dfrac{9}{4}+\dfrac{3}{4}\)
\(x=3\)
Vậy .............
f) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)
\(\dfrac{7}{8}:x=\dfrac{5}{2}\)
\(x=\dfrac{7}{8}:\dfrac{5}{2}\)
\(x=\dfrac{7}{20}\)
Vậy ................
g) \(x+\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{3}{4}\)
\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{1}{6}\)
\(x=\dfrac{7}{12}\)
Vậy .................
h) \(x+17,67=100-63,2\)
\(x+17,67=36,8\)
\(x=36,8-17,67\)
\(x=19,13\)
Vậy ................
i) \(x:0,01=10\)
\(x=10.0,01\)
\(x=0,1\)
Vậy ...............
k) \(8,01-x=1,99\)
\(x=8,01-1,99\)
\(x=6,02\)
Vậy ............
l) \(x.0,5=2,2\)
\(x=2,2:0,5\)
\(x=4,4\)
Vậy ............
m) \(x:7,5=3,7+4,1\)
\(x:7,5=7,8\)
\(x=7,8.7,5\)
\(x=58,5\)
Vậy ............
\(\Leftrightarrow\dfrac{1}{2}x+\dfrac{2}{3}x-x=-4\Leftrightarrow\dfrac{3x+4x-6x}{6}=-\dfrac{24}{6}\)
\(\Rightarrow x=-24\)
b) \(x.\left(\dfrac{1}{2}+\dfrac{2}{3}-1\right)=-4\)
\(x.\dfrac{-1}{6}=-4\)
\(x=-4:\dfrac{-1}{6}\)
\(x=-24\)