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`@` `\text {Ans}`
`\downarrow`
`x^3 = 125`
`=> x^3 = 5^3`
`=> x = 5`
Vậy, `x = 5`
_____
`4^5 \div 4^3`
`=`\(4^{5-3}\)
`= 4^2`
______
\(8^{12}\cdot8^3\div8^{15}\)
`=`\(8^{12+3-15}\)
`=`\(8^0=1\)
______
\(7^{15}\div7^{13}\)
`=`\(7^{15-13}\)
`=`\(7^2\)
_______
\(17^{15}\cdot17^{10}\div17^{22}\)
`=`\(17^{15+10-22}\)
`= 17^3`
_______
\(5^{10}\cdot25\div5^{11}\)
`=`\(5^{10}\cdot5^2\div5^{11}\)
`=`\(5^{10+2-11}\)
`= 5`
_______
\(15^{10}\cdot225\div15^{11}\)
`=`\(15^{10}\cdot15^2\div15^{11}\)
`=`\(15^{12}\div15^{11}\)
`= 15`
`@` `\text {Kaizuu lv uuu}`
\(x^3=125\)
\(\Rightarrow x^3=5^3\)
\(\Rightarrow x=3\)
___________________
\(4^5:4^3\)
\(=4^{5-3}\)
\(=4^2=16\)
___________________
\(8^{12}.8^3:8^{15}\)
\(=8^{15}:8^{15}\)
\(=1\)
___________________
\(7^{15}:7^{13}\)
\(=7^{15-13}\)
\(=7^2=49\)
___________________
\(17^{15}.17^{10}:17^{22}\)
\(=17^{25}:17^{22}\)
\(=17^3=4913\)
___________________
\(5^{10}.25:5^{11}\)
\(=5^{10}.5^2:5^{11}\)
\(=5^{12}:5^{11}\)
\(=5^1=5\)
___________________
\(15^{10}.225:15^{11}\)
\(=15^{10}.15^2:15^{11}\)
\(=15^{12}:15^{11}\)
\(=15^1=15\)
__________________
* an . am = an+m
an : am = an-m
Chúc bạn học tốt
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
a) \(\left(-25\right).8.\left(-4\right).\left(-125\right).3\)
\(=\text{ }\left[\left(-25\right).\left(-4\right)\right].\text{ }\left[8.\left(-125\right)\right].3\)
\(=100.\left(-1000\right).3\)
\(=-100000.3\)
\(=-300000\)
b) \(27.\left(-17\right)+17.\left(-73\right)\)
\(=\left(-27\right).17+17.\left(-73\right)\)
\(=17\left[\left(-27\right).\left(-73\right)\right]\)
\(=17.\left(-100\right)\)
\(=-1700\)
c) \(512\left(2-128\right)-128\left(-512\right)\)
\(=512.2-512.128-128\left(-512\right)\)
\(=512.2-128\left[512+\left(-512\right)\right]\)
\(=1024-128.0\)
\(=1024-0\)
\(=1024\)
a, (-25).8.(-4).(-125).3
= (-25).(-8).(-4).(-125).(-3)
= {(-25).(-4)}.{(-8).(-125)}.(-3)
= 100.1000.(-3)
= 100000.(-3)
= -300000
b, 27.(-17)+17.(-73)
= 27.17+17.73
= 27.(17+73)
=27.90
=2430
c, 512.(2-128)-128.(-512)
= 512.(-2) - 512.128 - (-128).512
= { 512.{(-2)-128-(-128)}
= 512.{(-2).256}
=512.(-512)
= -262144
a là x và y thuộc nhóm rỗng
b thì =-1+-1+-1+...+-1+2017=-1008+2017=1009
c là vì 4S+1 là 5^2016 chia hết cho 5^2016
vì 6(5+5^2+...+5^2014) chia hết cho 6 và bằng S
a: \(\left[\left(10-x\right)\cdot2+51\right]:3-2=3\)
=>\(\left[2\left(10-x\right)+51\right]:3=5\)
=>\(\left[2\left(10-x\right)+51\right]=15\)
=>\(2\left(10-x\right)=15-51=-36\)
=>10-x=-36/2=-18
=>\(x=10-\left(-18\right)=10+18=28\)
b: \(\left(x-12\right)-15=20-\left(17+x\right)\)
=>\(x-12-15=20-17-x\)
=>\(x-27=3-x\)
=>\(2x=30\)
=>\(x=\dfrac{30}{2}=15\)
c: \(720-\left[41-\left(2x-5\right)\right]=2^3\cdot5\)
=>\(720-\left[41-2x+5\right]=8\cdot5=40\)
=>\(\left[41-2x+5\right]=720-40=680\)
=>-2x+46=680
=>-2x=680-46=634
=>\(x=\dfrac{634}{-2}=-317\)
bài 1
a) 155 - 10.(x+1) = 55
=>10 .(x+1) = 100
=>x + 1 = 10
=>x = 9
còn lại tương tự
x-17=60
x=77
\(720:\left(x-17\right)=12\)
\(x-17=720:12\)
\(x-17=60\)
\(x=77\)
\(5^x.25^3=125^4\)
\(5^x.5^{2.3}=5^{3.4}\)
\(x+6=12\)
\(x=6\)