\(S_{ABCD}=\dfrac{1}{2}AC.\left(AB+AD\right)\)

Nếu ta đặt đọ dài cạnh AB=a

tam giác ABC vuông tại A có góc B=60⁰ nên nếu gọi AB=a thì \(AC=\sqrt{3}a\)

ta lại xét tam giác ADC vuông tại D có góc ACD dễ tính bằng 30 độ

bạn sẽ lại có:

\(AD=\dfrac{1}{2}AC=\dfrac{\sqrt{3}a}{2}\) đây là T/C của tam giác vuông có 1 góc = 30 độ

như vậy:

\(S_{ABCD}=\dfrac{1}{2}\sqrt{3}a.\left(a+\dfrac{\sqrt{3}a}{2}\right)=\dfrac{1}{2}a^2.\left(\sqrt{3}+\dfrac{1}{2}\right)\)

trong đó a=AB

\(\sqrt{x-6\sqrt{x}+9}-\sqrt{x+6\sqrt{x}+9}\left(x>=9\right)\\ =\sqrt{\left(\sqrt{x}-3\right)^2}-\sqrt{\left(\sqrt{x}+3\right)^2}\\ =\left|\sqrt{x}-3\right|-\left|\sqrt{x}+3\right|\\ =\sqrt{x}-3-\left(\sqrt{x}+3\right)=-6\)

\(cos\widehat{K}=\dfrac{HK}{KI}=\dfrac{19}{25}\Rightarrow\widehat{K}=arccos\dfrac{19}{25}\)

\(DI=\sqrt{KI^2-HK^2}=\sqrt{25^2-19^2}=2\sqrt{66}\left(cm\right)\)

\(\dfrac{1}{DH^2}=\dfrac{1}{HI^2}+\dfrac{1}{HK^2}=\dfrac{1}{264}+\dfrac{1}{361}=\dfrac{625}{95304}\)

\(\Rightarrow DH=\sqrt{\dfrac{95304}{625}}\left(cm\right)\)

Xét tam giác \(DHI\) vuông tại \(D\) đường cao \(DF\): 

\(DH^2=HF.HI\) (hệ thức trong tam giác vuông) 

Xét tam giác \(DHK\) vuông tại \(D\) đường cao \(DE\):

\(DH^2=HE.HF\) (hệ thức trong tam giác vuông) 

suy ra \(HE.HK=HF.HI\).

\(a^2+b^2=2\)

\(\Leftrightarrow\left(a+b\right)^2-2ab=2\)

\(\Leftrightarrow2ab=\left(a+b\right)^2-2\)

Theo đề: \(P=3\left(a+b\right)+ab\) 

\(\Leftrightarrow2P=6\left(a+b\right)+2ab\)

\(=6\left(a+b\right)+\left(a+b\right)^2-2\)

\(=\left(a+b\right)^2+2.3\left(a+b\right)+9-9-2\)

\(=\left[\left(a+b\right)+3\right]^2-11\)

\(P=\dfrac{1}{1}\left(a+b+3\right)^2-\dfrac{11}{2}\)

Ta có: \(\left(a+b+3\right)^2\ge0\forall a,b\in R\)

\(\Leftrightarrow\dfrac{1}{2}\left(a+b+3\right)^2-\dfrac{11}{2}\ge\dfrac{-11}{2}\forall a,b\in R\)

=> Giá trị nhỏ nhất \(P=-\dfrac{11}{2}\)

\(ĐK:x>0\)

\(E=\dfrac{1-\sqrt{x}+x}{\sqrt{x}}=\sqrt{x}-1+\dfrac{1}{\sqrt{x}}>=2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}-1=2\sqrt{1}-1=1\)

\(E>=1=>\sqrt{E}>=1;\sqrt{E}-1>=0\)

\(E-\sqrt{E}=\sqrt{E}\left(\sqrt{E}-1\right)>=0\\ =>E>=\sqrt{E}\)

\(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\left(ĐK:a>0;a\ne\left\{1;4\right\}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\dfrac{a-2}{a+2}\\ =\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right).\dfrac{a-2}{a+2}\\ =\dfrac{2\sqrt{a}}{\sqrt{a}}.\dfrac{a-2}{a+2}=\dfrac{2\left(a-2\right)}{a+2}\)

\(\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}-\dfrac{10}{3}>0\)

\(ĐK:x\ge0\)

