\(\frac{1212}{3131}=\frac{1212:101}{3131:101}=\frac{12}{31}\)

12/31 nhé

A = \(\frac{2^3.3}{2^2.3^2.5}\)

A = \(\frac{2^2.3.2}{2^2.3.3.5}\)

A = \(\frac{2}{3.5}\)

A = \(\frac{2}{15}\)

\(\frac{2^3.3}{2^23^2.5}=\frac{2}{3.5}=\frac{2}{15}\)

\(\frac{2^3.3}{2^2.3^2.5}=\frac{2}{3.5}=\frac{2}{15}\)

Thiếu dấu nhân ở chỗ \(2^2.3^2\)nha 

Ảnh hơi mờ 

\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=-\frac{5}{9}\)

-14 nha bạn

\(\frac{1}{15}\) và \(-7\)

Biết\(-7=\frac{-7}{1}\)nên MSC là : 15

Ta có:

\(\frac{1}{15}=\frac{1.1}{15.1}=\frac{1}{15}\)

\(\frac{-7}{1}=\frac{-7.15}{1.15}=\frac{-105}{15}\)

Vậy quy đồng mẫu số các phân số \(\frac{1}{15}\) và \(-7\) ta được \(\frac{1}{15}\) và \(\frac{-105}{15}\)

Đề bài:

\(A = \frac{1 2^{3} \cdot 12 1^{2} \cdot 5 - 2 2^{4} \cdot 3^{3}}{7 5^{2} \cdot 1 1^{4} - 3 0^{2} \cdot 1 1^{5}}\)

Bước 1: Biến đổi các lũy thừa

Ta rút gọn từng số:

• \(1 2^{3} = \left(\right. 3 \cdot 4 \left.\right)^{3} = 3^{3} \cdot 4^{3} = 3^{3} \cdot \left(\right. 2^{2} \left.\right)^{3} = 3^{3} \cdot 2^{6}\)
• \(12 1^{2} = \left(\right. 1 1^{2} \left.\right)^{2} = 1 1^{4}\)
• \(2 2^{4} = \left(\right. 2 \cdot 11 \left.\right)^{4} = 2^{4} \cdot 1 1^{4}\)
• \(3^{3} = 3^{3}\)
• \(7 5^{2} = \left(\right. 3 \cdot 5^{2} \left.\right)^{2} = 3^{2} \cdot 5^{4}\)
• \(1 1^{4} = 1 1^{4}\)
• \(3 0^{2} = \left(\right. 2 \cdot 3 \cdot 5 \left.\right)^{2} = 2^{2} \cdot 3^{2} \cdot 5^{2}\)
• \(1 1^{5} = 1 1^{5}\)

Bước 2: Thay vào biểu thức

Tử số:

\(1 2^{3} \cdot 12 1^{2} \cdot 5 - 2 2^{4} \cdot 3^{3} = \left(\right. 3^{3} \cdot 2^{6} \cdot 1 1^{4} \cdot 5 \left.\right) - \left(\right. 2^{4} \cdot 1 1^{4} \cdot 3^{3} \left.\right)\)

Rút chung:

\(= 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 2^{2} \cdot 5 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 4 \cdot 5 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 20 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot 19\)

Mẫu số:

\(7 5^{2} \cdot 1 1^{4} - 3 0^{2} \cdot 1 1^{5} = \left(\right. 3^{2} \cdot 5^{4} \cdot 1 1^{4} \left.\right) - \left(\right. 2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 1 1^{5} \left.\right)\)

Rút chung \(3^{2} \cdot 5^{2} \cdot 1 1^{4}\):

\(= 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 5^{2} - 2^{2} \cdot 11 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 25 - 4 \cdot 11 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 25 - 44 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. - 19 \left.\right)\)

Bước 3: Viết lại biểu thức đầy đủ

\(A = \frac{3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot 19}{3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. - 19 \left.\right)}\)

Rút gọn:

