\(A=\sqrt{1-4a+4a^2}-2a\) \(\left(a\ge0,5\right)\)

\(A=\sqrt{\left(1-2a\right)^2}-2a\)

\(A=\left|1-2a\right|-2a\)

\(A=2a-1-2a\)

\(A=-1\)

\(B=\sqrt{x-2+2\sqrt{x-3}}\) (ĐK: \(x\ge3\))

\(B=\sqrt{\left(\sqrt{x-3}\right)^2+2\sqrt{x-3}\cdot1+1^2}\)

\(B=\sqrt{\left(\sqrt{x-3}+1\right)^2}\)

\(B=\left|\sqrt{x-3}+1\right|\)

\(B=\sqrt{x-3}+1\)

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

\(C=\sqrt{\left(\sqrt{x}\right)^2-2\sqrt{x}\cdot1+1^2}+\sqrt{\left(\sqrt{x}\right)^2+2\sqrt{x}\cdot1+1^2}\) \(\left(x\ge0\right)\)

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

\(C=\left|\sqrt{x}-1\right|+\left|\sqrt{x}+1\right|\) 

\(C=\sqrt{x}-1+\sqrt{x}+1\)

\(C=2\sqrt{x}\) 

\(D=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\) \(\left(x\ge1\right)\)

\(D=\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}\cdot1^2+1^2}+\sqrt{\left(\sqrt{x-1}\right)^2-2\sqrt{x-1}\cdot1^2+1}\)

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

\(D=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

\(D=\sqrt{x-1}+1+\sqrt{x-1}-1\)

\(D=2\sqrt{x-1}\)

a) Đặt R là bán kính đường tròn tâm O

r là bán kính đường tròn tâm O'

Ta có:

OC = OA = R

∆OAC cân tại O

⇒ ∠OAC = ∠OCA

Mà ∠OAC = ∠O'AD (đối đỉnh)

⇒ ∠OCA = ∠O'AD (1)

Lại có:

O'A = OD = r

⇒ ∆O'AD cân tại O'

⇒ ∠O'AD = ∠O'DA (2)

Từ (1) và (2) suy ra ∠OCA = ∠O'DA

b) Sửa đề: chứng minh OC // O'D

Do ∠OCA = ∠O'DA (cmt)

Mà ∠OCA và ∠O'DA là hai góc so le trong

⇒ OC // O'D

c) Do CE là đường kính của đường tròn tâm O

A nằm trên đường tròn tâm O

⇒ ∆ACE vuông tại A

Hay AC ⊥ AE

Bài 3:

1: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)

\(M=\left(\dfrac{x+\sqrt{x}+10}{x-9}-\dfrac{1}{\sqrt{x}-3}\right):\dfrac{1}{\sqrt{x}-3}\)

\(=\left(\dfrac{x+\sqrt{x}+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{\sqrt{x}-3}{1}\)

\(=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{1}=\dfrac{x+7}{\sqrt{x}+3}\)

2: Thay x=25 vào M, ta được:

\(M=\dfrac{25+7}{\sqrt{25}+3}=\dfrac{32}{5+3}=\dfrac{32}{8}=4\)

3: \(M< \sqrt{x}+1\)

=>\(\dfrac{x+7}{\sqrt{x}+3}< \sqrt{x}+1\)

=>\(\dfrac{x+7-\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}< 0\)

=>\(x+7-x-4\sqrt{x}-3< 0\)

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

=>\(-4\sqrt{x}< -4\)

=>\(\sqrt{x}>1\)

=>x>1

kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>1\\x\ne9\end{matrix}\right.\)

bài 5:
1: Thay x=49 vào B, ta được:

\(B=\dfrac{1}{\sqrt{49}-1}=\dfrac{1}{7-1}=\dfrac{1}{6}\)

2:

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)

S=A-B

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

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

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

\(=\dfrac{-\sqrt{x}+1+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

3: \(S-\dfrac{1}{3}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}\)

