\(5x=6y\Rightarrow\frac{x}{6}=\frac{y}{5}=\frac{x+y}{6+5}=\frac{33}{11}=3\)

\(\Rightarrow x=18;y=15\)

Xét tg vuông ABH

\(\widehat{HAB}+\widehat{ABC}=\widehat{HAB}+\widehat{KBA}+\widehat{KBC}=90\)

Xét tg vuông BCK

\(\widehat{KBC}+\widehat{C}=90\Rightarrow\widehat{KBC}=90-\widehat{C}=90-65=25\)

\(\Rightarrow\widehat{HAB}+\widehat{KBA}=90-\widehat{KBC}=90-25=65\)

Cách 2:

Xét tg vuông BCK

\(\widehat{KBC}+\widehat{C}=90\) (1)

Xét tg vuông BIH

\(\widehat{KBC}+\widehat{BIH}=90\) (2)

Mà \(\widehat{BIH}=\widehat{AIK}\) (góc đối đỉnh) (3)

Từ (1) (2) (3) \(\Rightarrow\widehat{AIK}=\widehat{C}=65\)

Xét tg ABI

\(\widehat{AIK}=\widehat{HAB}+\widehat{KBA}=65\) (góc ngoài của 1 tam giác bằng tổng 2 góc trong không kề với nó)

\(=\sqrt{2^2.3}+\sqrt{3^3}-\sqrt{3}=2\sqrt{3}+3\sqrt{3}-\sqrt{3}=4\sqrt{3}\)

\(=\left(3^2\right)^4+\left(3^3\right)^3+3^6+\left(3.37\right)^2=\)

\(=3^8+3^9+3^6+3^2.37^2=3^6\left(3^2+3^3+1\right)+3^2.37^2\)

\(3^6.37+3^2.37^2=37\left(3^6+3^2.37\right)\) chia hết cho 37

thay \(ab=c^2\) vào\(\frac{a^2+c^2}{b^2+c^2}\)

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

Từ\(ab=c^2\Rightarrow ab=cc\Rightarrow\frac{a}{c}=\frac{c}{b}\)

Đặt \(\frac{a}{c}=\frac{c}{b}=k\Rightarrow\hept{\begin{cases}a=ck\\c=bk\end{cases}}\)

Khi đó : \(\frac{a}{b}=\frac{ck}{b}=\frac{b.k^2}{b}=k^2\)(1) ; 

\(\frac{a^2+c^2}{b^2+c^2}=\frac{c^2.k^2+c^2}{b^2.k^2+b^2}=\frac{c^2\left(k^2+1\right)}{b^2\left(k^2+1\right)}=\frac{c^2}{b^2}=\frac{b^2.k^2}{b^2}=k^2\)(2)

Từ (1) và (2) => đpcm

a) \(\left(-\frac{5}{2}\right)^2:\left(-15\right)-\left(-0,45+\frac{3}{4}\right).\left(-1\frac{5}{9}\right)\)

= \(-\frac{25}{4}:\left(-15\right)-\left(\frac{9}{20}+\frac{15}{20}\right).\left(-\frac{14}{9}\right)\)

=\(-\frac{25}{4}.\frac{1}{-15}-\frac{6}{5}.\left(-\frac{14}{9}\right)\)

= \(\frac{-5}{12}-\frac{8}{5}\)

= \(\frac{\left(-25\right)-96}{60}\)

= \(\frac{\left(-25\right)+\left(-96\right)}{60}\)

=\(\frac{121}{60}\)

b) \(\left(\frac{-1}{3}\right)-\left(\frac{-3}{5}\right)^0+\left(1-\frac{1}{2}\right)^2:2\)

= \(\left(\frac{-1}{3}\right)-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}\)

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

= \(\frac{-4}{3}+\frac{1}{8}\)=\(\frac{-32+3}{24}\)

=\(\frac{-29}{24}\)

c) E=\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

     =\(\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^{10}.3^8+6^8.20}\)

     =\(\frac{2^{10}.3^8-2.6^9}{2^{10}.3^8+6^8.20}\)

     =\(\frac{3}{5}\)

d)\(\frac{5^4.20^4}{25^5.4^5}\)

=\(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)

=\(\frac{100^4}{100^5}\)

=\(\frac{1}{100}\)

a) 12. \(\frac{4}{9}\)+\(\frac{4}{3}\)=\(\frac{16}{3}\)+\(\frac{4}{3}\)=\(\frac{20}{3}\)

b) (\(\frac{-5}{7}\)) . (12,5+1,5)= (\(\frac{-5}{7}\)).14=-10

a) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}=12.\frac{4}{9}+\frac{4}{3}=\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)

b) \(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)=-\frac{5}{7}.\left(12,5+1,5\right)=-\frac{5}{7}.14=-10\)

c) \(1:\left(\frac{2}{3}-\frac{3}{4}\right)^2=1:\left(-\frac{1}{12}\right)^2=1:\frac{1}{144}=1.144=144\)

d) \(15.\left(-\frac{2}{3}\right)^2-\frac{7}{3}=15.\frac{4}{9}-\frac{7}{3}=\frac{20}{3}-\frac{7}{3}=\frac{13}{3}\)

e) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(-1\right)^{2007}=\frac{1}{2}.8-\frac{2}{5}+\left(-1\right)=4-\frac{2}{5}-1=\frac{13}{5}\)