Các bạn giúp mn với ^^ mn k cho

Các bạn giúp mn với ^^ mn k cho

bn tên j

a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)

b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)

\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)

\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)

c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)

\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)

bạn ơi cho mk hỏi 1 bài làm giúp mk đc ko vậy ạ

2n  là số chẳn , n và n+1 n chẳn thì n+1 là lẻ và ngược lại nên A = -1

Câu a,  

2n là số chẵn nên    \((-1)^{2n}=1\)

n là số lẽ thì n+1 là chẵn và ngược lại nên \((-1)^n.(-1)^{n+1}1.\left(-1\right)=-1\)

Vậy nên    1 X  (- 1)  =  -1

\(log_5125=log_55^3=3\)

\(log_6216=log_66^3=3\)

\(log_{10}\dfrac{1}{10000}=log_{10}10^{-4}=-4\)

\(log\sqrt{1000}=log_{10}10^{\dfrac{3}{2}}=\dfrac{3}{2}\)

\(81^{log_35}=3^{3log_35}=3^{log_3125}=125\)

\(125^{log_52}=5^{3log_52}=5^{log_58}=8\)

\(\left(\dfrac{1}{49}\right)^{log_7\dfrac{1}{8}}=7^{-2log_7\dfrac{1}{8}}=7^{log_764}=64\)

\(\left(\dfrac{1}{625}\right)^{log_52}=5^{-4log_52}=5^{log_5\dfrac{1}{16}}=\dfrac{1}{16}\)

c)\(7^{2n}+7^{2n+2}=2450\)

⇒\(7^{2n}+7^{2n}.7^2=2450\)

⇒\(7^{2n}.50=2450\)

⇒\(7^{2n}=49\)\(=7^2\)

⇒2n=2

⇒n=1

a)\(\left(-\dfrac{1}{5}\right)^n=-\dfrac{1}{125}\)                   b)\(\left(-\dfrac{2}{11}\right)^m=\dfrac{4}{121}\)

\(\left(-\dfrac{1}{5}\right)^n=\left(-\dfrac{1}{5}\right)^3\)                    \(=\left(-\dfrac{2}{11}\right)^m=\left(-\dfrac{2}{11}\right)^2\)

⇒n=3                                          ⇒m=2

a) \(\dfrac{1}{10}=0,1\)

\(\dfrac{1}{100}=0,01\)

\(\dfrac{1}{1000}=0,001\)

\(\dfrac{1}{10000}=0,0001\)

b) \(\dfrac{84}{10}=8,4\)

\(\dfrac{225}{100}=2,25\)

\(\dfrac{6453}{100}=64,53\)

\(\dfrac{25789}{10000}=2,5789\)

a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)

b: 

\(B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+....+\dfrac{2}{99.101}\)

\(B=2.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)

\(B=\dfrac{2}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}\right)\)

\(B=1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(B=1.\left(1-\dfrac{1}{101}\right)\)

\(B=1.\dfrac{100}{101}\)

\(B=\dfrac{100}{101}\)

Câu 1:

a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)

\(\Rightarrow-\dfrac{2}{3x}+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2}{3}x+\dfrac{2}{3}x=\dfrac{1}{6}+\dfrac{1}{3}\)

\(\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{1}{2}\)

\(\Rightarrow x.\dfrac{4}{3}=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{3}{8}\)

lấy bài bd

2.

a) \(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{9999}{10000}\)

Ta có:

\(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{9999}{10000}< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.....\dfrac{10000}{10001}=B\)

\(\Rightarrow A.A^2< A.B\)

\(\Rightarrow A^2< \dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{9999}{10000}.\dfrac{10000}{10001}=\dfrac{1}{10001}\)

\(\Rightarrow A^2< \dfrac{1}{10001}< \dfrac{1}{10000}=\dfrac{1}{100}.\dfrac{1}{100}=\left(0,01\right)^2\)

\(\Rightarrow A< 0,01\)

Vậy A , 0,01