ƯCLN .
315 + [150+(-315)+(-50)]=
\(\frac{\left(315+372\right).3+\left(372+315\right).7}{26.13+74.14}\)
\(\frac{\left(315+372\right).3+\left(372+315\right).7}{26.13+74.48}\)=?
\(\frac{\left(315+372\right).3+\left(372+315\right).7}{26.13+74.48}\)
\(=\frac{\left(315+372\right).\left(3+7\right)}{\left(26.13\right)+\left(74.48\right)}\)
\(=\frac{687.10}{39+3552}\)
\(=\frac{6870}{3890}=\frac{687}{389}\)
B= \(\frac{\left(315+372\right).3+\left(372+315\right).7}{26.13+74.14}\)
Tính nhanh
\(B=\frac{\left(315+372\right).3+\left(372+315\right).7}{26.13+74.14}=\frac{315.3+372.3+372.7+315.7}{26.13+74.14}=\frac{315.\left(7+3\right)+372.\left(7+3\right)}{26.13+74.14}=\frac{3150+3720}{26.13+74.14}\)
đến đấy bó tay
Thực hiện phép tính sau khi bỏ dấu ngoặc: (- 315) – (2019 – 315)
Các bạn là chi tiết hộ mình nhé!
Thanks
(-315) - (2019-315)= (-315) - 1704 = -2019
(- 35 ) - (2019 - 315 ) = ( - 35 ) - 1704 = 2019
tui chẳng hiểu gì nha
A=\(2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)= ?
\(A=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
=\(\frac{1}{315}\cdot\frac{1}{651}+2\cdot\frac{1}{651}-\frac{1}{105}\cdot\left(4-\frac{1}{651}\right)-\frac{4}{315}\cdot\frac{1}{651}+\frac{4}{105}\)
=\(\frac{1}{315}\cdot\frac{1}{651}+2\cdot\frac{1}{651}-\frac{4}{105}+\frac{1}{105}\cdot\frac{1}{651}-\frac{4}{315}\cdot\frac{1}{651}+\frac{4}{105}\)
=\(\frac{1}{651}\cdot\left(\frac{1}{315}+\frac{1}{105}+2-\frac{4}{315}\right)\)+\(\frac{4}{105}-\frac{4}{105}\)
=\(\frac{2}{651}\)
Bạn sai dấu trừ ở trước số 4 phần 105 phải là cộng mình làm bài này rồi
giải cách này cũng được nè:
Đặt \(x\)=\(\frac{1}{315}\) ; \(y\)=\(\frac{1}{651}\) ; \(z\)=\(\frac{1}{105}\)
A=\(2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
=\(\left(2+\frac{1}{315}\right)\cdot\frac{1}{651}-\frac{1}{105}\cdot\left(4-\frac{1}{651}\right)-4\cdot\frac{1}{315}\cdot\frac{1}{651}+4\cdot\frac{1}{105}\)
=\(\left(2+x\right)y-z\left(4-y\right)-4xy+4z\)
=\(2y+xy-4z+zy-4xy+4z\)
=\(2y+zy-3xy\)
=\(2\cdot\frac{1}{651}+\frac{1}{105}\cdot\frac{1}{651}-3\cdot\frac{1}{315}\cdot\frac{1}{651}\)
=\(\frac{2}{651}+\left(\frac{1}{105}\cdot\frac{1}{651}-\frac{1}{105}\cdot\frac{1}{651}\right)\)
=\(\frac{2}{651}+0=\frac{2}{651}\)
92.4-27=\(\frac{x+350}{x}\)+315
Tìm y
\(92x4-27=\frac{y+350}{y}\)\(+315\)
\(92\cdot4-27=\frac{y+350}{y}+315\Leftrightarrow\frac{y+350}{y}=26\Leftrightarrow y+350=26y\Leftrightarrow350=25y\Leftrightarrow y=14\)
tính
\(2^{4^{2^{0^{315}}}}\)