c) \(\frac{3\cdot7-3\cdot19}{3\cdot4}=\frac{3\left(7-19\right)}{3\cdot4}=\frac{3\cdot\left(-12\right)}{3\cdot4}=\frac{3\cdot4\cdot\left(-3\right)}{3\cdot4}=-3\)

Vậy \(\frac{3\cdot7-3\cdot19}{3\cdot4}=-3\)

d,\(\frac{2^3\cdot9^4-2^4\cdot3^7}{2^4\cdot3^7}=\frac{2^3\cdot\left(3^2\right)^4-2^4\cdot3^7}{2^4\cdot3^7}=\frac{2^3\cdot3^8-2^4\cdot3^7}{2^2\cdot3^7}=\frac{2^3\cdot3^7\left(3-2\right)}{2^2\cdot3^7}=2\)

bạn làm đúng rồi nhé

chúc bạn học tốt@

phân số thứ 2 ở trên tử số\(5^3\) thành \(5^2\) nha

\(\dfrac{a}{b}=\dfrac{3^4\cdot2^2\cdot\left(2-1\right)}{3^5\cdot\left(3^2-5\right)}=\dfrac{2^2}{3}\cdot\dfrac{1}{4}=\dfrac{1}{3}\)

\(\dfrac{c}{d}=\dfrac{5^3\left(7-2\right)}{5^3\left(7-4\right)}=\dfrac{5}{3}\)

Do đó: a/b<c/d

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 !

a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)

b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)

\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)

A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)

B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)

B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)

B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)

X bất kì sao cho X \(\in\)N*

1 . M =-1/9

2. x= 3/7

3. 1/3

viết 7 phần -12 kiểu j vậy . dậy mk đi

gọi biểu thức cần CM là B

\(3B=1+\frac23+\frac{3}{2^2}+\frac{4}{3^3}+\cdots+\frac{100}{3^{99}}\)

=> \(3B-B=1+\left(\frac23-\frac13\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+\cdots+\left(\frac{100}{3^{99}}-\frac{99}{3^{99}}\right)-\frac{100}{3^{100}}\)

\(2B=1+\frac13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

đặt C= \(\frac13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{99}}\)

=> \(3C=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}\)

=> \(3C-C=\left(1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}\right)-\left(\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{99}}\right)\)

\(2C=1-\frac{1}{3^{99}}\)

=> \(C=\frac12-\frac{1}{2\cdot3^{99}}\)

\(2B=1+\frac12-\left(\frac{1}{2\cdot3^{99}}+\frac{100}{3^{100}}\right)\)

vì trong ngoặc lớn hơn 0

=> \(2B<\frac32\)

\(B<\frac34\left(đpcm\right)\)

\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\left(\frac{1}{4\cdot5}+.\ldots+\frac{1}{99\cdot100}\right)\)

\(A=\frac{7}{12}+\left(\frac{1}{4\cdot5}+\cdots+\frac{1}{99\cdot100}\right)\)

Mà \(\left(\frac{1}{4\cdot5}+\cdots+\frac{1}{99\cdot100}\right)>0\)

=> A>\(\frac{7}{12}\)

mặt khác ta có: \(A=1-\frac12+\frac13-\frac14+\frac14-\frac15+\cdots+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\left(\frac12-\frac13\right)-\left(\frac14-\frac15\right)-.\ldots-\left(\frac{1}{98}-\frac{1}{98}\right)-\frac{1}{100}\)

\(A=\frac56-\left(\frac14-\frac15\right)-.\ldots-\left(\frac{1}{98}-\frac{1}{98}\right)-\frac{1}{100}\)

=> \(A<\frac56\)

Vậy \(\frac{7}{12}<A<\frac56\)