2.3^x + 3^{x - 1} = 7 . (3² + 2 . 6²)

=> 2.3^x + 3^{x - 1} = 567

=> 7 . 3^{x - 1} = 567

=> 3^{x - 1} = 567 : 7 = 81

=> x - 1 = 4

=> x = 5

a)2*3^x+3^x-1=7(3²+2*6²)

2*3^x+3^x-1=7(9+72)=7*81

2*3^x+3^x/3=567

2*3^x+3^x*1/3=567

(2+1/3)*3^x=567

7/3*3^x=567

3^x=567:7/3

3^x=243=3⁵

=>x=5

b) mk ko hiểu đề mấy, cái chỗ 7^x+2 là nhân vs 2 ak

\(9^x:3^x=3^7\)

\(\Rightarrow9:3^x=3^7\)

\(\Rightarrow3^x=3^7\)

\(\Rightarrow x=7\)

9^x : 3^x = 3⁷

=> 9 : 3^x = 3⁷

=> 3^x = 3⁷

=> x = 7

9^x ÷ 3^x = 3⁷

=> 9 ÷ 3^x = 3⁷

=> 3^x = 3⁷

=> x = 7

\(\left(\frac{4}{9}\right)^n=\left(\frac{2}{3}\right)^5\)

<=>\(\left(\frac{2}{3}\right)^{\frac{n}{2}}=\left(\frac{2}{3}\right)^5\)

<=>\(\frac{n}{2}=5\)

<=>n=10

\(\left(\frac{4}{9}\right)^n=\left(\frac{2}{3}\right)^5\)

\(\Rightarrow\left(\frac{2}{3}\right)^{2n}=\left(\frac{2}{3}\right)^5\)

\(\Rightarrow2n=5\Rightarrow n=\frac{5}{2}\)

Vậy n = 5/2 

\(\frac{1}{2}\cdot2^x+2^x\cdot2^2=2^8+2^5\)

\(2^x\left(\frac{1}{2}+4\right)=2^8+2^5\)

\(2^x\cdot\frac{9}{2}=288\)

\(2^x=64\)

\(2^x=2^6\)

\(x=6\)

\(9^x:3^x=3^7\)

\(3^{2x}:3^x=3^7\)

\(3^x=3^7\)

\(x=7\)

\(7^{x+2}+2\cdot7^{x-1}=345\)

\(7^x\cdot7^2+2\cdot7^x:7=345\)

\(7^x\left(7^2+\frac{2}{7}\right)=345\)

\(7^x\cdot\frac{345}{7}=345\)

\(7^x=7\)

\(x=1\)

a) 1/2.2^x + 2^x+2 = 256 + 32

1/2.2^x + 2^2.2^x=288

2^x(1/2+4)= 288

2^x.4,5=288

2^x= 288:4,5

2^x=64=2^6

x=6

a.\(3^{-2}.3^2.27^x=\frac{1}{3}\)

\(\Rightarrow3^{-2+2}.\left(3^3\right)^x=\frac{1}{3}\)

\(\Rightarrow3^0.3^{3x}=3^{-1}\)

\(\Rightarrow3^{3x}=3^{-1}\)

=> 3x=-1

=> x=\(-\frac{1}{3}\)

b.\(7^{x+2}+2.7^{x-1}=345\)

\(\Rightarrow7^{x-1}.\left(7^3+2\right)=345\)

\(\Rightarrow7^{x-1}.345=345\)

=> 7^x-1=345 : 345

=> 7^x-1=1

=> 7^x-1=7⁰

=> x-1=0

Vậy x=1.

c.\(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Rightarrow\left(2x-1\right)\in\left\{-1;0;1\right\}\)

=> 2x-1=-1               hoặc 2x-1=0                         hoặc 2x-1=1

=> 2x=0                   hoặc 2x=1                            hoặc 2x=2

=> x=0                     hoặc x=\(\frac{1}{2}\)                           hoặc x=1

Vậy \(x\in\left\{0;\frac{1}{2};1\right\}\)

b/100x+(1+2+3+...+100)=205550

100x+5050=205550

100x=205550-5050

100x=200500

x=200500/100

x=2005

d/(3x-2⁴).7⁵=2.7⁶.1/2009⁰

(3x-2⁴).7⁵=2.7⁶.1

(3x-2⁴)=2.7⁶:7⁵

(3x-2⁴)=2.7

3x-16=14

3x=14+16

3x=30

x=30/10=3

b) ( \(x\)+ 1 ) + ( \(x\)+ 2 ) + ( \(x\)+ 3 ) + ... + ( \(x\)+ 100 ) = 205550

\(x\)x 100 + ( 1 + 2 + 3 + ... + 100 ) = 205550

\(x\)x 100 + 5050 = 205550

 \(x\)x 100 = 205550 - 5050 

\(x\)x 100 = 200500

\(x\)= 200500 : 100

\(x\)= 2005

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

\(1+\frac{1}{3}+\frac{1}{6}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\frac{2}{6}+\frac{2}{12}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)\right]=4\)

\(\Leftrightarrow1+2\left(\frac{1}{2}-\frac{1}{\left(x+1\right)}\right)=4\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{4-1}{2}=\frac{3}{2}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2}-\frac{3}{2}=-1\)

\(\Leftrightarrow x=-1+1=-2\)

Vậy x = -2 

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{2.6}+\frac{2}{2.10}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)

\(\Leftrightarrow2\left(1-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)

\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=1\frac{1991}{1993}\div2\)

\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=\frac{1992}{1993}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=1-\frac{1992}{1993}=\frac{1}{1993}\)

\(\Leftrightarrow x+1=1993\)

\(\Leftrightarrow x=1992\)