a,

\(\left(25^6-15^6-10^6\right):5^6\\ =\left[\left(5\cdot5\right)^6-\left(3\cdot5\right)^6-\left(2\cdot5\right)^6\right]:5^6\\ =\left(5^6\cdot5^6-3^6\cdot5^6-2^6\cdot5^6\right):5^6\\ =5^6\left(5^6-3^6-2^6\right):5^6\\ =5^6-3^6-2^6\\ =15625-729-64\\ =14896-64\\ =14832\)

b,

\(1+2+2^2+...+2^{100}\\ =1\cdot\left(1+2+2^2+...+2^{100}\right)\\ =\left(2-1\right)\left(1+2+2^2+...+2^{100}\right)=\left(2-1\right)\cdot1+\left(2-1\right)\cdot2+\left(2-1\right)\cdot2^2+...+\left(2-1\right)\cdot2^{100}\\ =2-1+2^2-2+2^3-2^2+...+2^{101}-2^{100}\\ =2^{101}-1\)

B) 1+2+2²+2³+....+2

Tính B vs B=1+2+2²+2³+....+2⁹⁹+2¹⁰⁰

=>2B=2²+2³+2⁴+...+2¹⁰⁰+2¹⁰¹

=.2B-B=(2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰

B=2¹⁰¹-1

=>B=2¹⁰¹-1

1. \(MSC=BCNN(5,9)=45\)

Quy đồng :

\(\frac{4}{5}=\frac{4\cdot9}{5\cdot9}=\frac{36}{45}\)

\(\frac{6}{9}=\frac{6\cdot5}{9\cdot5}=\frac{30}{45}\)

Mà \(36>30\Rightarrow\frac{36}{45}>\frac{30}{45}\). Vậy : \(\frac{4}{5}>\frac{6}{9}\)

2. \(\frac{9}{7}+\frac{6}{7}=\frac{15}{7}\)

\(\frac{7}{9}+\frac{7}{6}=\frac{42}{54}+\frac{63}{54}=\frac{105}{54}=\frac{35}{18}\)

Tự so sánh :v

Kết quả là 2870.

Giải nhanh bằng công thức tổng bình phương:

\(1^{2} + 2^{2} + \hdots + n^{2} = \frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}\)

Với \(n = 20\):

\(\frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870.\)

Ta có biểu thức:

\(1\times1+2\times2+3\times3+\ldots+20\times20=1^2+2^2+3^2+\ldots+20^2\)

Đây là tổng các số chính phương từ 1 đến 20.

Áp dụng công thức tổng bình phương:

\(1^2+2^2+3^2+\ldots+n^2=\frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}\)

Thay \(n = 20\):

\(\frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870\)

Bài 1:

a) Ta có: \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)

\(=\frac{1}{5}+\frac{-1}{5}-\frac{3}{5}\)

\(=\frac{-3}{5}\)

b) Ta có: \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)

\(=\frac{-7}{12}+\frac{-2}{12}-\frac{-1}{12}\)

\(=\frac{-9+1}{12}=\frac{-8}{12}=\frac{-2}{3}\)

c) Ta có: \(\left(\frac{1}{6}+\frac{-4}{13}\right)-\left(-\frac{17}{6}-\frac{30}{13}\right)\)

\(=\frac{1}{6}+\frac{-4}{13}+\frac{17}{6}+\frac{30}{13}\)

\(=3+2=5\)

d) Ta có: \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)

\(=\frac{5}{4}+\frac{7}{4}+\frac{11}{7}+\frac{2}{7}\)

\(=3+\frac{13}{7}=\frac{21}{7}+\frac{13}{7}=\frac{34}{7}\)

e) Ta có: \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)

\(=-1+1+\frac{-7}{9}\)

\(=-\frac{7}{9}\)

f) Ta có: \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)

\(=\frac{-2-\left(-11\right)}{9}+\frac{5-6}{7}\)

\(=1+\frac{-1}{7}=\frac{7}{7}+\frac{-1}{7}=\frac{6}{7}\)

329534849

\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)

\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)

\(=\frac{7}{12}+\frac{31}{12}\)

\(=\frac{38}{12}=\frac{19}{6}\)

\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)

\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)

\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)

\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)

\(=\frac{-5}{9}\cdot\frac{2}{13}\)

\(=\frac{-10}{117}\)

\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)

\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)

\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)

\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)

\(=\frac{-8}{5}-\frac{7}{5}\)

\(=-3\)

\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)

\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)

\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)

\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)

\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)

\(=\frac{4}{5}\cdot\frac{13}{7}\)

\(=\frac{52}{35}\)

a)7/12.6/11+7/12.5/11-2.7/12

=7/12(6/11+5/11-2)

=7/12(1-2)

=7/12.(-1)

=-7/12

Ta có: \(\left(5^7+7^5\right)\cdot\left(6^8+8^6\right)\left(2^4-4^2\right)\)

\(=\left(5^7+7^5\right)\left(6^8+8^6\right)\left(16-16\right)\)

=0

Ta có: \(\left(\right. 5^{7} + 7^{5} \left.\right) \cdot \left(\right. 6^{8} + 8^{6} \left.\right) \left(\right. 2^{4} - 4^{2} \left.\right)\)

\(= \left(\right. 5^{7} + 7^{5} \left.\right) \left(\right. 6^{8} + 8^{6} \left.\right) \left(\right. 16 - 16 \left.\right)\)

=0

kb với em với

108/91

a)56+48=104

b)343-216-125=2

c)1296-32*27=1296-864=432

d)=0(vì các số *với 0 đều =0(\(2^4\)-4\(^2\)=0)

a) \(2^3.7+3^2.6=8.7+9.6\)

                          \(=56+54\) 

                          \(=110\)

b) \(7^3-6^3-5^3=343-216-125\)

                            \(=2\)   

c) \(6^4-2^5.3^3=1296-32.27\)

                        \(=1296-864\)

                        \(=432\)

d) \(\left(7^9-9^7\right)\left(6^8-8^6\right)\left(3^5-5^3\right)\left(2^4-4^2\right)\)

\(=\left(7^9-9^7\right)\left(6^8-8^6\right)\left(3^5-5^3\right).0\)

\(=0\)

NHỚ K CHO MÌNH NHÉ !