Tìm x biết:
1/ \(\dfrac{3}{5}-x=0,2\)
2/ \(\dfrac{x}{3}-\dfrac{1}{8}=\dfrac{5}{8}\)
3/ \(3\dfrac{1}{3}.x-6\dfrac{3}{4}=3\dfrac{1}{4}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{3}{2}x-0,2=\dfrac{3}{5}\)
\(\dfrac{3}{2}x-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\dfrac{3}{2}x=\dfrac{3}{5}+\dfrac{1}{5}\)
\(\dfrac{3}{2}x=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}:\dfrac{3}{2}\)
\(x=\dfrac{4}{5}\cdot\dfrac{2}{3}\)
\(x=\dfrac{8}{15}\)
\(\dfrac{1}{3}+x=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{1}{3}\)
\(x=\dfrac{9}{12}-\dfrac{4}{12}\)
\(x=\dfrac{5}{12}\)
\(1\dfrac{1}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{3}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{3}{2}x=\dfrac{1}{4}+\dfrac{2}{5}\)
\(\dfrac{3}{2}x=\dfrac{13}{20}\)
\(x=\dfrac{13}{20}:\dfrac{3}{2}\)
\(x=\dfrac{13}{20}\cdot\dfrac{2}{3}\)
\(x=\dfrac{13}{30}\)
\(\dfrac{11}{8}-\dfrac{3}{8}\cdot x=\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{11}{8}-\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}:\dfrac{3}{8}\)
\(x=\dfrac{5}{4}\cdot\dfrac{8}{3}\)
\(x=\dfrac{10}{3}\)
a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)
hay \(x=-\dfrac{17}{21}\)
Vậy: \(x=-\dfrac{17}{21}\)
b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)
Vậy: \(x=\dfrac{4}{5}\)
c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)
\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)
hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)
Vậy: \(x=-\dfrac{5}{7}\)
f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)
\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)
\(\Leftrightarrow-x-\dfrac{9}{60}=0\)
\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)
hay \(x=-\dfrac{3}{20}\)
Vậy: \(x=-\dfrac{3}{20}\)
g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)
a: =>6/x=x/24
=>x^2=144
=>x=12 hoặc x=-12
b: =>x(1-7/12+3/8)=5/24
=>x*19/24=5/24
=>x=5/24:19/24=5/19
c: =>(x-1/3)^2=1+3/4+1/2=9/4
=>x-1/3=3/2 hoặc x-1/3=-3/2
=>x=11/6 hoặc x=-7/6
d: =>(x-3)^2=16
=>x-3=4 hoặc x-3=-4
=>x=-1 hoặc x=7
e: =>9/x=-1/3
=>x=-27
f: =>x-1/2=0 hoặc -x/2-3=0
=>x=1/2 hoặc x=-6
\(x-\dfrac{1}{2}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}+\dfrac{1}{2}\)
\(x=\dfrac{5}{4}\)
\(x+\dfrac{7}{8}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{7}{8}\)
\(x=\dfrac{-1}{8}\)
\(\dfrac{1}{2}\cdot x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\dfrac{1}{2}\cdot x=\dfrac{-1}{2}+\dfrac{1}{4}\)
\(\dfrac{1}{2}\cdot x=\dfrac{-1}{4}\)
\(x=\dfrac{-1}{4}\div\dfrac{1}{2}\)
\(x=\dfrac{-1}{2}\)
Câu D ko bt
a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27
b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4
tick cho mik nhé
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)
\(\Leftrightarrow5x+1-2x+4=4\)
\(\Leftrightarrow3x=-1\)
hay \(x=-\dfrac{1}{3}\)
2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
\(\Leftrightarrow9x+27+12-36x=-2x+2\)
\(\Leftrightarrow-27x+2x=2-39\)
hay \(x=\dfrac{37}{25}\)
3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x+6-10x=4-4x\)
\(\Leftrightarrow-7x+4x=4-6=-2\)
hay \(x=\dfrac{2}{3}\)
4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
\(\Leftrightarrow5x-15-x-1=2x-4\)
\(\Leftrightarrow4x-2x=-4+16=12\)
hay x=6
5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
\(\Leftrightarrow12x+3-9x+5+4x-8=0\)
\(\Leftrightarrow7x=0\)
hay x=0
a, \(\dfrac{3}{7}\)\(x\) - 0,4 = - \(\dfrac{17}{35}\)
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{17}{35}\) + 0,4
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{3}{35}\)
\(x\) = - \(\dfrac{3}{35}\): \(\dfrac{3}{7}\)
\(x\) = - \(\dfrac{1}{5}\)
b, 0,2.(\(x\) - 3) +2,4 = 10
0,2.(\(x\) - 3) = 10 - 2,4
0,2.(\(x\) - 3) = 7,6
\(x\) - 3 = 7,6:0,2
\(x\) - 3 = 38
\(x\) = 38 + 3
\(x\) = 41
a)\(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\)
\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\)
\(\dfrac{2}{3}x=\dfrac{5}{4}+\dfrac{5}{6}\)
\(\dfrac{2}{3}x=\dfrac{25}{12}\)
\(x=\dfrac{25}{12}:\dfrac{2}{3}\)
=>\(x=\dfrac{25}{8}\)
a) \(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\) b) \(2\dfrac{1}{3}-\dfrac{4}{5}:x=0,2\)
\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\) \(\dfrac{7}{3}-\dfrac{4}{5}:x=\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{5}{4}-\dfrac{5}{6}\) \(\dfrac{4}{5}:x=\dfrac{7}{3}-\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{30}{24}-\dfrac{20}{24}\) \(\dfrac{4}{5}:x=\dfrac{35}{15}-\dfrac{3}{15}\)
\(\dfrac{2}{3}x=\dfrac{5}{12}\) \(\dfrac{4}{5}:x=\dfrac{32}{15}\)
\(x=\dfrac{5}{12}:\dfrac{2}{3}\) \(x=\dfrac{4}{5}:\dfrac{32}{15}\)
\(x=\dfrac{5}{12}:\dfrac{8}{12}\) \(x=\dfrac{4}{5}.\dfrac{15}{32}\)
\(x=\dfrac{5}{12}.\dfrac{12}{8}=\dfrac{5}{8}\) \(x=\dfrac{4.15}{5.32}\)
\(x=\dfrac{1.3}{1.8}=\dfrac{3}{8}\)
d)\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\dfrac{-8}{27}\)
\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\left(\dfrac{-2}{3}\right)^3\)
\(\Rightarrow\dfrac{4}{3}-\dfrac{1}{4}x=\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{1}{4}x=\dfrac{4}{3}-\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{1}{4}x=2\)
\(\Rightarrow x=2:\dfrac{1}{4}\)
\(\Rightarrow x=2.4=8\)
1: =>x=3/5-1/5=2/5
b: =>x/3=5/8+1/8=3/4
=>x=9/4
3: =>10/3x=3+1/4+6+3/4=10
=>x=10:10/3=3