\(\dfrac{5}{6x^2y}+\dfrac{7}{12xy^2}=\dfrac{10y+7x}{12x^2y^2}\)

Ta có \(\Delta CAB\sim\Delta CDB\left(g.g\right)\Rightarrow\dfrac{CB}{CD}=\dfrac{CA}{CB}\)

\(\Rightarrow CB^2=CD\cdot CA\)

Ta có \(CA=CD+DA=16\left(cm\right)\)

\(\Rightarrow CB^2=9\cdot16=144\Rightarrow CB=12\left(cm\right)\)

Ta có \(\dfrac{DB}{BA}=\dfrac{CB}{CA}=\dfrac{12}{16}=\dfrac{3}{4}\)

\(\Rightarrow DB=\dfrac{3}{4}AB=18\left(cm\right)\)

được 1 lần trong đời đẹp z thoi á e :)))

\(c,\) Ta có \(\widehat{ABM}=\widehat{ADM}\left(\Delta AMB=\Delta AMD\right)\)

\(\Rightarrow180^0-\widehat{ABM}=180^0-\widehat{ADM}\\ \Rightarrow\widehat{MBH}=\widehat{MDC}\)

Mà \(\left\{{}\begin{matrix}\widehat{BMH}=\widehat{CMD}\left(đđ\right)\\MD=MB\end{matrix}\right.\Rightarrow\Delta MBH=\Delta MDC\left(g.c.g\right)\)

Áp dụng tc dtsbn:

\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=\dfrac{-2x+y+5z}{8-7+15}=\dfrac{2x-3y-6z}{-8+21-18}\\ \Rightarrow A=\dfrac{8-7+15}{-8+21-18}=\dfrac{16}{-5}=-\dfrac{16}{5}\)

\(n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\ a,PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ b,n_{H_2}=1,5.n_{Al}=0,3(mol)\\ \Rightarrow V_{H_2}=0,3.22,4=6,72(l)\\ c,n_{H_2SO_4}=n_{H_2}=0,3(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,3.98}{19,6\%}=150(g)\\ n_{Al_2(SO_4)_3}=0,5.n_{Al}=0,1(mol)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{5,4+150-0,3.2}.100\%=22,09\%\)

\(\dfrac{1}{a+1}+\dfrac{1}{b+1}=\dfrac{1}{2}\left(a,b\ne-1\right)\\ \Rightarrow2\left(a+b+2\right)=\left(a+1\right)\left(b+1\right)\\ \Rightarrow2a+2b+4=ab+a+b+1\\ \Rightarrow a+b-ab+3=0\\ \Rightarrow\left(b-1\right)-a\left(b-1\right)=-4\\ \Rightarrow\left(a-1\right)\left(b-1\right)=4=1\cdot4=2\cdot2\)

Vậy \(\left(a;b\right)=\left(2;5\right);\left(5;2\right);\left(3;3\right)\)