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a, (1200+60) :12= 1200: 12+ 60:12= 100 +5= 105
b, (2100-42) :21= 2100: 21-42: 21= 100- 2=98
Mình có cách khác nhé :
a ) ( 1200 + 60 ) : 12
= ( 12 x 100 + 12 x 5 ) : 12
= 12 x ( 100 + 5 ) : 12
= 12 x 105 : 12
= 105
b ) ( 2100 - 42 ) : 21
= ( 21 x 100 - 21 x 2 ) : 21
= 21 x ( 100 - 2 ) : 21
= 21 x 98 : 21
= 98

A = 18:26+(-5):27+(-22):86+12:39+(-32):43 = 9:13+(-5):27+(-11):43+4:13+(-32):43 = (9:13+4:13)+[(-11):43+(-32):43]+(-5):27 = 1+(-1)+5:27 = -5:27
B =(-10):12+8:15+(-19):56+3:(-18)+28:60 = (-5):6+8:15+(-19):56+1:(-6)+7:15 = [(-5):6+1:(-6)]+(8:15+7:15)+(-19):56 = (-1)+1+(-19):56 = (-19) :56

Câu 8:
a:Sửa đề: \(4+4^2+\cdots+4^{2025}\)
Ta có: \(4+4^2+\cdots+4^{2025}\)
\(=\left(4+4^2+4^3\right)+\left(4^4+4^5+4^6\right)+\cdots+\left(4^{2023}+4^{2024}+4^{2025}\right)\)
\(=4\left(1+4+4^2\right)+4^4\left(1+4+4^2\right)+\cdots+4^{2023}\left(1+4+4^2\right)\)
\(=21\left(4+4^4+\cdots+4^{2023}\right)\) ⋮21
b: \(5+5^2+5^3+5^4+\cdots+5^{2024}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{2023}+5^{2024}\right)\)
\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+\cdots+5^{2022}\left(5+5^2\right)\)
\(=30\left(1+5^2+\cdots+5^{2022}\right)\) ⋮30
Câu 7:
a: \(A=2+2^2+2^3+\cdots+2^{99}\)
=>\(2A=2^2+2^3+\cdots+2^{100}\)
=>\(2A-A=2^2+2^3+\cdots+2^{100}-2-2^2-\cdots-2^{99}\)
=>\(A=2^{100}-2\)
b: \(B=1-7+7^2-7^3+\cdots+7^{48}-7^{49}\)
=>\(7B=7-7^2+7^3-7^4+\cdots+7^{49}-7^{50}\)
=>\(7B+B=7-7^2+7^3-7^4+\cdots+7^{49}-7^{50}+1-7+7^2-7^3+\cdots+7^{48}-7^{49}\)
=>\(8B=-7^{50}+1\)
=>\(B=\frac{-7^{50}+1}{8}\)
Câu 4:
a: \(x^3=125\)
=>\(x^3=5^3\)
=>x=5
b: \(11^{x+1}=121\)
=>\(11^{x+1}=11^2\)
=>x+1=2
=>x=2-1=1
c: \(\left(x-5\right)^3=27\)
=>\(\left(x-5\right)^3=3^3\)
=>x-5=3
=>x=3+5=8
d: \(4^5:4^{x}=16\)
=>\(4^{x}=4^5:16=4^5:4^2=4^3\)
=>x=3
e: \(5^{x-1}\cdot8=1000\)
=>\(5^{x-1}=1000:8=125=5^3\)
=>x-1=3
=>x=3+1=4
f: \(2^{x}+2^{x+3}=72\)
=>\(2^{x}+2^{x}\cdot8=72\)
=>\(2^{x}\cdot9=72\)
=>\(2^{x}=\frac{72}{9}=8=2^3\)
=>x=3
g: \(\left(3x+1\right)^3=343\)
=>\(\left(3x+1\right)^3=7^3\)
=>3x+1=7
=>3x=6
=>x=2
h: \(3^{x}+3^{x+2}=270\)
=>\(3^{x}+3^{x}\cdot9=270\)
=>\(10\cdot3^{x}=270\)
=>\(3^{x}=\frac{270}{10}=27=3^3\)
=>x=3
i: \(25^{2x+4}=125^{x+3}\)
=>\(\left(5^2\right)^{2x+4}=\left(5^3\right)^{x+3}\)
=>\(5^{4x+8}=5^{3x+9}\)
=>4x+8=3x+9
=>x=1
Câu 6:
1 giờ=3600 giây
Số tế bào hồng cầu được tạo ra sau mỗi giờ là:
\(25\cdot10^5\cdot3600=25\cdot36\cdot10^7=900\cdot10^7=9\cdot10^9\) =9 tỉ (tế bào)
câu 5:
a. \(16^{16}=\left(2^4\right)^{16}=2^{64}\)
\(64^{11}=\left(2^6\right)^{11}=2^{66}\)
vì \(2^{66}>2^{64}\) nên \(64^{11}>16^{16}\)
b. \(625^5=\left(5^4\right)^5=5^{20}\)
\(125^7=\left(5^3\right)^7=5^{21}\)
\(5^{20}<5^{21}\Rightarrow625^5<125^7\)
c. \(3^{36}=\left(3^3\right)^{12}=27^{12}\)
\(5^{24}=\left(5^2\right)^{12}=25^{12}\)
\(27^{12}>25^{12}\Rightarrow3^{36}>5^{24}\)

\(24^4\cdot3^4-32^{12}:16^{12}\)
\(=\left(24:3\right)^4-\left(32 :16\right)^{12}\)
\(=8^4-2^{12}\)
\(=8^4-\left(2^3\right)^4\)
\(=8^4-8^4=0\)

\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54.57.60}\)
\(P=4.\left(\frac{3}{1.4.7}+\frac{3}{4.7.10}+\frac{3}{7.10.13}+...+\frac{3}{54.57.60}\right)\)
\(P=4\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}\right)\)
\(P=4.\left(\frac{1}{4}-\frac{1}{3420}\right)\)
\(P=4.\frac{427}{1710}\)
\(P=\frac{854}{855}\)

1/a) 12 - x= 1-(-5)
12 - x = 6
x= 12-6
x=6
b)| x+4|= 12
x+4 = \(\pm\)12
*x+4=12
x=8
*x+4= -12
x=-16
2/Tìm n
\(n-5⋮n+2\)
=> \(n+2-7⋮n+2\)
mà \(n+2⋮n+2\)
=> 7\(⋮\)n+2
=> n+2 \(\varepsilon\)Ư(7)= {1;-1;7;-7}
n+2 | 1 | -1 | 7 | -7 |
n | -1 | -3 | 5 | -9 |
3/a)4.(-5)2 + 2.(-12)
= 2.2.(-5)2 + 2.(-12)
=2[2.25.(-12)]
=2.(-600)
=-1200

Bài 4:
a; 12 - (10 - 19)
= 12 - (-9)
= 12 + 9
= 21
b; (-27) - (13 - 19)
= 27 - 16
= 11
Bài 5:
a; 321 + (-15) + 30 + (-321)
= [321 - 321] + (30 - 15)
= 0 + 15
= 15
b; 2018 + 432 + 168 + (-2018)
= [2018 - 2018] + (432 + 168)
= 0 + 600
= 600
(1200 + 60) :12 = 1200 : 12 + 60 : 12 = 100 + 5 = 105