9.5 x 3.7+ 9.5 x 1.3+ 9.5:0.25+9.5
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\(2,5\) x \(9,5\) x \(4+20\) x \(9,5\) x \(0,5\) +\(9,5\) x \(8\) x \(1,25\)
\(=9,5\) x ( \(2,5\) x \(4+20\) x \(0,5\) + \(8\) x \(1,25\) )
\(=9,5\) x \(30\)
\(=285\)
`2,5xx9,5xx4+20xx9,5xx0,5+9,5xx8xx1,25`
`=9,5xx10+9,5xx10+9,5xx10`
`=9,5xx(10+10+10)`
`=9,5xx30`
`=285`
Đặt P= \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\) : \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
Có : \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\)
= \(\dfrac{\left(2.5-9\right).5^{21}}{\left(5^2\right)^{10}}\)= \(\dfrac{\left(10-9\right).5^{21}}{5^{20}}\)=\(\dfrac{5^{21}}{5^{20}}\)= 5 (1)
Có: \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
= \(\dfrac{5.\left[7^{14}.\left(3.7-19\right)\right]}{\left[7^{15}.\left(3+7\right)\right]}\)=\(\dfrac{5.7^{14}.2}{7^{15}.10}\)=\(\dfrac{10.7^{14}}{7^{15}.10}\)=\(\dfrac{1}{7}\) (2)
Từ (1) và (2) suy ra:
A= 5:\(\dfrac{1}{7}\)=5.7=35
Vậy A=35 hay \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\):\(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)= 35
\(-\dfrac{5}{9}\times\dfrac{5}{11}+\dfrac{6}{9}\times\dfrac{5}{11}+\dfrac{5}{9}\)
\(=\dfrac{5}{11}\times\left(-\dfrac{5}{9}+\dfrac{6}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{5}{11}\times\dfrac{6}{9}\)
\(=\dfrac{30}{99}\)
\(=\dfrac{10}{33}\)
Cho mình sửa lại ạ:
\(-\dfrac{5}{9}\times\dfrac{5}{11}+\dfrac{6}{9}\times\dfrac{5}{11}+\dfrac{5}{9}\)
\(=\dfrac{5}{11}\times\left(-\dfrac{5}{9}+\dfrac{6}{9}\right)+\dfrac{5}{9}\)
\(=\dfrac{5}{11}\times\dfrac{1}{9}+\dfrac{5}{9}\)
\(=\dfrac{5}{99}+\dfrac{5}{9}\)
\(=\dfrac{60}{99}\)
\(=\dfrac{20}{33}\)
\(9,5\times3,7+9,5\times1,3+9,5:0,25+9,5\)
\(=9,5\times3,7+9,5\times1,3+9,5\times4+9,5\)
\(=9,5\times\left(3,7+1,3+4+1\right)\)
\(=9,5\times10\)
\(=95\)