3-x=15-(-5)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)
b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)
=9/17+8/17=1
c: =>x-3/10=7/15*1/5=7/75
=>x=7/75+3/10=59/150
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x , biết :
|x-2| = -3
Vô lí vì |x-2| phải lớn hơn hoặc bằng 0
=> không có giá trị x nào thỏa mãn đề bài trên .
|x-7| - 5 + 3 = 1
<=> |x-7| - 5 = -2
<=> |x-7| = 3
<=> x - 7 = 3 hoặc x - 7 = -3
<=> x = 10 hoặc x = 4
Vậy x€ { 10 ; 4 }
|x+5| - 5 = 4 - (-3)
<=> |x+5| - 5 = 7
<=> |x+5| = 7 + 5
<=> |x +5| = 12
<=> x + 5 = 12 hoặc x + 5 =-12
<=> x = 7 hoặc x = -17
Vậy x € { 7 ; -17 }
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\dfrac{5\times11\times15\times18}{6\times15\times22\times10}\)
= \(\dfrac{5\times11\times15\times6\times3}{6\times15\times11\times2\times2\times5}\)
= \(\dfrac{5\times11\times15\times6}{5\times11\times15\times6}\) \(\times\) \(\dfrac{3}{2\times2}\)
= 1 \(\times\) \(\dfrac{3}{4}\)
= \(\dfrac{3}{4}\)
b, \(\dfrac{7}{12}\) \(\times\) \(\dfrac{3}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(\dfrac{5}{12}\)
= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{7}{12}+\dfrac{5}{12}\))
= \(\dfrac{3}{5}\) \(\times\) \(\dfrac{12}{12}\)
= \(\dfrac{3}{5}\) \(\times\) 1
= \(\dfrac{3}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, thiếu đề
b, -5 ( 2 - x ) + 4 ( x - 3 ) = 10x - 15
<=> -10 + 5x + 4x - 12 = 10x - 15
<=> -x = 7 <=> x = 7
c,-7(5-x)-2(x-10)=15
<=> -35 + 7x - 2x + 20 = 15 <=> 5x = 30 <=> x = 6
d, -4(x+1) + 89x - 3 = 24
<=> 85x = 31 <=> x = 31/85
e, 5(x-30-2(x+6)) = 9
<=> 5 (-x-49) = 9 <=> -5x = 254 <=> x = -254/5
b: =>-10+5x+4x-12=10x-15
=>9x-10x=-15+22
=>-x=7
hay x=-7
c: =>-35+7x-2x+20=15
=>5x-15=15
=>x=6
d: =>-4x-4+72x-24=24
=>68x-32=24
hay x=14/17
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{5}{3}\cdot\dfrac{7}{10}=\dfrac{5}{10}\cdot\dfrac{7}{3}=\dfrac{14}{3}>2\)
\(\dfrac{5}{7}:\dfrac{7}{10}=\dfrac{5}{7}\cdot\dfrac{10}{7}=\dfrac{50}{49}< 2\)
=>\(\dfrac{5}{3}\cdot\dfrac{7}{10}>\dfrac{5}{7}:\dfrac{7}{10}\)
b: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(\dfrac{8}{9}:\dfrac{4}{3}=\dfrac{8}{9}\cdot\dfrac{3}{4}=\dfrac{8}{4}\cdot\dfrac{3}{9}=\dfrac{2}{3}\)
Do đó: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{8}{9}:\dfrac{4}{3}\)
c: \(\dfrac{5}{11}\cdot\dfrac{33}{15}=\dfrac{5}{15}\cdot\dfrac{33}{11}=\dfrac{3}{3}=1\)
\(\dfrac{6}{17}\cdot\dfrac{34}{25}=\dfrac{34}{17}\cdot\dfrac{6}{25}=2\cdot\dfrac{6}{25}=\dfrac{12}{25}< 1\)
=>\(\dfrac{5}{11}\cdot\dfrac{33}{15}>\dfrac{6}{17}\cdot\dfrac{34}{25}\)
d: \(\dfrac{15}{19}\cdot\dfrac{38}{5}=\dfrac{15}{5}\cdot\dfrac{38}{19}=3\cdot2=6\)
\(\dfrac{15}{16}:\dfrac{3}{8}=\dfrac{15}{16}\cdot\dfrac{8}{3}=\dfrac{15}{3}\cdot\dfrac{8}{16}=\dfrac{5}{2}\)<6
=>\(\dfrac{15}{19}\cdot\dfrac{38}{5}>\dfrac{15}{16}:\dfrac{3}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x + 3/5 = -1/15
=> x = -1/15 - 3/5
=> x = -2/3
b) x/14 = 1/7 - 3/14
=> x/14 = -1/14
=> x = -1
c) -1/2x + 1/5 = 3/10
=> -1/2x = 3/10 - 1/5
=> -1/2x = 1/5
=> x = 1/5 : (-1/2)
=> x = -2/5
d) 4/5 - (2/15 + x) = 1/15
=> 4/5 - 2/15 - x = 1/15
=> 2/3 - x = 1/15
=> x = 2/3 - 1/15
=> x = 3/5
3 - x = 15 - (-5)
3 - x = 15 + 5
3 - x = 20
x = 3 - 20
x = - 17
3 - x = 15 - (-5)
3 - x = 15 +5
3 - x = 20
x = 3 - 20
x = - (20 - 3)
x = -17