\(\Leftrightarrow\dfrac{9\sqrt{x}+24-10\sqrt{x}-20}{3\left(\sqrt{x}+2\right)}>0\)

\(\Leftrightarrow\dfrac{-\sqrt{x}+4}{3\left(\sqrt{x}+2\right)}>0\)

\(\Leftrightarrow-\sqrt{x}+4>0\) ( vì \(3\left(\sqrt{x}+2\right)>0\) )

\(\Leftrightarrow\sqrt{x}-4< 0\)

\(\Leftrightarrow\sqrt{x}< 4\)

\(\Leftrightarrow x< 16\)

Vậy \(S=\left\{x|0\le x< 16\right\}\)

\(ĐK:x>=0\)

\(\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}-\dfrac{10}{3}>0\\ < =>\dfrac{3\left(3\sqrt{x}+8\right)-10\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}+2\right)}>0\\ < =>\dfrac{9\sqrt{x}+24-10\sqrt{x}-20}{3\left(\sqrt{x}+2\right)}>0\\ < =>\dfrac{-\sqrt{x}+4}{3\left(\sqrt{x}+2\right)}>0\)

Vì : \(3\left(\sqrt{x}+2\right)>0\forall x>=0\)

\(=>-\sqrt{x}+4>0\\ < =>-\sqrt{x}>-4\\< => \sqrt{x}< 4\\< =>0\le x< 16\)

`@`Xét tam giác ABC vuông A, đường cao AH:

\(BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10\left(cm\right)\)

\(BH=\dfrac{AB^2}{BC}=\dfrac{6^2}{10}=3,6\left(cm\right)\)

\(CH+BC-BH=10-3,6=6,4\left(cm\right)\)

\(AH=\sqrt{BH.CH}=\sqrt{3,6.6,4}=4,8\left(cm\right)\)

`@`Xét tam giác AHB vuông H, đường cao HM:

\(\dfrac{1}{HM^2}=\dfrac{1}{BH^2}+\dfrac{1}{AH^2}\)

\(\Leftrightarrow\dfrac{1}{HM^2}=\dfrac{1}{3,6^2}+\dfrac{1}{4,8^2}\)

\(\Leftrightarrow HM=2,88\left(cm\right)\)

\(BM=\sqrt{BH^2-HM^2}=\sqrt{3,6^2-2,88^2}=2,16\left(cm\right)\)

`@`Xét tam giác AHC vuông H, đường cao HN:

\(\dfrac{1}{HN^2}=\dfrac{1}{HC^2}+\dfrac{1}{AH^2}\)

\(\Leftrightarrow\dfrac{1}{HN^2}=\dfrac{1}{6,4^2}+\dfrac{1}{4,8^2}\)

\(\Leftrightarrow HN=3,84\left(cm\right)\)

\(NC=\sqrt{HC^2-HN^2}=\sqrt{6,4^2-3,84^2}=5,12\left(cm\right)\)

`@`Ta có: \(\left\{{}\begin{matrix}\widehat{B}+\widehat{MHB}=90^o\\\widehat{C}+\widehat{NHC}=90^o\\\widehat{B}+\widehat{C}=90^o\end{matrix}\right.\) \(\rightarrow\widehat{MHN}=90^o\)

Ta có: 

\(S_{BHM}=\dfrac{1}{2}.MB.MH=\dfrac{1}{2}.2,16.2,88=3,1104\left(cm^2\right)\)

\(S_{NHC}=\dfrac{1}{2}.NH.NC=\dfrac{1}{2}.3,84.5,12=9,8304\left(cm^2\right)\)

\(S_{MHN}=\dfrac{1}{2}.MH.NH=\dfrac{1}{2}.2,88.3,84=5,5296\left(cm^2\right)\)

\(S_{BCNM}=S_{BHM}+S_{NHC}+S_{MHN}\)

             \(=3,1104+9,8304+5,5296\)

            \(=18,4704\left(cm^2\right)\)

\(\left\{{}\begin{matrix}a+b\le10\\b\le7\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}b\le7\\a\le3\end{matrix}\right.\) ⇔  P ≤ 3² + 7² = 58

P(max)= 58  ⇔ a =3; b = 7

a ≥ 0; b ≥ 0 ⇔P = a² + b²≥ 0 + 0 = 0 ⇔ P(min) = 0 ⇔ a=b=0