Bài 4: 

2: Xét ΔBAK vuông tại A có AD là đường cao ứng với cạnh huyền BK

nên \(BD\cdot BK=BA^2\left(1\right)\)

Xét ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC

nên \(BH\cdot BC=BA^2\left(2\right)\)

Từ (1) và (2) suy ra \(BD\cdot BK=BH\cdot BC\)

Bài nào nhỉ

Bài 5: 

\(\cos\alpha=\sqrt{1-\dfrac{1}{25}}=\dfrac{2\sqrt{6}}{5}\)

\(\tan\alpha=\dfrac{1}{5}:\dfrac{2\sqrt{6}}{5}=\dfrac{\sqrt{6}}{12}\)

\(\cot\alpha=2\sqrt{6}\)

Bài 5: 

\(\cos\alpha=\sqrt{1-\dfrac{1}{25}}=\dfrac{2\sqrt{6}}{5}\)

\(\tan\alpha=\dfrac{1}{5}:\dfrac{2\sqrt{6}}{5}=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{24}\)

\(\cot\alpha=1:\dfrac{\sqrt{6}}{24}=4\sqrt{6}\)

Câu c bài 3 ạ

\(B=\dfrac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ B=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\\ c,P=B:A=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\\ P=\dfrac{\sqrt{x}+2-5}{\sqrt{x}+2}=1-\dfrac{5}{\sqrt{x}+2}\)

Ta có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\dfrac{5}{\sqrt{x}+2}\le\dfrac{5}{2}\)

\(\Leftrightarrow P=1-\dfrac{5}{\sqrt{x}+2}\ge1-\dfrac{5}{2}=-\dfrac{3}{2}\\ P_{min}=-\dfrac{3}{2}\Leftrightarrow x=0\)

\(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}+2}+\dfrac{9\sqrt{x}-10}{4-x}\)

\(=\dfrac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)

\(P=\dfrac{B}{A}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}:\dfrac{\sqrt{x}-2}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}.\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=1-\dfrac{5}{\sqrt{x}+2}\)

Do \(\sqrt{x}+2\ge2\Rightarrow\dfrac{5}{\sqrt{x}+2}\le\dfrac{5}{2}\Rightarrow-\dfrac{5}{\sqrt{x}+2}\ge-\dfrac{5}{2}\)

\(\Rightarrow P=1-\dfrac{5}{\sqrt{x}+2}\ge1-\dfrac{5}{2}=-\dfrac{3}{2}\)

\(minP=-\dfrac{3}{2}\Leftrightarrow x=0\)

Câu 4 số mol h2so4 dùng để trung hòa 500g là n=1.0,02.10=0,2mol

2M+2H2O->2MOH+H2

a                     a

M2O+H2O->2MOH

b                     2b

2MOH+H2SO4-> M2SO4+2H2O

n(MOH)=2n(H2SO4)=0,2.2=0,4mol

Ta có hệ a+2b=0,4

aM+b(2M+16)=10,8

<=> (a+2b)M +16b=10,8

0,4M+16b=10,8

<=>M+40b=27

Ta có M<27 và b<0,2

=> M chỉ có thể là Na(M=23)

b=0,1 ; a=0,2

Lần sau em đăng tách ra 1-2 bài cho 1 câu hỏi nha

\(5,\\ K=\sqrt{5x-9+6\sqrt{5x-9}+9}+\sqrt{5x-9-6\sqrt{5x-9}+9}\\ K=\sqrt{\left(\sqrt{5x-9}+3\right)^2}+\sqrt{\left(\sqrt{5x-9}-3\right)^2}\\ K=\sqrt{5x-9}+3+\sqrt{5x-9}-3=2\sqrt{5x-9}\ge0,\forall x\\ K_{min}=0\Leftrightarrow\sqrt{5x-9}=0\Leftrightarrow x=\dfrac{9}{5}\)

\(3,\\ 1,A=\dfrac{1,44+7}{\sqrt{1,44}}=\dfrac{7,44}{1,2}=\dfrac{31}{5}\\ 2,B=\dfrac{x-3\sqrt{x}+\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(3,S=\dfrac{1}{B}+A=\dfrac{\sqrt{x}+3}{\sqrt{x}}+\dfrac{x+7}{\sqrt{x}}=\dfrac{x+\sqrt{x}+10}{\sqrt{x}}\\ S=\sqrt{x}+1+\dfrac{10}{\sqrt{x}}\ge2\sqrt{\sqrt{x}\cdot\dfrac{10}{\sqrt{x}}}+1=2\sqrt{10}+1\left(BĐT.cosi\right)\)

Dấu \("="\Leftrightarrow x=10\)

\(1,HC=BC-HB=6\left(cm\right)\) 

Áp dụng HTL:

\(\left\{{}\begin{matrix}AB^2=BH\cdot BC=16\\AC^2=CH\cdot BC=48\\AH^2=BH\cdot HC=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}AB=4\left(cm\right)\\AC=4\sqrt{3}\left(cm\right)\\AH=2\sqrt{3}\left(cm\right)\end{matrix}\right.\)

\(2,\widehat{ADB}=\widehat{AHB}\left(=90^0\right)\) nên \(ADHB\) nội tiếp

\(\Rightarrow\widehat{HDB}=\widehat{HAB}\left(cùng.chắn.HB\right)\)

Mà \(\widehat{HAB}=\widehat{ACB}\left(cùng.phụ.\widehat{HAC}\right)\)

\(\Rightarrow\widehat{HDB}=\widehat{ACB}\)

\(\left\{{}\begin{matrix}\widehat{HDB}=\widehat{ACB}\left(cm.trên\right)\\\widehat{KBC}.chung\end{matrix}\right.\Rightarrow\Delta BHD\sim\Delta BKC\left(g.g\right)\\ \Rightarrow\dfrac{BD}{BC}=\dfrac{BH}{BK}\Rightarrow BD\cdot BK=BH\cdot BC\)

\(c,\) Áp dụng công thức tính diện tích hình tam giác bằng