4 8/3 và 3\(\sqrt2\)

4\(\frac83\) = \(\frac{20}{3}\) = \(\sqrt{\frac{400}{3}}\)

3\(\sqrt2\) = \(\sqrt{9.2}\) = \(\sqrt{18}\) = \(\sqrt{\frac{54}{3}}\)

Vì 400/3 > 54/3 nên \(\sqrt{\frac{400}{3}}\) > \(\sqrt{\frac{54}{3}}\)

Vậy: 4\(\frac83\) > 3\(\sqrt2\)

So sánh:

5\(\sqrt{\left(-10\right)^2}\) và 10\(\sqrt{\left(-5\right)^2}\)

5\(\sqrt{\left(-10\right)^2}\) = \(\sqrt{100.5^2}\) = \(\sqrt{2500}\)

10\(\sqrt{\left(-5\right)^2}\) = \(\sqrt{10^2.25}\) = \(\sqrt{2500}\)

Vậy 5\(\sqrt{\left(-10\right)^2}\) = 10\(\sqrt{\left(-5\right)^2}\)

các bạn giúp mk để mk ăn tết cho zui

luong thuy anh giúp mk vs

\(a)\)

Ta có : 

\(1-\frac{2}{3}=\frac{1}{3};1-\frac{4}{5}=\frac{1}{5};1-\frac{7}{8}=\frac{1}{8};1-\frac{3}{4}=\frac{1}{4}\)

\(1-\frac{9}{10}=\frac{1}{10};1-\frac{8}{9}=\frac{1}{9};1-\frac{5}{6}=\frac{1}{6};1-\frac{6}{7}=\frac{1}{7}\)

Do \(\frac{1}{3}>\frac{1}{4}>\frac{1}{5}>\frac{1}{6}>\frac{1}{7}>\frac{1}{8}>\frac{1}{9}>\frac{1}{10}\)

\(\Rightarrow1-\frac{1}{3}< 1-\frac{1}{4}< 1-\frac{1}{5}< 1-\frac{1}{6}< 1-\frac{1}{7}< 1-\frac{1}{8}< 1-\frac{1}{9}< 1-\frac{1}{10}\)

\(\Rightarrow\frac{2}{3}< \frac{3}{4}< \frac{4}{5}< \frac{5}{6}< \frac{6}{7}< \frac{7}{8}< \frac{8}{9}< \frac{9}{10}\)

Nếu \(\frac{a}{b}\)là 1 số thuộc dãy trên thì số tiếp theo là : 

\(\frac{a+1}{b+1}\)

\(b)\) 

Ta có : 

\(a\left(a+2\right)=a^2+2a\)

\(b\left(a+1\right)=ab+b\)

Sorry , đến bước này mik chịu 

~ Ủng hộ nhé 

Phần b) Ý bạn là so sánh \(\frac{a}{b}\)và \(\frac{a+1}{b+2}\)

ta có A= \(\frac{8^{18}+1}{8^{19} +1}\)=> 8A=\(\frac{8^{19}+8}{8^{19}+1}\)= \(\frac{\left(8^{19}+1\right)+7}{8^{19}+1}\)=\(\frac{8^{19}+1}{8^{19} +1}\)+\(\frac{7}{8^{19}+1}\) =1+\(\frac{7}{8^{19}+1}\) =\(\frac{7}{8^{19}+1}\) 

         B= \(\frac{8^{23}+1}{8^{24}+1}\)=> 8B=\(\frac{8^{24}+8}{8^{24}+1}\)= \(\frac{\left(8^{24}+1\right)+7}{8^{24}+1}\)=\(\frac{8^{24}+1}{8^{24}+1}\)+\(\frac{7}{8^{24}+1}\) =1+\(\frac{7}{8^{24} +1}\)=\(\frac{7}{8^{24}+1}\)

       vì  \(8^{19}\)<\(8^{24}\)=> \(8^{19}\)+1 >\(8^{24}\)+1 => \(\frac{7}{8^{19}+1}\)<\(\frac{7}{8^{24}+1}\)=> A<B

a) ta có \(8A=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\\ 8B=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)

Vì \(8^{24}+1>8^{19}+1\\\frac{7}{8^{24}+1}< \frac{7}{8^{19}+1} \)

vậy 8A>8B nên A>B