a)=\(\frac{7}{30}\)-\(\frac{17}{12}\)=\(\frac{-71}{60}\)                                                                                                                                        b) = \(\frac{29}{20}\)+\(\frac{-31}{30}\)+\(\frac{-2}{3}=\frac{-1}{4}\)

a)\(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}\cdot\frac{1}{3}=4+\frac{1}{27}=\frac{108}{27}+\frac{1}{27}=\frac{109}{27}\)

b)\(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[\left(-8\right)\cdot2\right]=8+3-16=-5\)

a/ \(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}:3=5-1+\frac{1}{27}=4+\frac{1}{27}=\frac{109}{27}\)

b/ \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[-8:\frac{1}{2}\right]=11+-16=-5\)

a) 31 . 65 + 31 . 35 - 500

=31(65+35)-500

=31..100-500

=3100-500

=2600

a, 31.56+31.35-500

=31(65+35)-500

=3100-500

=2600

a) \(3^{7} \cdot 27^{5} \cdot 81^{3}\)

• Đổi về cơ số 3: \(27 = 3^{3} , \textrm{ }\textrm{ } 81 = 3^{4}\).

\(3^{7} \cdot \left(\right. 3^{3} \left.\right)^{5} \cdot \left(\right. 3^{4} \left.\right)^{3} = 3^{7} \cdot 3^{15} \cdot 3^{12} = 3^{7 + 15 + 12} = 3^{34} .\)

b) \(100^{6} \cdot 1000^{5} \cdot 10 \textrm{ } 000^{3}\)

• Đổi về cơ số 10: \(100 = 10^{2} , \textrm{ }\textrm{ } 1000 = 10^{3} , \textrm{ }\textrm{ } 10 \textrm{ } 000 = 10^{4}\).

\(\left(\right. 10^{2} \left.\right)^{6} \cdot \left(\right. 10^{3} \left.\right)^{5} \cdot \left(\right. 10^{4} \left.\right)^{3} = 10^{12} \cdot 10^{15} \cdot 10^{12} = 10^{39} .\)

c) \(\frac{36^{5}}{18^{5}}\)

• Do cùng số mũ:

\(\left(\left(\right. \frac{36}{18} \left.\right)\right)^{5} = 2^{5} = 32.\)

d) \(24 \cdot 5^{2} + 5^{2} \cdot 5^{3}\)

\(= 24 \cdot 25 + 25 \cdot 125 = 600 + 3125 = 3725.\)

e) \(\frac{125^{4}}{5^{8}}\)

• Đổi về cơ số 5: \(125 = 5^{3}\).

\(\left(\right. 5^{3} \left.\right)^{4} : 5^{8} = 5^{12} : 5^{8} = 5^{12 - 8} = 5^{4} = 625.\)

TÓM lại là a) \(3^{34}\)
b) \(10^{39}\)
c) \(32\)
d) \(3725\)
e) \(625\)

bạn muốn chép đáp án hay sem cách làm///??

a: \(3^7\cdot27^5\cdot81^3\)

\(=3^7\cdot\left(3^3\right)^5\cdot\left(3^4\right)^3\)

\(=3^7\cdot3^{15}\cdot3^{12}=3^{7+15+12}=3^{34}\)

b: \(100^6\cdot1000^5\cdot10000^3\)

\(=\left(10^2\right)^6\cdot\left(10^3\right)^5\cdot\left(10^4\right)^3\)

\(=10^{12}\cdot10^{15}\cdot10^{12}=10^{15+12+12}=10^{39}\)

c: \(36^5:18^5=\left(\frac{36}{18}\right)^5=2^5=32\)

d: \(24\cdot5^2+5^2\cdot5^3\)

\(=5^2\left(24+5^3\right)\)

\(=25\cdot\left(24+125\right)=25\cdot149=3725\)

e: \(125^4:5^8=\left(5^3\right)^4:5^8=5^{12}:5^8=5^{12-8}=5^4\)

a) \(22\frac{1}{2}\cdot\frac{7}{9}+50\%-1,25\)

\(=\frac{45}{2}\cdot\frac{7}{9}+\frac{50}{100}-\frac{125}{100}\)

\(=\frac{5}{2}\cdot\frac{7}{1}+\frac{1}{2}-\frac{5}{4}\)

\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}=18-\frac{5}{4}=\frac{67}{4}\)

b) \(1,4\cdot\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)

\(=\frac{7}{5}\cdot\frac{15}{49}-\frac{22}{15}:\frac{11}{15}\)

\(=\frac{1}{1}\cdot\frac{3}{7}-\frac{22}{15}\cdot\frac{15}{11}\)

\(=\frac{3}{7}-2=\frac{3-14}{7}=\frac{-11}{7}\)

c) \(\left(-\frac{1}{2}\right)^2-\frac{7}{16}:\frac{7}{4}+75\%\)

\(=\frac{1}{4}-\frac{7}{16}\cdot\frac{4}{7}+\frac{75}{100}\)

\(=\frac{1}{4}-\frac{1}{4}+\frac{3}{4}=\frac{3}{4}\)

Bài 2  Bạn tự làm nhé

1.a,\(22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)

\(=\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)

\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)

\(=\frac{67}{4}\)

b,Các phép tính khác làm tương tự

Đổi các số ra hết thành phân số,có ngoặc thì lm ngoặc trc,Xoq đến nhân chia trước dồi mới cộng trừ

c,tương tự

2.

a,\(1\frac{3}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)

\(\frac{8}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)

\(\frac{7}{12}\div x=\frac{-77}{20}\)

Đến đây dễ bạn tự làm

b,\(\left(2\frac{4}{5}.x+50\right)\div\frac{2}{3}=-51\)

\(\left(\frac{14}{5}x+50\right)\div\frac{2}{3}=-51\)

\(\frac{14}{5}x+50=-34\)

\(\frac{14}{5}x=-84\)

Tự làm tiếp

c,\(\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)\(\Rightarrow\left|\frac{3}{4}x-\frac{1}{2}\right|=\varnothing\)

a) 22 + 3 . 2³ = 22 + 3 . 8 = 22 + 24 = 46

b) 41 . 67 + 33 . 41 = 41 . ( 67 + 33 ) = 41 . 100 = 4100

c) 322 . 5 - [ 622 : 4 + 2 . ( 11 - 8) ] = 322 . 5 - [ 622 : 4 + 2 . 3 ] = 322 . 5 - [ 155,5 + 6 ] = 322 . 5 - 161,5 = 1610 + 161,5 = 1448,5

a58b chia hết cho 2 và 5 => b=0 => a58b = a580 

a580 chia hết cho 9 thì cũng chia hết cho 3 => để a580 chia hết cho 9 => a+5+8 = 13+a phải chia hết cho 9 => a=5

\(=\frac{\left(2^2\right)^3.3^5-\left(3^2\right)^2.4^3}{\left(2^2\right)^3.9^2-\left(3^2\right)^2.\left(2^3\right)^2}=\frac{4^3.3^5-3^4.4^3}{2^6.9^2-9^2.2^6}=\frac{4^3.\left(3^5-3^4\right)}{9^2.\left(2^6-2^6\right)}=\frac{162}{0}\)

phps tính không hợp lệ vì không tồn taij phân số có mẫu bằng 0

a) \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right]\)

\(=8+3.1+4:\frac{1}{2}\)

\(=8+3+8=19\)

b)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}\)\(=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)

c) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)

\(=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)

d) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)