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![](https://rs.olm.vn/images/avt/0.png?1311)
\(4,7\div0,25+5,3\times4\)
\(=18,8+21,2\)
\(=40\)
\(3\times\left(a-2\right)+150=240\)
\(3\times\left(a-2\right)=90\)
\(a-2=30\)
\(a=32\)
\(\dfrac{1}{9}+a+\dfrac{7}{12}=\dfrac{17}{18}\)
\(\dfrac{1}{9}+a=\dfrac{13}{36}\)
\(a=\dfrac{1}{4}\)
\(\left(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}+\dfrac{1}{6}\times\dfrac{1}{7}+\dfrac{1}{7}\times\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)
\(\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\right)\times a=\dfrac{9}{16}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)
\(\dfrac{3}{8}\times a=\dfrac{9}{16}\)
\(a=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{3}\)x \(\frac{5}{8}\)x \(\frac{8}{15}\)= \(\frac{2}{3}\)x \(\frac{1}{15}\)= \(\frac{2}{45}\)
\(\frac{22}{5}\)x \(12\)x \(\frac{20}{40}\)= \(\frac{22}{5}\)x \(12\)x \(\frac{1}{2}\)= \(\frac{22}{5}\)x 6 = \(\frac{122}{5}\)
\(\frac{7}{2}\)x \(\frac{26}{7}\)x \(\frac{4}{13}\)= \(\frac{91}{7}\)x \(\frac{4}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1+2-3-4+5+6-7-8+9+9+10-11-12+12+...229-300+301+302
=(302+1)-(301+2)-(300+3)+(299+4)-...
=303-303-303-303-....
=0-0-0-0-0-....
=0
1+2-3-4+5+6-7-8+9+10-11-12+...-229-300+301+302
= (1+2-3)+(4+5+6-7)+...+(228-229-300+301)+302
= 0 + 0 +...+ 0 + 302
= 302
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{1.3+3.5+5.7+7.9}{3.6+9.10+15.14+21.18}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.2.3+3.5.2.3+5.7.2.3+7.9.2.3}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.6+3.5.6+5.7.6+7.9.6}\)
= \(\frac{1.3+3.5+5.7+7.9}{6.\left(1.3+3.5+5.7+7.9\right)}=\frac{1}{6}\)
Dấu "." là dấu nhân cấp 2
b) \(\frac{1.2+2.3+3.4+4.5}{3.6+6.9+9.12+12.15}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.3.3+2.3.3.3+3.4.3.3+4.5.3.3}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.9+2.3.9+3.4.9+4.5.9}\)
= \(\frac{1.2+2.3+3.4+4.5}{9.\left(1.2+2.3+3.4+4.5\right)}=\frac{1}{9}\)
Dấu "." là dấu nhân cấp 2
c) \(\frac{0,3+\frac{3}{7}+\frac{3}{11}}{0,4+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{\frac{3}{10}+\frac{3}{7}+\frac{3}{11}}{\frac{4}{10}+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{3.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}{4.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}=\frac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,x/12-1/3=3/4
x/12 =3/4+1/3
x/12 =13/12
vậy x =13
b,3/4-x=5/12
x=3/4-5/12
x=4/12=1/3
c,x nhân 2/3-1/3=1/4
x nhân 2/3 =1/4+1/3
x nhân 2/3 =7/12
x =7/12 : 2/3
x =21/24=7/8
d,5/7-(x-1/2)=1/7
x-1/2 =5/7-1/7
x -1/2 =4/7
x =4/7+1/2
x =15/14
bạn nhớ k cho mình nhé!
a) (-52+1)-9.3
=26-27
=-1
b)80-(4.52-3.23)
=80-(100-24)
=80-76
=4
c) [(-18)+(-7)]-15
=-25-15
=-40
d) (-219)-(-229)+12.5
=10+60
=70