1+1/(x+2) = 12/(8-x^3)
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\(a,x-\dfrac{7}{12}x=\dfrac{5}{24}-\dfrac{3}{8}x\)
\(\Leftrightarrow\dfrac{5}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)
\(\Leftrightarrow\dfrac{19}{24}x=\dfrac{5}{24}\Leftrightarrow x=\dfrac{5}{19}\)
Vậy x = 5/19
\(b,\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\-3-\dfrac{x}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
Vậy x = 1/2 hoặc x = -6
\(c,\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
Vậy x = 7 hoặc x = -1
\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{13}{24}\)
\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)
\(=\dfrac{1}{5}\)
a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(x=\dfrac{-13}{24}\)
vậy x =....
b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=\dfrac{17}{12}\)
\(x=\dfrac{3}{4}-\dfrac{17}{12}\)
\(x=\dfrac{-2}{3}\)
vậy x =....
ban lam sai rui de mk lam lai nhe.
\(12.\left(x-1\right):3=4^3-2^3\)
\(12.\left(x-1\right):3=64-8\)
\(12.\left(x-1\right):3=56\)
\(12.\left(x-1\right)=56.3\)
\(12.\left(x-1\right)=168\)
\(x-1=168:12\)
\(x-1=14\)
\(x=15\)
Bài giải:
Câu 1: a, \(\left(-2\right).4.5.38.\left(-25\right)\)
\(=\left[\left(-2\right).5\right].\left[4.\left(-25\right)\right].38\)
\(=\left(-10\right).\left(-100\right).38\)
\(=1000.38=38000\)
b,\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}\)
\(=\left(\frac{1}{3}+\frac{3}{8}\right)-\frac{7}{12}\)
\(=\frac{17}{24}-\frac{7}{12}=\frac{1}{8}\)
c, \(\frac{-5}{8}.\frac{5}{12}+\frac{-5}{8}.\frac{7}{12}+2\frac{1}{8}\)
\(=\frac{-5}{8}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{17}{8}\)
\(=\frac{-5}{8}.1+\frac{17}{8}\)
\(=\frac{3}{2}\)
Câu 2: a, \(x-\frac{2}{5}=0,24\)
\(x-0,4=0,24\)
\(x=0,24+0,4\)
\(\Rightarrow x=0,64\left(\frac{16}{25}\right)\)
b,\(\frac{2}{3}.x+\frac{1}{12}=\frac{1}{10}\)
\(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{12}\)
\(\frac{2}{3}.x=\frac{1}{60}\)
\(x=\frac{1}{60}:\frac{2}{3}\)
\(\Rightarrow x=\frac{1}{40}\)
c, \(\left(3\frac{1}{2}-2x\right).1\frac{1}{3}=7\frac{1}{3}\)
\(\frac{7}{2}-2x=\frac{22}{3}:\frac{4}{3}\)
\(\frac{7}{2}-2x=\frac{11}{2}\)
\(2x=\frac{7}{2}-\frac{11}{2}\)
\(2x=-2\)
\(\Rightarrow x=-2:2\)
\(x=-1\)
`(x+2)-2=0`
`=>x+2=0+2`
`=>x+2=2`
`=>x=2-2`
`=>x=0`
__
`(x+3)+1=7`
`=>x+3=7-1`
`=>x+3=6`
`=>x=6-3`
`=>x=3`
__
`(x+3)+4=12`
`=>x+3=12-4`
`=>x+3=8`
`=>x=8-3`
`=>x=5`
__
`(5x+4)-1=13`
`=>5x+4=13+1`
`=>5x+4=14`
`=>5x=14-4`
`=>5x=10`
`=>x=10:5`
`=>x=2`
__
`(4x-8)+3=12`
`=>4x-8=12-3`
`=>4x-8=9`
`=>4x=9+8`
`=>4x=17`
`=> x=17/4`
__
`3+(x-5)=14`
`=>x-5=14-3`
`=>x-5=11`
`=>x=11+5`
`=>x=16`
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
ĐK: \(x\ne\left\{1;3;8;20\right\}\)
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow\)\(\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-8}-\frac{1}{x-3}+\frac{1}{x-20}-\frac{1}{x-8}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}=\frac{3}{4}\)
\(\Rightarrow\)\(x-1=\frac{4}{3}\)
\(\Leftrightarrow\)\(x=\frac{7}{3}\)(t/m)
Vậy...
\(\dfrac{11}{8}:x-\dfrac{2}{5}+-\dfrac{1}{6}=-\dfrac{1}{5}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{5}-\left(-\dfrac{1}{6}\right)\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{5}+\dfrac{1}{6}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{6}{30}+\dfrac{5}{30}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{30}\\ =>\dfrac{11}{8}:x=-\dfrac{1}{30}+\dfrac{2}{5}\\ =>\dfrac{11}{8}:x=-\dfrac{1}{30}+\dfrac{12}{30}\\ =>\dfrac{11}{8}:x=\dfrac{11}{30}\\ =>x=\dfrac{11}{8}:\dfrac{11}{30}\\ =>x=\dfrac{11}{8}.\dfrac{30}{11}\\ =>x=\dfrac{30}{8}\\ =>x=\dfrac{15}{4}\\ \dfrac{4}{7}x-\dfrac{1}{3}x+\left(-\dfrac{16}{21}\right)=-\dfrac{2}{3}\\ =>\left(\dfrac{4}{7}-\dfrac{1}{3}\right)x=-\dfrac{2}{3}-\left(-\dfrac{16}{21}\right)\\ =>\left(\dfrac{12}{21}-\dfrac{7}{21}\right)x=-\dfrac{2}{3}+\dfrac{16}{21}\\ =>\dfrac{5}{21}x=-\dfrac{14}{21}+\dfrac{16}{21}\\ =>\dfrac{5}{21}x=\dfrac{2}{21}\\ =>x=\dfrac{2}{21}:\dfrac{5}{21}\)
\(=>x=\dfrac{2}{21}.\dfrac{21}{5}\\ =>x=\dfrac{2}{5}\\ -\dfrac{11}{12}x+\dfrac{15}{2}\left(x+-\dfrac{1}{5}\right)=\dfrac{67}{8}\\ =>-\dfrac{11}{12}x+\dfrac{15}{2}.x-\dfrac{1}{5}=\dfrac{67}{8}\\ =>\left(-\dfrac{11}{12}+\dfrac{15}{2}\right)x=\dfrac{67}{8}+\dfrac{1}{5}\\ =>\left(-\dfrac{11}{12}+\dfrac{90}{12}\right)x=\dfrac{335}{40}+\dfrac{8}{40}\\ =>\dfrac{79}{12}x=\dfrac{343}{40}\\ =>x=\dfrac{343}{40}:\dfrac{79}{12}\\ =>x=\dfrac{343}{40}.\dfrac{12}{79}\\ =>x=\dfrac{343.12}{40.79}\\ =>x=\dfrac{343.3}{10.79}\\ =>x=\dfrac{1029}{790}\)
1) Ta có : 2x2 + 3x - 5
= 2x2 - 2x + 5x - 5
= 2x(x - 1) + 5(x - 1)
= (x - 1) (2x + 5)
3) x2 + x - 6
= x2 + 2x - 3x - 6
= x(x + 2) - (3x + 6)
= x(x + 2) - 3(x + 2)
= (x - 3)(x + 2)