• \(3^{3} / 3^{2} = 3\)
• \(1 1^{4}\) triệt tiêu
• \(19 / \left(\right. - 19 \left.\right) = - 1\)

Còn lại:

\(A = \frac{3 \cdot 2^{4}}{5^{2}} \cdot \left(\right. - 1 \left.\right) = \frac{3 \cdot 16}{25} \cdot \left(\right. - 1 \left.\right) = \frac{48}{25} \cdot \left(\right. - 1 \left.\right) = \boxed{- \frac{48}{25}}\)

✅ Kết quả cuối cùng:

\(\boxed{A = - \frac{48}{25}}\)

Ta có: \(12^3\cdot121^2\cdot5-22^4\cdot3^3\)

\(=\left(2^2\cdot3\right)^3\cdot\left(11^2\right)^2\cdot5-11^4\cdot2^4\cdot3^3\)

\(=2^6\cdot3^3\cdot11^4\cdot5-11^4\cdot2^4\cdot3^3=11^4\cdot3^3\cdot2^4\left(2^2\cdot5-1\right)\)

\(=11^4\cdot3^3\cdot2^4\cdot19\)

Ta có: \(75^2\cdot11^4-30^2\cdot11^5\)

\(=\left(3\cdot5^2\right)^2\cdot11^4-\left(2\cdot3\cdot5\right)^2\cdot11^5\)

\(=3^2\cdot5^4\cdot11^4-2^2\cdot3^2\cdot5^2\cdot11^5\)

\(=3^2\cdot5^2\cdot11^4\left(5^2-2^2\cdot11\right)=3^2\cdot5^2\cdot11^4\cdot\left(-19\right)\)

Ta có: \(A=\frac{12^3\cdot121^2\cdot5-22^4\cdot3^3}{75^2\cdot11^4-30^2\cdot11^5}\)

\(=\frac{2^4\cdot3^3\cdot11^4\cdot19}{3^2\cdot5^2\cdot11^4\cdot\left(-19\right)}=\frac{2^4\cdot3}{5^2\cdot\left(-1\right)}=\frac{48}{-25}\)

\(A=\frac{2.6.10+4.12.20+6.18.30+...+20.60.100}{1.2.3+2.4.6+3.6.9+...+10.20.30}\)

=> \(A=\frac{2^3.1.3.5+4^3.1.3.5+6^3.1.3.5+...+20^3.1.3.5}{1.2.3+2^3.1.2.3+3^3.1.2.3+...+10^3.1.2.3}\)

=> \(A=\frac{1.3.5\left(2^3+4^3+6^3+...+20^3\right)}{1.2.3\left(1+2^3+3^3+...+10^3\right)}=\frac{1.5.2^3.\left(1+2^3+3^3+...+10^3\right)}{1.2.\left(1+2^3+3^3+...+10^3\right)}\)

=> \(A=5.2^2=20\)

Đáp số: A=20

\(\frac{2\cdot6\cdot10+4\cdot12\cdot20+...+20\cdot60\cdot100}{1\cdot2\cdot3+2\cdot4\cdot6+...+10\cdot20\cdot30}=\frac{10\cdot2\left(1\cdot2\cdot3+2\cdot4\cdot6+...+10\cdot20\cdot30\right)}{1\cdot2\cdot3+2\cdot4\cdot6+...+10\cdot20\cdot30}\)

                                                                                   \(=20\)

\(\frac{-5^3\cdot40\cdot4^3}{135\cdot\left(-2\right)^{14}\left(-100\right)^0}=\frac{-125\cdot2^3\cdot5\cdot\left(2^2\right)^3}{5\cdot27\cdot2^{14}\cdot1}=\frac{-125\cdot2^6}{27\cdot2^{11}}=\frac{-125}{27\cdot2^5}=\frac{-125}{864}\)

\(\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}\)\(=\frac{\left(-5\right)^3.5.2^3.2^6}{3^3.5.2^{14}.1}\)\(=\frac{-125}{864}\)