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\(\left|2\dfrac{1}{5}-x\right|\)\(+\left|x-\dfrac{1}{5}\right|\)\(+8\dfrac{1}{5}\)\(=1,2\)
\(\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=\dfrac{6}{5}-\dfrac{41}{5}\)
\(\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=\dfrac{-36}{5}\) (vô lý vì \(\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|\ge0\))
Vậy: Không tìm được giá trị x thoả mãn.
\(\dfrac{x-2}{-1,2}=\dfrac{-5}{2}\Rightarrow x=\dfrac{-5.\left(-1,2\right)}{2}+2=\dfrac{6}{2}+2=3+2=5\\ \dfrac{-6}{x+1}=\dfrac{1,8}{9}\Rightarrow x=\dfrac{-6.9}{1,8}-1=\dfrac{-54}{1,8}-1=-30-1=-31\\ \dfrac{-3}{x}=\dfrac{x}{-12}\Rightarrow x=\sqrt{\left(-12\right).\left(-3\right)}=\sqrt{36}=\sqrt{\left(\pm6\right)^2}=\pm6\)
\(\dfrac{x-4}{x-1}=\dfrac{3}{5}\\ \Rightarrow5\left(x-4\right)=3\left(x-1\right)\\ \Leftrightarrow5x-20=3x-3\\ \Leftrightarrow5x-3x=-3+20\\ \Leftrightarrow2x=17\\ \Leftrightarrow x=\dfrac{17}{2}\\ ---\\ \dfrac{1,12}{-10}=\dfrac{11,2}{x}\Rightarrow x=\dfrac{11,2.\left(-10\right)}{1,12}=\dfrac{10.1,12.\left(-10\right)}{1,12}=-100\)
\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)
\(a,=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{271}{105}\\ b,=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-\dfrac{5}{4}=1-1-\dfrac{1}{8}=-\dfrac{1}{8}\)
1a) \(\dfrac{5}{3}-\dfrac{2}{7}+1,2=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{175}{105}-\dfrac{30}{105}+\dfrac{126}{105}=\dfrac{107-30+126}{105}=\dfrac{203}{205}\)
a, - \(\dfrac{1}{10}\) + \(\dfrac{2}{5}\)\(x\) + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\)\(x\) = \(\dfrac{1}{10}\) - \(\dfrac{7}{20}\) + \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\) \(x\) = - \(\dfrac{3}{20}\)
\(x\) = - \(\dfrac{3}{20}\): \(\dfrac{2}{5}\)
\(x\) = - \(\dfrac{3}{8}\)
b, \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{8}{15}\)
\(x\) = \(\dfrac{1}{2}\): (- \(\dfrac{8}{15}\))
\(x\) = - \(\dfrac{15}{16}\)
\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)
\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)
a) (31 6/13 + 5 9/41) - 36 6/13
= 409/13 + 214/41 - 474/13
= (409/13 - 474/13) + 214/41
= -5 + 214/41
= 9/41
b) 5/3 + (-2/7) - (-1,2)
= 5/3 - 2/7 + 6/5
= 29/21 + 6/5
= 271/105
c) 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)
= 1/4 + 3/5 - 1/8 + 2/5 - 5/4
= (1/4 - 5/4) + (3/5 + 2/5) - 1/8
= -1 + 1 - 1/8
= -1/8
\(\dfrac{3}{16}\) - (\(x\) - \(\dfrac{5}{4}\)) - ( \(\dfrac{3}{4}\) - \(\dfrac{7}{8}\) - 1) = 2\(\dfrac{1}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) + 1 = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + ( \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\)) + (\(\dfrac{7}{8}\) + 1) = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\) = \(\dfrac{5}{2}\)
( \(\dfrac{3}{16}\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\)) - \(x\) = \(\dfrac{5}{2}\)
\(\dfrac{41}{16}\) - \(x\) = \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{41}{16}\) - \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{1}{16}\)
2, \(\dfrac{1}{2}\).( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\)) = \(\dfrac{1}{5}\) - \(x\) + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\))
\(\dfrac{1}{2}\).(-\(\dfrac{11}{15}\)) = \(\dfrac{1}{5}\) - \(x\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{5}\)
- \(\dfrac{11}{30}\) = ( \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{15}\)) - \(x\)
- \(\dfrac{11}{30}\) = \(\dfrac{7}{15}\) - \(x\)
\(x\) = \(\dfrac{7}{15}\) + \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{5}{6}\)
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=-\dfrac{2}{6}\)
\(\Rightarrow x=-\dfrac{2}{6}-\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{2}{3}\)
b) \(\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)
\(\Rightarrow\left(\dfrac{15}{40}-\dfrac{8}{40}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)
\(\Rightarrow\dfrac{7}{40}+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)
\(\Rightarrow\dfrac{5}{8}-x=\dfrac{1}{5}-\dfrac{7}{40}\)
\(\Rightarrow\dfrac{5}{8}-x=\dfrac{1}{40}\)
\(\Rightarrow x=\dfrac{5}{8}-\dfrac{1}{40}\)
\(\Rightarrow x=\dfrac{3}{5}\)
`@` `\text {Ans}`
`\downarrow`
`1,`
`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`
`=> 3/16 - x + 19/8 = 2 1/2`
`=> 3/16 - x = 2 1/2 - 19/8`
`=> 3/16 - x =1/8`
`=> x = 3/16 - 1/8`
`=> x = 1/16`
Vậy, `x = 1/16`
`2,`
`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`
`=> 1/2 * (-11/15) = 1/5 - x + 4/15`
`=> -11/30 = x + 1/5 - 4/15`
`=> x + (-1/15) = -11/30`
`=> x = -11/30 + 1/15`
`=> x = -3/10`
Vậy, `x = -3/10.`
\(\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+8\dfrac{1}{5}=1,2\\\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=1,2-8\dfrac{1}{5}\\ \Rightarrow \left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=-7\)
Nhận xét:
\(\left\{{}\begin{matrix}\left|2\dfrac{1}{5}-x\right|\ge0,\forall x\\\left|x-\dfrac{1}{5}\right|\ge0,\forall x\end{matrix}\right.\\ \Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|\ge0,\forall x\)
Mà \(-7< 0\) nên:
Không tìm được giá trị \(x\) thỏa mãn đề bài
Vậy...