a, \(x.\dfrac{1}{4}=\dfrac{1}{2}-\dfrac{4}{5}\)
b, \(\dfrac{2}{3}-x=\dfrac{1}{2}:\dfrac{3}{8}\)
a) \(2\dfrac{1}{2}\) - x + \(\dfrac{4}{5}\) = \(\dfrac{2}{3}\) - (\(-\dfrac{4}{7}\) )
b) \(-\dfrac{4}{7}\) - x = \(\dfrac{3}{5}\) - 2x
c) (\(\dfrac{3}{8}\) - \(\dfrac{1}{5}\) ) + (\(\dfrac{5}{8}\) - x) = \(\dfrac{1}{5}\)
\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)
a,\(\dfrac{2}{3}\)x\(\dfrac{5}{2}\):\(\dfrac{9}{5}\)
b,\(\dfrac{1}{3}\)x\(\dfrac{1}{4}\)+\(\dfrac{5}{6}\)
c,\(\dfrac{1}{2}\)-\(\dfrac{7}{8}\):\(\dfrac{7}{4}\)
d,\(\dfrac{6}{5}\)-\(\dfrac{4}{5}\)x\(\dfrac{3}{2}\)
Bài 2:Tìm x biết:
a)\(\dfrac{1}{7}+x=\dfrac{-2}{3}\)
b)\(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
c)\(\left\{\dfrac{3}{5}-2x\right\}.\dfrac{5}{8}=1\)
d)\(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
e)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
f)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
g)\(\left|X+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
h)\(\left(\dfrac{1}{32}\right)^x.8^{2x}=512\)
i)\(5,3x+\left(-3,3\right)x+1,7=-4,9\)
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: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)
Bài 3: (Đề 2) Tìm y
a) \(2\dfrac{2}{5}:\) y x \(1\dfrac{3}{4}=\dfrac{7}{8}\) b)\(3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\) c) \(\dfrac{12}{5}-2\dfrac{2}{5}x\) y \(=1\dfrac{1}{4}\)
\(a,2\dfrac{2}{5}:y\times1\dfrac{3}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y\times\dfrac{7}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y=\dfrac{7}{8}:\dfrac{7}{4}\\ \dfrac{12}{5}:y=\dfrac{1}{2}\\ y=\dfrac{12}{5}:\dfrac{1}{2}=\dfrac{24}{5}\\ b,3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\\ \dfrac{17}{5}:y:\dfrac{5}{4}=\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{5}:\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{13}\\ y=\dfrac{17}{13}\times\dfrac{5}{4}=\dfrac{85}{52}\)
\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)
a, 2\(\dfrac{2}{5}\): y \(\times\)1\(\dfrac{3}{4}\) = \(\dfrac{7}{8}\)
\(\dfrac{12}{5}\) : y \(\times\dfrac{7}{4}\) = \(\dfrac{7}{8}\)
\(\dfrac{12}{5}\) : y = \(\dfrac{7}{8}\) : \(\dfrac{7}{4}\)
\(\dfrac{12}{5}\) : y = \(\dfrac{1}{2}\)
y = \(\dfrac{12}{5}\) : \(\dfrac{1}{2}\)
y = \(\dfrac{24}{5}\)
b, 3\(\dfrac{2}{5}\): y : 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)
\(\dfrac{17}{5}\): y: \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)
\(\dfrac{17}{5}\):y = \(\dfrac{13}{5}\times\dfrac{5}{4}\)
\(\dfrac{17}{5}\) : y = \(\dfrac{13}{4}\)
y = \(\dfrac{17}{5}\) : \(\dfrac{13}{4}\)
y = \(\dfrac{68}{65}\)
c, \(\dfrac{12}{5}\) - 2\(\dfrac{2}{5}\)\(\times y\) = 1\(\dfrac{1}{4}\)
\(\dfrac{12}{5}\) - \(\dfrac{12}{5}\)\(\times\)y = \(\dfrac{5}{4}\)
\(\dfrac{12}{5}\times y\) = \(\dfrac{12}{5}\) - \(\dfrac{5}{4}\)
\(\dfrac{12}{5}\) \(\times\) y = \(\dfrac{23}{20}\)
\(y\) = \(\dfrac{23}{20}\): \(\dfrac{12}{5}\)
y = \(\dfrac{23}{48}\)
\(\dfrac{2}{5}\) x \(\dfrac{3}{4}-\dfrac{1}{8}\) \(= \)
\(\dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\)
\(\dfrac{9}{20}-\dfrac{3}{5}\) x \(\dfrac{1}{4}\)\(= \)
