Ê, hình như bài này sai đề

ko bt. Cô ra đề ntn thì lm thôi.Chiều hỏi cô

\(2n-1+5n-2=\frac{7}{32}\)

                   \(7n-3=\frac{7}{32}\)

                           \(7n=\frac{7}{32}+3\)

                           \(7n=\frac{103}{32}\)

                              \(n=\frac{103}{32}:7\)

                              \(n=\frac{103}{224}\)

c, \(\frac{-32}{-2^n}=4\)

\(\Rightarrow-2^n=-32:4\)

\(\Rightarrow-2^n=-8\)

\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)

d, \(\frac{8}{2^n}=2\)

\(\Rightarrow2^n=8:2\)

\(\Rightarrow2^n=4\)

\(\Rightarrow2^n=2^2\Rightarrow n=2\)

e, \(\frac{25^3}{5^n}=25\)

\(\Rightarrow5^n=25^3:25\)

\(\Rightarrow5^n=25^2\)

\(\Rightarrow5^n=5^4\Rightarrow n=4\)

i , \(8^{10}:2^n=4^5\)

\(\Rightarrow2^n=8^{10}:4^5\)

\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)

\(\Rightarrow2^n=2^{30}:2^{10}\)

\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)

k, \(2^n.81^4=27^{10}\)

\(\Rightarrow2^n=27^{10}:81^4\)

\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)

\(\Rightarrow2^n=3^{30}:3^{16}\)

\(\Rightarrow2^n=3^{14}\)

\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn 

a) \(\frac{1}{9}.27^n=3^n\)

\(\Leftrightarrow3^{-2}.3^{3n}=3^n\)

\(\Leftrightarrow3^{3n-2}=3^n\)

\(\Leftrightarrow3n-2=n\)

\(\Leftrightarrow2n=2\)

\(\Leftrightarrow n=1\)

b)\(3^{-2}.3^4.3^n=3^7\)

\(\Leftrightarrow3^{2+n}=3^7\)

\(\Leftrightarrow2+n=7\)

\(\Leftrightarrow n=5\)

1. \(\left(\frac{1}{2}\right)^n=\frac{1}{32}\)

\(\left(\frac{1}{2}\right)^n=\frac{1^5}{2^5}\)

\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^5\)

Vậy \(n=5\)

2. \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)

\(\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)

\(\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)

Vậy \(n=3\)

3. \(\frac{16}{2^n}=2\)

\(2^n=\frac{16}{2}\)

\(2^n=8=2^3\)

Vậy \(n=3\)

1. (1/2)² = 1/32 <=> (2¹)ⁿ = (2⁵)ⁿ <=> 1.n = 5.1 <=> n = 5

=> n = 5

2) 343/125 = (7/5)ⁿ <=> (7/5)³ = (7/5)ⁿ <=> 3 = n

=> n = 3

3) 16/2ⁿ = 2 <=> 16.2ⁿ <=> 2ⁿ = 2/16 <=> 2ⁿ = 1/8 <=> 2ⁿ = 8 <=> 2ⁿ = 2³ <=> n = 3

=> n = 3

\(A=1+3+3^2+3^3+...+3^{101}\)

\(3A=3+3^2+3^3+3^4+...+3^{101}\)

\(3A-A=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)

\(2A=3^{101}-1\)

\(A=\left(3^{101}-1\right):2\)

Thu gọn tổng sau:

A=1+3+3²+3³+...+3¹⁰⁰ 

B= 2¹⁰⁰-2⁹⁹-2⁹⁸-2⁹⁷-...-2²-2

C= 3¹⁰⁰-3⁹⁹+3⁹⁸-3⁹⁷-...+3²-3+1 

a)

\(\left(\frac{1}{3}\right)^n\cdot27^n=3^n\)

\(\Rightarrow\left(\frac{1}{3}\cdot27\right)^n=3^n\)

\(\Rightarrow9^n=3^n\)

\(\Rightarrow\left(3^2\right)^n=3^n\)

\(\Rightarrow3^{2n}=3^n\)

\(\Rightarrow2n=n\)

\(\Leftrightarrow n=0\)

Vậy \(n=0\)

d) Ta có:

\(6^{3-n}=216\)

\(\Rightarrow6^{3-n}=6^3\)

\(\Rightarrow3-n=3\)

\(\Rightarrow n=3-3\)

\(\Rightarrow n=0\)

Vậy \(n=0\)\(\text{ }\)

Viết lại thành : \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Dựa theo tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

-> x = \(12.\dfrac{3}{2}=18\)

y =\(12.\dfrac{4}{3}=16\)

z =\(12.\dfrac{5}{4}\) = 15

Câu a:

\(\frac{2x}{3}\) = \(\frac{3y}{4}\) = \(\frac{4z}{5}\) và \(x+y+z\) = 49

\(\frac{2x}{3}\).\(\frac{1}{12}\) = \(\frac{3y}{4}.\frac{1}{12}\) = \(\frac{4z}{5}\).\(\frac{1}{12}\)

\(\frac{x}{18}\) = \(\frac{y}{16}\) = \(\frac{z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{18}\) = \(\frac{y}{16}\) = \(\frac{z}{15}\) = \(\frac{x+y+z}{18+16+15}\) = \(\frac{49}{49}\) = 1

\(x=18.1\) = 18

\(y=16.1\) = 16

z = 15 .1 = 15

Vậy các cặp số \(x;y;z\) thỏa mãn đề bài là:

(\(x;y;z\)) = (18; 16; 15)