\(B=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)\)

\(=\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)...\left(\frac{n.\left(n+2\right)+1}{n.\left(n+2\right)}\right)\)

\(=\left(\frac{2^2}{1.3}\right).\left(\frac{3^2}{2.4}\right).\left(\frac{4^2}{3.5}\right)...\left(\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\right)\)

\(=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)

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

\(=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}=2.\frac{n+1}{n+2}< 2\)(vì \(\frac{n+1}{n+2}< 1\))

Vậy B < 2

Ta có:

\(1+\frac{1}{1.3}=\frac{4}{1.3}=\frac{2^2}{1.3}\)

\(1+\frac{1}{2.4}=\frac{9}{2.4}=\frac{3^2}{2.4}\)

\(1+\frac{1}{3.5}=\frac{16}{3.5}=\frac{4^2}{3.5}\)

...

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

=>

\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2^2.3^2.4^2...\left(n+1\right)^2}{1.2.3^2.4^2...\left(n+1\right)\left(n+2\right)}=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}\)

\(=\frac{2\left(n+2\right)-2}{n+2}=2-\frac{2}{n+2}< 2\)

Vậy B < 2 

| 4x - 1 | + 4x =5

TH1: \(4x-1\ge0\)

\(4x-1+4x=5\)

\(4x+4x=5+1\)

\(8x=6\)

\(x=\frac{3}{4}\)( thỏa mãn \(4x-1\ge0\))

TH2: 4x - 1 <0

\(-\left(4x-1\right)+4x=5\)

\(-4x+1+4x=5\)

\(1=5\) ( vô lí)

Vậy \(x=\frac{3}{4}\)

1. A = 100

2. B = 2098

mik ko biết có đúng ko đâu nhé vì mình nhowfbanj làm cho rồi viết vô đây mà ahihi

\(\Leftrightarrow\left(1-3x\right)^3=\left(-5\right)^3\)

\(\Leftrightarrow1-3x=-5\)

\(\Leftrightarrow3x=1-\left(-5\right)\)

\(\Leftrightarrow3x=6\)

\(\Rightarrow x=2\)

a, ta có (-5)³=-125

=>1-3x=-5

3x=1+5

3x=6

x=2

b,3⁴-x=27

81-x=27

x=54

nha

a) \(2^{x-1}.3=48\)

\(2^{x-1}=16\)

\(2^{x-1}=2^4\)

\(x-1=4\)

\(x=5\)

b.

\(\frac{\left(-3\right)^x}{81}=-27\)

\(\frac{\left(-3\right)^x}{3^4}=\left(-3\right)^3\)

\(\frac{\left(-3\right)^x}{\left(-3\right)^4}=\left(-3\right)^3\)

\(\left(-3\right)^x=\left(-3\right)^4.\left(-3\right)^3\)

\(\left(-3\right)^x=\left(-3\right)^7\)

x=7

\(=\frac{6^{80}}{9^{80}}=\left(\frac{6}{9}\right)^{80}=\left(\frac{2}{3}\right)^{80}\)

\(2x+\frac{1}{5}=\frac{3}{2}x-1\)

\(\Leftrightarrow2x-\frac{3}{2}x=-1-\frac{1}{5}\)

\(\Leftrightarrow\frac{1}{2}x=-\frac{6}{5}\)

\(\Leftrightarrow x=\frac{-6}{5}\times2\)

\(\Leftrightarrow x=-\frac{12}{5}\)