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\(a)x+30\%x=-1,31\)
\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)
\(\Leftrightarrow10x+3x=-13,1\)
\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)
\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=9\)
\(\Leftrightarrow6x-3=18\)
\(\Leftrightarrow x=\frac{7}{2}\)
c) pt <=> \(x-\frac{21}{5}=\frac{23}{7}< =>x=\frac{23}{7}+\frac{21}{5}=\frac{262}{35}\)
vậy x = \(\frac{262}{35}\)
d) \(x-\frac{3}{4}=\frac{51}{8}< =>x=\frac{51}{8}+\frac{3}{4}=\frac{57}{8}\)
vậy x = \(\frac{57}{8}\)
e) pt <=> \(\frac{7}{8}:x=\frac{7}{2}< =>\frac{7}{8}.\frac{1}{x}=\frac{7}{2}< =>\frac{7}{8x}=\frac{7}{2}< =>56x=14< =>x=\frac{14}{56}=\frac{1}{4}\)
vậy x = \(\frac{1}{4}\)
a) pt <=> \(x+\frac{11}{4}=\frac{17}{3}< =>x=\frac{17}{3}-\frac{11}{4}=\frac{35}{12}\)
vậy x = \(\frac{35}{12}\)
b) pt <=> \(\frac{x.7}{2}=\frac{19}{4}< =>x=\frac{19.2}{4.7}=\frac{38}{28}=\frac{19}{14}\)
vậy x = \(\frac{19}{14}\)
a) \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x=-\dfrac{5}{6}+\dfrac{7}{3}=\dfrac{3}{2}\\ =>x=\dfrac{3}{2}:\dfrac{-1}{4}=-6\)
b) \(\left|x-\dfrac{1}{6}\right|+-\dfrac{5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\\ < =>\left|x-\dfrac{1}{6}\right|=\left(\dfrac{4}{7}.\dfrac{14}{48}\right)-\left(-\dfrac{5}{12}\right)=\dfrac{1}{6}+\dfrac{5}{12}=\dfrac{7}{12}\\ \)
Xảy ra 2 trường hợp:
+) TH1: \(x-\dfrac{1}{6}=\dfrac{7}{12}\\ =>x=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{3}{4}->\left(a\right)\)
+) TH2" \(-\left(x-\dfrac{1}{6}\right)=\dfrac{7}{12}\\ < =>-x+\dfrac{1}{6}=\dfrac{7}{12}\\ < =>-x=\dfrac{7}{12}-\dfrac{1}{6}=\dfrac{5}{12}\\ =>x=-\dfrac{5}{12}->\left(b\right)\)
Từ (a) và (b) => \(x\in\left\{-\dfrac{5}{12};\dfrac{3}{4}\right\}\)
a, \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{-1}{4}x=\dfrac{-5}{6}+\dfrac{7}{3}\)
\(\Rightarrow\dfrac{-1}{4}x=\dfrac{3}{2}\Rightarrow x=-6\)
Vậy \(x=-6\)
b, \(\left|x-\dfrac{1}{6}\right|+\dfrac{-5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|-\dfrac{5}{12}=\dfrac{1}{6}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|=\dfrac{7}{12}\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{-7}{12}\\x-\dfrac{1}{6}=\dfrac{7}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{12}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-5}{12};\dfrac{3}{4}\right\}\)
Chúc bạn học tốt!!!
Bài 1
a, 23 + ( x - 32 ) = 1
x - 32 = 1 - 23 = -7
x = -7 + 32
x = 2
b, 5 . (x+7) -10 = 40
5 . (x+7) = 50
x+7 = 50 :5 =10
x = 10 - 7
x = 3
a) x-1/3=5/14.(-7/6)
x-1/3=-5/12
x=-5/12+1/3
x= -1/12
b) 3/4+1/4x=0,2
1/4x=0,2-3/4
1/4x=-11/20
x=-11/20:1-4
x=-11/5
c) 1/12.x^2=1.1/2
1/12.x^2=1/2
x^2= 1/2:1/12
x^2=6
=> x=căn bậc của 6
Chúc bn hok tốt
a)\(x-\frac{1}{3}=\frac{5}{14}\cdot\left(\frac{-7}{6}\right)\)
\(x-\frac{1}{3}=\frac{-5}{12}\)
\(x=\frac{-5}{12}+\frac{1}{3}\)
\(x=\frac{-5}{12}+\frac{4}{12}\)
\(x=\frac{-1}{12}\)
b)\(\frac{3}{4}+\frac{1}{4}\cdot x=0,2\)
\(\frac{3}{4}+\frac{1}{4}\cdot x=\frac{1}{5}\)
\(\frac{1}{4}\cdot x=\frac{1}{5}-\frac{3}{4}\)
\(\frac{1}{4}\cdot x=\frac{4}{20}-\frac{15}{20}\)
\(\frac{1}{4}\cdot x=\frac{-11}{20}\)
\(x=\frac{-11}{20}\cdot4\)
\(x=-\frac{11}{5}\)
c)\(\frac{1}{12}\cdot x^2=1\cdot\frac{1}{3}\)
\(\frac{1}{12}\cdot x^2=\frac{1}{3}\)
\(x^2=\frac{1}{3}\cdot12\)
\(x^2=4=\left(\pm2\right)^2\)
\(x=\pm2\)
a) x − 2 3 = − − 1 5 + 3 4 ⇔ x − 2 3 = − 1 5 + 3 4 = 11 20 ⇔ x − 2 3 = 11 20 x − 2 3 = − 11 20 ⇔ x = 73 60 x = 7 60
b) x − 3 = 1 5 + 1 7 − 1 14 ⇔ x − 3 = 1 5 + 1 14 = 19 70 ⇔ x − 3 = 19 70 x − 3 = − 19 70 ⇔ x = 229 70 x = 191 70