\(\dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}\)\(=\)
\(\dfrac{2}{5}\times\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{1}{5}\times\dfrac{3}{2}-\dfrac{1}{8}=\dfrac{3}{10}-\dfrac{1}{8}=\dfrac{24}{80}-\dfrac{10}{80}=\dfrac{14}{80}=\dfrac{7}{40}\\ \dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{3}-\dfrac{1}{5}=\dfrac{25}{15}-\dfrac{3}{15}=\dfrac{22}{15}\\ \dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\\ \dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{2}{8}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{2}{8}+\dfrac{1}{3}\times\dfrac{5}{2}=\dfrac{2}{8}+\dfrac{5}{6}=\dfrac{1}{4}+\dfrac{5}{6}=\dfrac{6}{24}+\dfrac{20}{24}=\dfrac{26}{24}=\dfrac{13}{12}\)
1/ \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
2/ \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
3/ \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
4/ \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
5/ \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
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
\(\dfrac{3}{2}X-0,2=\dfrac{3}{5}\)
\(\dfrac{1}{3}+x=\dfrac{3}{4}\)
\(1\dfrac{1}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{11}{8}-\dfrac{3}{8}.x=\dfrac{1}{8}\)
giúp với
\(\dfrac{3}{2}x-0,2=\dfrac{3}{5}\)
\(\dfrac{3}{2}x-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\dfrac{3}{2}x=\dfrac{3}{5}+\dfrac{1}{5}\)
\(\dfrac{3}{2}x=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}:\dfrac{3}{2}\)
\(x=\dfrac{4}{5}\cdot\dfrac{2}{3}\)
\(x=\dfrac{8}{15}\)
\(\dfrac{1}{3}+x=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{1}{3}\)
\(x=\dfrac{9}{12}-\dfrac{4}{12}\)
\(x=\dfrac{5}{12}\)
\(1\dfrac{1}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{3}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{3}{2}x=\dfrac{1}{4}+\dfrac{2}{5}\)
\(\dfrac{3}{2}x=\dfrac{13}{20}\)
\(x=\dfrac{13}{20}:\dfrac{3}{2}\)
\(x=\dfrac{13}{20}\cdot\dfrac{2}{3}\)
\(x=\dfrac{13}{30}\)
\(\dfrac{11}{8}-\dfrac{3}{8}\cdot x=\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{11}{8}-\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}:\dfrac{3}{8}\)
\(x=\dfrac{5}{4}\cdot\dfrac{8}{3}\)
\(x=\dfrac{10}{3}\)
Bài 4. Tìm x biết:
a. \(\dfrac{x}{5}=\dfrac{2}{5},\dfrac{3}{8}=\dfrac{6}{x},\dfrac{1}{9}=\dfrac{x}{27}\)
b. \(\dfrac{4}{x}=\dfrac{8}{6},\dfrac{3}{x-5}=\dfrac{-4}{x+2},\dfrac{x}{-2}=\dfrac{-8}{x}\)
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) x= 2, x= 8.(6 : 3) = 16, x= 1. (27 : 9)= 3
b) x= 6 : (8 : 4) = 3, x= -1, x= -2 . -8 = x.x => 16 = x2 => 42 = x2 => x=4
Tick cho mình đi
Câu 10:
a) \(-3\dfrac{1}{4}.x-75\%+\dfrac{3x}{2}=-1,2:-\dfrac{9}{10}-1\dfrac{1}{4}\)
b) \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)(x thuộc N sao)
a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)
\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)
\(\Leftrightarrow-7x=\dfrac{10}{3}\)
hay \(x=-\dfrac{10}{21}\)
b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)
\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)
\(\Leftrightarrow85x+85=84x+168\)
\(\Leftrightarrow x=83\)