3) 5,16 2x 5,7 2,3 .( 0,3)
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\(\dfrac{1}{7}-\left(\dfrac{5}{21}-x\right)=\dfrac{7}{3}\)
\(\Rightarrow\dfrac{5}{21}-x=\dfrac{1}{7}-\dfrac{7}{3}\)
\(\Rightarrow\dfrac{5}{21}-x=-\dfrac{46}{21}\)
\(\Rightarrow x=\dfrac{5}{21}+\dfrac{46}{21}\)
\(\Rightarrow x=\dfrac{17}{7}\)
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\(5,16-2x=\left(5,7+2,3\right)\cdot\left(-0,3\right)\)
\(\Rightarrow5,16-2x=8\cdot\left(-0,3\right)\)
\(\Rightarrow5,16-2x=-2,4\)
\(\Rightarrow2x=5,16+2,4\)
\(\Rightarrow2x=7,56\)
\(\Rightarrow x=3,78\)
a: =>5/21-x=1/7-7/3=3/21-49/21=-46/21
=>x=5/21+46/21=51/21=17/7
b: =>5,16-2x=8*(-0,3)=-2,4
=>2x=5,16+2,4=7,56
=>x=3,78
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42
2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57
3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165
4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315
5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5
![](https://rs.olm.vn/images/avt/0.png?1311)
mk bảo tính bằng cách thuận tiện,nếu tính bằng máy tính thì mk đã tính rùi
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\Leftrightarrow2x^2=72\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm6\)
Vậy ...
\(b,\Leftrightarrow\dfrac{3}{5}x-0,75=2\dfrac{4}{5}.\dfrac{3}{7}=\dfrac{6}{5}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{6}{5}+0,75=\dfrac{39}{20}\)
\(\Leftrightarrow x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{13}{4}\)
Vậy ...
\(c,\Leftrightarrow2x=1\dfrac{5}{6}.\dfrac{6}{11}-\dfrac{3}{10}=\dfrac{7}{10}\)
\(\Leftrightarrow x=\dfrac{7}{10}:2=\dfrac{7}{20}\)
Vậy ...
\(d,\Leftrightarrow\dfrac{1}{x-7\dfrac{1}{3}}=1.5:2\dfrac{1}{4}=\dfrac{2}{3}\)
\(\Leftrightarrow x-7\dfrac{1}{3}=\dfrac{3}{2}\)
\(\Leftrightarrow x=\dfrac{3}{2}+7\dfrac{1}{3}=\dfrac{53}{6}\)
Vậy ...
a) 2x2 - 72 = 0
\(\Rightarrow\) 2x2 = 72
\(\Rightarrow\) x2 = 36 = 62 = (- 6)2
\(\Rightarrow\) x = 6 hoặc x = - 6
Vậy x = 6 hoặc x = - 6
b) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(2\dfrac{4}{5}\)
\(\Rightarrow\) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(\dfrac{14}{5}\)
\(\Rightarrow\) \(\dfrac{3}{5}\)x - \(\dfrac{3}{4}\) = \(\dfrac{6}{5}\)
\(\Rightarrow\) \(\dfrac{3}{5}\)x = \(\dfrac{39}{20}\)
\(\Rightarrow\) x = \(\dfrac{13}{4}\)
Vậy x = \(\dfrac{13}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,0,75-\dfrac{2}{3}-0,5=\dfrac{3}{4}-\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{9}{12}-\dfrac{8}{12}-\dfrac{1}{2}=\dfrac{1}{12}-\dfrac{1}{2}\)
\(=\dfrac{2}{24}-\dfrac{12}{24}=\dfrac{-10}{24}=\dfrac{-5}{12}\)
\(2,\dfrac{1}{5}-0,125-\dfrac{5}{4}=\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{5}{4}=\dfrac{8}{40}-\dfrac{5}{40}-\dfrac{5}{4}=\dfrac{3}{40}-\dfrac{5}{4}\)
\(=\dfrac{3}{40}-\dfrac{50}{40}=\dfrac{-47}{40}\)
\(3,1,25-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{5}{4}-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{2}{4}+\dfrac{4}{3}=\dfrac{6}{12}+\dfrac{16}{12}=\dfrac{22}{12}=\dfrac{11}{6}\)
\(4,0,15-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{5}{20}+\dfrac{2}{5}=\dfrac{-2}{20}+\dfrac{2}{5}\)
\(=\dfrac{-2}{20}+\dfrac{8}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)
\(5,5-3,4+\dfrac{1}{5}=\dfrac{5}{1}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25}{5}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25-17+1}{5}=\dfrac{9}{5}\)
\(6,\dfrac{1}{4}-0,3+\dfrac{4}{3}=\dfrac{1}{4}-\dfrac{3}{10}+\dfrac{4}{3}=\dfrac{10}{40}-\dfrac{12}{40}+\dfrac{4}{3}=\dfrac{-2}{40}+\dfrac{4}{3}\)
\(=\dfrac{-1}{20}+\dfrac{4}{3}=\dfrac{-3}{60}+\dfrac{80}{60}=\dfrac{77}{60}\)
\(7,0,2-3,25+4,7=\dfrac{1}{5}-\dfrac{13}{4}+\dfrac{47}{10}=\dfrac{4}{20}-\dfrac{65}{20}+\dfrac{47}{10}=\dfrac{-61}{20}+\dfrac{47}{10}\)
\(=\dfrac{-61}{20}+\dfrac{94}{20}=\dfrac{33}{20}=1,65\)
\(8,5,4+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{27}{5}+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{81}{15}+\dfrac{-35}{15}-\dfrac{-5}{7}\)
\(=\dfrac{46}{15}-\dfrac{-5}{7}=\dfrac{322}{105}-\dfrac{-75}{105}=\dfrac{397}{105}\)
\(9,\dfrac{-4}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{-12}{6}+\dfrac{2}{6}-\dfrac{1}{4}=\dfrac{-10}{6}-\dfrac{1}{4}=\dfrac{-5}{3}-\dfrac{1}{4}\)
\(=\dfrac{-20}{12}-\dfrac{3}{12}=\text{ }\dfrac{-23}{12}\)
\(10,5,4-1,5-\left(7,2-1\right)=3,9-6,2=-2,3\)
\(11,4,9-\left(1,5-7,7+3\right)=4,9-\left(-3,2\right)=8,1\)
\(12,7,8-4,7+\left(5,3-1,4\right)=3,1+3,9=7\)
\(14,\dfrac{1}{2}-0,4+\dfrac{1}{5}\text{=}0,5-0,4+0,2=0,3\)
\(15,4,2-\dfrac{4}{5}+\dfrac{1}{2}=4,2-0,8+0,5=3,9\)