Minh goi y thoi nhe muon roi mik chuan bi di ngu bn thong cam

a) ban dung dinh nghia tiep tuyen la xong

b) cm O la truc tam tam giac BAN roi dung yeu to // la ok

c) mik nghi la : de thay Samn=1/2 Sabc (t/c trung tuyen.....)

thi Samn Min <=> Sabc min

Cau c) mik ko chac lam co cau a,b ban cu lam theo mik kieu gi cung ra

Co gi de mai mik ngu day mik lam cho

a)  Xet \(\Delta ADE\) co AO=DO=EO=R => DE la duong kinh (O)

Ta co MD cat (O) duy nhat tai D=> MD la tiep tuyen (O)=> \(MD\perp DE\)

Tuong tu NE la tiep tuyen (O) =>\(NE\perp DE\) 

Suy ra MD//NE ( Quan he tu vuong goc den song song)

b) Noi NO , Goi F la trung diem OH

Xet \(\Delta AHC\) co OH=OA ( gt) , HN=NC (gt)

=> ON la duong trung binh => ON//AC

ma AB \(AB\perp AC\left(\Delta ABCvuong\right)\)

Suy ra \(NO\perp AB\)

Xet tam giac ABN co \(AH\perp BN\left(gt\right),NO\perp AB\left(cmt\right)\) => O la truc tam tam giac ABN

=> \(BO\perp AN\)     (1)

Xet tam giac giac BHO co M la trung diem BH (gt) , F la trung diem OH ( gt)

=> MF la duong trung binh => MF//BO  (2)

Tu (1) va (20 suy ra \(MF\perp AN\) (quan he tu vuong goc den song song)

Xet tam giac AMN co \(\hept{\begin{cases}AH\perp MN\left(gt\right)\\MF\perp AN\left(cmt\right)\end{cases}\Rightarrow}\) Trung diem F cua OH la truc tam \(\Delta AMN\)

c) Xet tam giac ABH co AM la duong trung tuyen =>Samh =Sabm ( t/c trug tuyen chia cat doi dien thang 2 phan co dien h bag nhau)

=> Samh=1/2Sabh

tuong tu ta cung co Sahn = 1/2 Sahc

Suy ra Samh+Sahn =1/2 (Sabh +Sahc) 

           <=>  Samn=1/2 Sabc

=> Samn min <=> Sabc min

Theo minh thi tam giac ABC can co dk la dien h tam giac ABC nho nhat thi Samn dat gtnn

Mik ko chac cau c) lam dau. Neu sai mong cac bn thong cam ma sua ho mik,Mik cam on.

Chuc ban hoc tot

ĐKXĐ: \(x\ge0\)

\(x-\sqrt{x}-1\)

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{5}{4}\ge\frac{-5}{4}\forall x\ge0\)

Dấu"=" xảy ra<=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{4}\)

Vậy.......

Ban tu ve hinh nha

( tam giac ABC vuong tai A , duong cao AH)

Xet tam giac HAB va tam giac HCA Co

\(\hept{\begin{cases}\widehat{AHB}=\widehat{CHA}=90\\\widehat{HBA}=\widehat{HAC}\left(phu\widehat{HAB}\right)\end{cases}=>\Delta HAB}\) dong dang voi \(\Delta HCA\left(G-G\right)\)

Suy ra\(\frac{AB}{AC}=\frac{HB}{HA}=\frac{HA}{HC}=\frac{4}{9}\left(gt\right)\) =>\(HB=\frac{4HA}{9},HC=\frac{9HA}{4}\) 

=>\(\frac{HB}{HA}=\frac{\frac{4HA}{9}}{\frac{9HA}{4}}=\frac{4HA}{9}.\frac{4}{9HA}=\frac{16}{81}\)

Suy ra ti so hinh chieu cua hai canh goc vuong do tren canh huyen =16/81

Chuc ban hoc tot

giải thử thôi nha

a) \(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{\sqrt{x}\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)\left(\sqrt{x}-1\right)}-\frac{\left(6\sqrt{x}-4\right)\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x^2-2x\sqrt{x}+2\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x^2-2x\sqrt{x}+2\sqrt{x}-1}{\left(x-1\right)^2}\)

a, ĐKXĐ: \(x\ne1;x\ge0\)\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{3\left(\sqrt{x}-1\right)}{x-1}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}\)

\(=\frac{x-2\sqrt{x}+1}{x-1}\)

\(=\frac{\left(\sqrt{x}-1\right)^2}{x-1}\)\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v