a) \(\frac{17}{30}>\frac{51}{92}\)

b) \(\frac{-45}{47}>\frac{31}{-30}\)

c) \(\frac{22}{67}< \frac{51}{152}\)

d) \(-\frac{17}{39}< -\frac{17}{41};\frac{18}{-39}< -\frac{17}{41}\)

Xét: \(B=\frac{3}{4}-3\sqrt{x-2}\)\(\left(Đk:x\ge2\right)\)

Vì \(\sqrt{x-2}\ge0\Rightarrow-3\sqrt{x-2}\le0\Rightarrow B=\frac{3}{4}-3\sqrt{x-2}\le\frac{3}{4}\)

\(B=\frac{3}{4}\Leftrightarrow\sqrt{x-2}=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(tmđk\right)\)

Vậy: \(maxB=\frac{3}{4}\Leftrightarrow x=2\)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\Rightarrow\left(a^2+b^2\right)cd=\left(c^2+d^2\right)ab\)

=>\(a^2cd+b^2cd=c^2ab+d^2ab\)

=>\(a^2cd+b^2cd-c^2ab-d^2ab=0\)

=>\(ac\left(ad-bc\right)+bd\left(bc-ad\right)=0\)

=>\(ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)

=>\(\left(ac-bd\right)\left(ad-bc\right)=0\)

=>\(\orbr{\begin{cases}ac-bd=0\\ad-bc=0\end{cases}\Rightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}\Rightarrow}\orbr{\begin{cases}\frac{a}{b}=\frac{d}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}}\) (đpcm)

Theo đề ta có :  5^x . ( 5^2 + 1 ) = 5^x. 5^2 + 5^x       

 nên cm xong cha rùi

\(5^x.5^2+5^x.1=5^x.\left(5^2+1\right)\)

\(5^x.\left(5^2+1\right)=5^x.\left(5^2+1\right)\)

\(5^x.26=5^x.26\left(dpcm\right)\)

\(M=2^{2010}-2^{2009}-2^{2008}-...-2^1-2^0\)

\(-M=-\left(2^{2010}-2^{2009}-2^{2008}-...-2^1-2^0\right)\)

\(-M=2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\)

\(-2M=2.\left(2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\right)\)

\(-2M=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)

\(-M=2^{2011}+2^{2010}+...+2^2+2^1-\left(2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\right)\)

\(-M=2^{2011}-1=>M=-2^{2011}+1\)

tại sao lại có dấu ''-'' vậy bạn mình không hiểu lắm.