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a) \(\left(-28\right).\left(-3\right).\left(+4\right).\left(-7\right)=\left(+84\right).\left(+4\right).\left(-7\right)=\left(+336\right).\left(-7\right)=\left(-2352\right)\)
b) \(2.8.\left(-14\right).\left(-3\right)=16.\left(-14\right).\left(-3\right)=\left(-224\right).\left(-3\right)=672\)
Bài 1 : Ta có:
\(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}\)
= \(\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}\)
= \(\frac{7}{9}\)
Bài 2 :
\(\frac{x}{2}+\frac{3x}{4}+\frac{5x}{6}=\frac{10}{24}\)
=> \(\frac{12x+18x+20x}{24}=\frac{10}{24}\)
=> 50x = 10
=> x = 10 : 50
=> x = 1/5
Bài 3 : Để A nhận giá trị nguyên thì 3 \(⋮\)x + 3
<=> x + 3 \(\in\)Ư(3) = {1; -1; 3; -3}
Lập bảng :
x + 3 | 1 | -1 | 3 | -3 |
x | -2 | -4 | 0 | -6 |
Vậy
a.\(12^7:6^7=\left(12:6\right)^7=2^7\)
b. \(27^5:81^3=\left(3^3\right)^5:\left(3^4\right)^3=3^{15}:3^{12}=3^3\)
c. \(18^3:9^3=\left(18:9\right)^3=2^3\)
d. \(125^3:25^4=\left(25^3\right)^3:25^4=25^9:25^4=25^5\)
12^7:6^7=(12:6)^7=2^7
27^5:81^3=(3^3)^5:(3^4)^3=3^15:3^12=3^3
18^3:9^3=(18:9)^3=2^3
125^3:25^4=(5^3)^3:(5^2)^4=5^9:5^8=5
-5/12 . 2/7 + 7/12 . -3/4
= -10/84 + -21/48
= -5/42 + -7/16
= -40/336 + -147/336
= -187/336
\(\frac{-5}{12}\).\(\frac{2}{7}\)+\(\frac{7}{12}\).\(\frac{-3}{4}\)
=\(\frac{5}{12}\).\(\frac{-2}{7}\)+\(\frac{7}{12}\).\(\frac{-3}{4}\)
=\(\frac{5}{12}\).(\(\frac{-2}{7}\)+\(\frac{-3}{4}\))
\(\frac{5}{12}\).\(\frac{-29}{28}\)
=\(\frac{-145}{336}\)
((nho k cho mink nhe)
7/-78+9/65=-35/390+54/390=-19/390
-3/16+5/18=-27/144+40/144=13/144
23/-18+19/-57=-437/342+(-114/342)=-29/18
(-23).(-3).(+4).(-7)
= [(-23).(-3)].[(+4).(-7)]
= 69.(-28) = - (69. 28) = -1932