\(\dfrac14+\dfrac56\)

\(=\dfrac{3}{12}+\dfrac{10}{12}\)

\(=\dfrac{3+10}{12}\)

\(=\dfrac{13}{12}\)

\(\frac14+\frac56\)

\(=\) \(\frac{6}{24}+\frac{20}{24}\)

\(=\) \(\frac{6+20}{24}\)

\(=\) \(\frac{26}{24}\) \(=\) \(\frac{13}{8}\)

\(\frac67+\frac54\)

= \(\frac{24}{28}+\) \(\frac{35}{28}\)

= \(\frac{24+35}{28}\)

= \(\frac{59}{28}\)

\(\dfrac67+\dfrac54\)

\(=\dfrac{24}{28}+\dfrac{35}{28}\)

\(=\dfrac{24+35}{28}\)

\(=\dfrac{59}{28}=2\dfrac{3}{28}\)

\(\frac23+\frac24\)

\(=\frac{8}{12}+\frac{6}{12}\)

\(=\frac{14}{12}=\frac76\)

\(\frac23+\frac34\)

\(=\) \(\frac{8}{12}+\) \(\frac{9}{12}\)

\(=\) \(\frac{8+9}{12}\)

\(=\) \(\frac{17}{12}\)

 \(5^{2} \cdot \left(\right. 16 \cdot 10 - 140 + 4^{4} + 1000 \left.\right)\)

\(=25\cdot\left(160-140+256+1000\right)\)

\(=25\cdot1276\)

\(=31900\)

\(5^2\cdot\left(16\cdot10-140+4^4+1000\right)\)

\(=25\cdot\left(160-140+256+1000\right)\)

\(=25\cdot1276\)

\(=31900\)

\(\left(x+3\right)+\left(3x+5\right)+\left(4x+7\right)=23\)

\(x+3+3x+5+4x+7=23\)

\(\left(x+3x+4x\right)+\left(3+5+7\right)=23\)

\(8x+15=23\)

\(8x=8\)

\(x=1\)

Vậy \(x=1\)

gọi số có năm chữ số đó là \(\overline{abcde}\) (a khác 0)

khi thêm chữ số 7 trước số có năm chữ số ta được \(\overline{7abcde}\)

Thì khi đó mới hơn số đã cho số đơn vị là

\(\overline{7abcde}-\overline{abcde}\) =700000(đơn vị)

Đáp số 700000(đơn vị)

bất lực-ing

\(a,\left(x-3,025\right)\cdot2,5=27,75\)

\(x-3,025=27,75:2,5\)

\(x-3,025=11,1\)

\(x=11,1+3,025\)

\(x=14,125\)

-------

\(b,x-17,4\cdot2,1=88,76\)

\(x-36,54=88,76\)

\(x=88,76+36,54\)

\(x=125,3\)

a)

\(\left(x-3,025\right)\times2,5=27,75\)

\(x-3,025=11,1\)

\(x=14,125\)

Vậy \(x=14,125\)

b)

\(x-17,4\times2,1=88,76\)

\(x=88,76+17,4\times2,1\)

\(x=125,3\)

Vậy \(x=125,3\)

14kg

\(\frac75\) yến = 10kg x \(\frac75\) = 14kg

Vậy \(\frac75\) yến = 14kg

\(=\frac{13}{3}\)

um

\(\dfrac67-\dfrac13\cdot\dfrac{-5}{7}-\dfrac32\)

\(=\dfrac67+\dfrac{5}{21}-\dfrac32\)

\(=\dfrac{36}{42}+\dfrac{10}{42}-\dfrac{63}{42}\)

\(=-\dfrac{17}{42}\)