\(a,21.\left(-15\right)+67.\left(-15\right)+\left(-15\right)+\left(-15\right).12\)

\(=21.\left(-15\right)+67.\left(-15\right)+\left(-15\right).1+\left(-15\right).12\)

\(=-15.\left(21+67+1+12\right)\)\(=-15.101=-1515\)

\(b,\left(-30\right)+76+30\)\(=\left(-30+30\right)+76=0+76=76\)

\(d,15.\left\{22-\left[400:\left(140+12.5\right)\right]\right\}\)

\(=15.\left\{22-\left[400:\left(140+60\right)\right]\right\}\)

\(=15.\left[22-\left(400:200\right)\right]\)

\(=15.\left(22-2\right)=15.20=600\)

a. (-15) x 101 = -1515 

b. (-30+30) +76 = 76

a) 21.(-15)+67.(-15)+(-15)+(-15).12

= (-15).(21+67+1+12)

= (-15).101

= -1515

b) (-30)+76+30

= [(-30)+30]+76

= 0+76 = 76

c) 15.{22-[400:(140+12.5)]}

= 15.{22-[400:(140+60)]}

= 15.{22-[400:200]}

= 15.{22-2}

= 15.20

= 300

\(3^5\)

A

A.15 là ước chung của 15 và 30.

A

\(x^2=900\Leftrightarrow x^2=30^2\Rightarrow x=30\)

Chọn A

mik nghĩ A

\(\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}=\left(\frac{2}{7}\right)^{30}\)

\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}=\left(\frac{3}{7}\right)^{30}\)

Vì \(\frac{2}{7}<\frac{3}{7}\Rightarrow\left(\frac{2}{7}\right)^{30}<\left(\frac{3}{7}\right)^{30}\)

Vậy \(\frac{4^{15}}{7^{30}}<\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\).

Bài 1:

a) \(24 - (-15) - 2\)

\(=39-2\)

\(=37\)

b) \((-85) + 10 - (-85) - 50\)

\(=[(-85)-(-85)]+10-50\)

\(=0+10-50\)

\(=10-50\)

\(=-40\)

c) \(71 - (-30) - (+18) + (-30)\)

\(=[(-30)-(-30)]+71-(+18)\)

\(=0+71-18\)

\(=71-18\)

\(=53\)

d) \(-(30) - (+37) + (+37) + (-85)\)

\(=[-(+37)+(+37)]-(30)+(-85)\)

\(=0-(30)+(-85)\)

\(=(-30)+(-85)\)

\(=-115\)

e) \((35-815) - (795-65)\)

\(=(-780)-730\)

\(=-1510\)

g) \((2002-79+15) + (-79+15)\)

\(=1938+(-64)\)

\(=1874\)

Bài 2:

a) \(25 - (30+x) = x - (27-8)\)

\(25-30-x=x-27+8\)

\(x+x=25-30+27-8\)

\(2x=14\)

\(x=14\div2\)

\(x=7\)

b) \((x-12) - 15 = (20-17) - (18+x) \)

\(x-12-15=13-18-x\)

\(x-27=-5-x\)

\(x+x=-5+27\)

\(2x=22\)

\(x=22\div2\)

\(x=11\)

c) \(15 - x = 7 - (-2)\)

\(15-x=9\)

\(x=15-9\)

\(x=6\)

d) \(x - 35 = (-12) - 3\)

\(x-35=-15\)

\(x=-15+35\)

\(x=20\)

e) \(\left|5-x\right|-26=-15\)

\(\left|5-x\right|=-15+26\)

\(\left|5-x\right|=11\)

Từ đây ta có:

*Nếu \(5-x=11\)

\(x=5-11\)

\(x=-6\)

*Nếu \(5-x=-11\)

\(x=5-(-11)\)

\(x=16\)

Vậy \(x=-6;x=16\)

a: \(x\in\left\{45;60\right\}\)

b: \(x\in\varnothing\)

D

D

d?