\(B=\left\{\left(-3\right)^n|1\le n\le4,n\inℕ\right\}\)

\(B=\left\{3^n|1\le n\le4,n\inℕ\right\}\)

\(B=\left\{3^n|1\le n\le4,n\in N\right\}\)

Vì b,d>0 => b+d>0 nên các phép nhân,chia 2 vế BĐT cho b,d hay (b+d) sẽ không đổi dấu BĐT.

\(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< bc\)

Xét \(\frac{a}{b}< \frac{a+c}{b+d}\Leftrightarrow a\left(b+d\right)< \left(a+c\right)b\Leftrightarrow ab+ad< ab+bc\Leftrightarrow ad< bc\)---> Đúng

Xét \(\frac{a+c}{b+d}< \frac{c}{d}\Leftrightarrow\left(a+c\right)d< \left(b+d\right)c\Leftrightarrow ad+cd< bc+cd\Leftrightarrow ad< bc\)---> Lại đúng

Vậy ta có đpcm :))

\(B=\left\{4x|0\le x\le4,x\in N\right\}\)

\(B=\left\{4n|n\le4,n\inℕ\right\}\)

a, \(\frac{12\times3\times4}{24\times25\times26}\)=\(\frac{12\times3\times2\times2}{12\times2\times25\times13\times2}\)=\(\frac{1\times3\times1\times1}{1\times1\times25\times13\times1}\)=\(\frac{3}{25\times13}\)=\(\frac{3}{325}\)

b, \(\frac{5}{8}\)\(\times\)\(\frac{3}{7}\)\(+\)\(\frac{5}{8}\)\(\times\)\(\frac{4}{7}\)= \(\frac{5}{8}\)\(\times\)\((\)\(\frac{3}{7}\)\(+\)\(\frac{4}{7}\)\()\)

                                                                       = \(\frac{5}{8}\)\(\times\)\(1\)=\(\frac{5}{8}\)

thanks

a) \(b>0,d>0\) nên \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< bc\Leftrightarrow\hept{\begin{cases}ab+ad< bc+ab\\cd+ad< bc+cd\end{cases}\Leftrightarrow\hept{\begin{cases}a\left(b+d\right)< b\left(a+c\right)\\d\left(a+c\right)< c\left(b+d\right)\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{a}{b}< \frac{a+c}{b+d}\\\frac{a+c}{b+d}< \frac{c}{d}\end{cases}}\)----> ĐPCM

b) \(\frac{1}{3}=\frac{4}{12},\frac{1}{4}=\frac{4}{16}\)Vậy 3 số hữu tỉ cần tìm là \(\frac{4}{13},\frac{4}{14},\frac{4}{15}\)

Ta có :

\(2^{2^{-1}}=\sqrt{2}\)

\(\Leftrightarrow2^{2^{-1^2}}=4\)

\(\Leftrightarrow2^{2^1}=4\)

\(\Leftrightarrow2^2=4\Leftrightarrow4=4\)

=> Điều phải chứng minh 

Nhầm cho sửa chỗ \(2^{2^{-1^2}}\) thành \(2^{2^{\left(-1\right)^2}}\)

đk \(x-2\ge0\Leftrightarrow x\ge2\)

\(\sqrt{x-2}< 5\)

\(\Leftrightarrow\left(\sqrt{x-2}\right)^2< 25\)

\(\Leftrightarrow x-2< 25\)

\(\Leftrightarrow x< 27\)

vậy \(2\le x< 27\)

\(\sqrt{x-2}< 5\)

ĐKXĐ : \(x\ge2\)

Bình phương hai vế

\(\Leftrightarrow\left(\sqrt{x-2}\right)^2< 5^2\)

\(\Leftrightarrow x-2< 25\)

\(\Leftrightarrow x< 27\)

Kết hợp ĐKXĐ => \(2\le x< 27\)

\(x\approx0,14779887;\frac{1}{2}\)