\(\frac{31}{2}\times\frac{32}{2}\times\frac{33}{2}\times...\times\frac{60}{2}=\frac{1\times2\times...\times60}{2^{30}\times\left(1\times2\times...\times30\right)}\)

\(=\frac{1\times2\times...\times60}{\left(2\times1\right)\times\left(2\times2\right)\times...\times\left(2\times30\right)}=\frac{\left(2\times4\times...\times60\right)\times\left(1\times3\times...\times59\right)}{2\times4\times...\times60}=1\times3\times...\times59\)

a) \(\frac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}=\frac{\left(-3\right)^{10}.3^5.5^5}{\left(5^2\right)^3.\left(-3^2\right)^7}=\frac{\left(-3\right)^{10}.3^5.5^5}{5^6.\left(-3\right)^{14}}=\frac{3^5}{5.\left(-3\right)^4}=\frac{-3}{5}\)

\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)

\(=\frac{100}{2}\)

\(=50\)

Bài 8a:

a) \(\left(5x+1\right)^2=\frac{36}{49}\)

Ta có: \(\frac{36}{49}=\left(\frac{6}{7}\right)^2=\left(\frac{-6}{7}\right)^2\)

\(\Rightarrow\left(5x+1\right)\in\left\{\frac{6}{7};\frac{-6}{7}\right\}\)

\(\Rightarrow TH1:5x+1=\frac{6}{7}\)

\(\Rightarrow5x=\frac{6}{7}-1\)

\(\Rightarrow5x=-\frac{1}{7}\)

\(\Rightarrow x=\frac{-1}{7}:5\)

\(\Rightarrow x=\frac{-1}{35}\)

\(TH2:5x+1=\frac{-6}{7}\)

\(\Rightarrow5x=\frac{-6}{7}-1\)

\(\Rightarrow5x=\frac{-13}{7}\)

\(\Rightarrow x=\frac{-13}{7}:5\)

\(\Rightarrow x=\frac{-13}{35}\)

Vậy, \(x\in\left\{\frac{-1}{35};\frac{13}{35}\right\}\)

Bài 8b:

b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\Rightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^{2.3}\)

\(\Rightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)

\(\Rightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)3\)

\(\Rightarrow x-\frac{2}{9}=\frac{4}{9}\)

\(\Rightarrow x=\frac{4}{9}+\frac{2}{9}\)

\(\Rightarrow x=\frac{6}{9}=\frac{2}{3}\)

Vậy, \(x=\frac{2}{3}\)

T.ick và chọn câu trả lời của mình nhé!

Học tốt!

\(\left|x\right|+\left|x-1\right|+\left|x-2\right|=2^2\)

Trường hợp 1: \(x>2\)

\(\Leftrightarrow x+x-1+x-2=4\)

\(\Leftrightarrow3x=7\)

\(\Leftrightarrow x=\frac{7}{3}\left(TM\right)\)

Trường hợp 2: \(2>x>1\)

\(\Leftrightarrow x+x-1-x+2=4\)

\(\Leftrightarrow x=3\)(Loại)

Trường hợp 3: \(x< 1\)

\(\Leftrightarrow-x-x+1-x+2=4\)

\(\Leftrightarrow-3x=1\)

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

|x|+|x−1|+|x−2|=2^2

x=-1/3, x=7/3

nha bạn chúc bạn học tốt nha 

a. ta có 

\(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)vậy ta có đpcm

b. ta có 

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..+\frac{1}{49^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{48.49}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{48}-\frac{1}{49}=1-\frac{1}{49}< 1\)

vậy ta có đpcm