\(=\dfrac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)

\(=-\dfrac{\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)

=>\(S< \dfrac{1}{3}\)

Gọi K là trung điểm của BD

Xét ΔDBH có

K,I lần lượt là trung điểm của DB,DH

=>KI là đường trung bình của ΔDBH

=>KI//BH

Ta có: KI//BH

AH\(\perp\)BH

Do đó: KI\(\perp\)AH

Xét ΔAKH có

KI,HD là các đường cao

KI cắt HD tại I

Do đó: I là trực tâm

=>AI\(\perp\)HK

Ta có: ΔABC cân tại A

mà AH là đường cao

nên H là trung điểm của BC

Xét ΔBDC có

K,H lần lượt là trung điểm của BD,BC

=>KH là đường trung bình

=>KH//DC

Ta có: KH//DC
AI\(\perp\)KH

Do đó: AI\(\perp\)DC

b: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}\)

\(=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{-\sqrt{6}\left(\sqrt{8}-\sqrt{3}\right)}=-\dfrac{2}{\sqrt{6}}\)

\(=-\dfrac{2\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{3}\)

d: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)

\(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}{\left(2+\sqrt{3}\right)^2}}\)

\(=\sqrt{\dfrac{4-3}{\left(2+\sqrt{3}\right)^2}}=\sqrt{\dfrac{1}{\left(2+\sqrt{3}\right)^2}}\)

\(=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}\)

f: \(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{\sqrt{5}-1}\)

\(=\dfrac{\sqrt{5}-1-\left(3-\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)}\)

\(=\dfrac{\sqrt{5}-1-3+\sqrt{5}}{3\sqrt{5}-3-5+\sqrt{5}}\)

\(=\dfrac{-4+2\sqrt{5}}{-8+4\sqrt{5}}=\dfrac{-4+2\sqrt{5}}{2\left(-4+2\sqrt{5}\right)}=\dfrac{1}{2}\)

h: \(\dfrac{1}{5+3\sqrt{2}}+\dfrac{1}{5-3\sqrt{2}}\)

\(=\dfrac{5-3\sqrt{2}+5+3\sqrt{2}}{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}\)

\(=\dfrac{10}{25-18}=\dfrac{10}{7}\)

a: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}\)

\(=\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}\)

\(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}=\sqrt{20}=2\sqrt{5}\)

c: \(\dfrac{\sqrt{2}}{\sqrt{5}-\sqrt{3}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)

\(=\dfrac{\sqrt{10}+\sqrt{6}}{5-3}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)

e: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{6-1}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right)\cdot\left(\sqrt{6}+11\right)\)

\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

=6-121

=-115

g: \(\dfrac{1}{\sqrt{5}+1}+\dfrac{1}{\sqrt{5}-2}-\dfrac{1}{3-\sqrt{5}}-\sqrt{5}\)

\(=\dfrac{\sqrt{5}-1}{4}+\dfrac{\sqrt{5}+2}{1}-\dfrac{1\left(3+\sqrt{5}\right)}{4}-\sqrt{5}\)

\(=\dfrac{1}{4}\sqrt{5}-\dfrac{1}{4}+\sqrt{5}+2-\dfrac{3}{4}-\dfrac{\sqrt{5}}{4}-\sqrt{5}\)

\(=-1+2=1\)

\(V=\left(\dfrac{x-\sqrt{x}}{x-2\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1}{\sqrt{x}-1}\left(x\ge0;x\ne1\right)\)

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}-\dfrac{1}{\sqrt{x}+1}\right]\cdot\dfrac{\sqrt{x}-1}{x+1}\)

\(=\left[\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right]\cdot\dfrac{\sqrt{x}-1}{x+1}\)

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}-1}{x+1}\)

\(=\dfrac{x+\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{x+1}\)

\(=\dfrac{x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\)

\(=\dfrac{1}{\sqrt{x}+1}\)

#\(Toru\)

Đề nghị em viết đề có dấu tiếng Việt!