a) khỏi bàn

b) Ta có: \(\widehat{DOK}=\widehat{DEK}\left(=\frac{1}{2}sđ\widebat{DK}\right)\left(1\right)\)

\(\widehat{DEK}=\widehat{DBC}=\left(\frac{1}{2}sđ\widebat{DC}\right)\left(2\right)\)

Mà OD=OB \(\Rightarrow\Delta ODB\)cân tại O

\(\Rightarrow\widehat{DBC}=\widehat{BDO}\left(3\right)\)

Từ (1), (2) và (3) \(\Rightarrow\widehat{DOK}=\widehat{BDO}\)Mà 2 góc này ở vị trí so le trong

\(\Rightarrow OK//DB\)

Xét tam giác CBH có: OK//CH  ; O là trung điểm của BC

=> K là trung điểm của CH

c từ từ nha chiều làm sau

Đặt \(A=\sqrt{7+4\sqrt{3}}=\sqrt{7+2.2\sqrt{3}}\)

\(=\sqrt{4+2.2\sqrt{3}+3}=\sqrt{\left(2+\sqrt{3}\right)^2}=\left|2+\sqrt{3}\right|=2+\sqrt{3}\)

Đặt \(B=\sqrt{4+\sqrt{15}}\)

\(\sqrt{2}B=\sqrt{8+2\sqrt{15}}=\sqrt{8+2\sqrt{5}\sqrt{3}}\)

\(=\sqrt{5+2\sqrt{5}\sqrt{3}+3}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\left|\sqrt{5}+\sqrt{3}\right|\)

\(=\sqrt{5}+\sqrt{3}\Rightarrow B=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{10}+\sqrt{6}}{2}\)

a) Δ AHC ~ Δ CAB (g.g) 

vì: \(\hept{\begin{cases}\widehat{C}:chung\\\widehat{AHC}=\widehat{BAC}=90^0\end{cases}}\)

=> \(\frac{HC}{AC}=\frac{AC}{BC}\Leftrightarrow HC\cdot BC=AC^2\Rightarrow b^2=ab'\)

b) Δ AHB ~ Δ CHA (g.g)

vì: \(\hept{\begin{cases}\widehat{ACH}=\widehat{BAH}=\left(90^0-\widehat{HAC}\right)\\\widehat{AHC}=\widehat{AHB}=90^0\end{cases}}\)

=> \(\frac{BH}{AH}=\frac{AH}{HC}\Leftrightarrow AH^2=HB\cdot HC\Rightarrow h^2=b'c'\)

c) \(S_{ABC}=\frac{1}{2}AH\cdot BC=\frac{1}{2}AB\cdot AC\)

\(\Rightarrow AH\cdot BC=AB\cdot AC\Rightarrow ah=bc\)

d) \(\frac{1}{b^2}+\frac{1}{c^2}=\frac{b^2+c^2}{b^2c^2}=\frac{a^2}{a^2h^2}=\frac{1}{h^2}\) (theo c)

chị lớp 9 em  có lớp 7 thui em k biết nha

đề là rút gọn hả bạn:

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

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

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

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

\(\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

\(1:x< 0\left(B\right)\)

\(2:\left(D\right)\)

\(3:x< 2021\left(C\right)\)

\(4:x\ge15\left(D\right)\)

\(5:\)để pt có nghĩa thì 2x-5>0

\(2x>5< =>x>\frac{5}{2}\)

chọn (C)

\(6:\frac{1}{2}\sqrt{20}-\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(\frac{1}{2}\sqrt{20}-\sqrt{5}+2\)

\(\sqrt{5}-\sqrt{5}+2=2\)

chọn (B)

\(7:\frac{6xy^2}{x^2-y^2}\sqrt{\frac{\left(x-y\right)^2}{\left(3xy^2\right)^2}}\)

\(\frac{6xy^2}{x^2-y^2}\frac{x-y}{3xy^2}\)

\(\frac{2}{x+y}\)

chọn (B)

\(8:\left(1+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(\frac{3+\sqrt{3}}{\sqrt{3}+1}-1\right)\)

\(\left(1+\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right)\left(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-1\right)\)

\(\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)

\(\sqrt{3}^2-1^2=3-1=2\)

chọn (D)

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

\(M=\sqrt{3}-1+\sqrt{3}-1\)

\(M=2\sqrt{3}-2\)

chọn (A)

\(10:\sqrt{4+\sqrt{x^2-1}}=2\)

\(4+\sqrt{x^2-1}=2^2=4\)

\(\sqrt{x^2-1}=0\)

\(x^2-1=0< =>x=1\)

chọn (A)

1 B 

2 D 

3 C 

4 D 

5 C 

6 B 

7 B 

8 D 

9 D 

10 B 

\(\sqrt{x^2-4x+4}-\sqrt{x^2+2x+1}=-3\)

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

\(\Leftrightarrow\left|x-2\right|-\left|x+1\right|=-3\)(1)

Có: \(\left|x-2\right|-\left|x+1\right|=\left|x-2\right|-\left|x-2+3\right|\ge\left|x-2\right|-\left(\left|x-2\right|+3\right)=-3\)

Dấu \(=\)khi \(3\left(x-2\right)\ge0\Leftrightarrow x\ge2\).

Do đó nghiệm của (1) là \(x\ge2\).

Vậy nghiệm phương trình đã cho là \(x\ge2\).

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

<=>\(|x-2|-|x+1|=-3\)(1)

nếu \(\hept{\begin{cases}x-2\ge0\\x+1\ge0\end{cases}=>\hept{\begin{cases}x\ge2\\x\ge-1\end{cases}=>x\ge}2}\)

(1)<=> x-2-x-1+3=0

<=>0x=0(đúng với mọi x)

=>x\(\in\left\{x|x\ge2\right\}\)

nếu \(\hept{\begin{cases}x-2< 0\\x+1< 0\end{cases}< =>\hept{\begin{cases}x< 2\\x< -1\end{cases}< =>x< -1}}\)

(1)<=>2-x+x+1+3=0

<=>0x=-3(vô lí)

vậyphương trình đã cho có tập nghiêm là \(x\in\left\{x|x\ge2\right\}\)

\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

áp dụng bunhia - cốpxki

\(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)

\(=6\left(a+b+c\right)\)

\(=6.2021=12126< =>P=\sqrt{12126}\)

vậy MAX P=\(\sqrt{12126}\)

\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

\(\Rightarrow P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)

Áp dụng BĐT Bunyakovsky ta có:

\(P^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=6\left(a+b+c\right)=6\cdot2021\)

\(\Rightarrow P\le\sqrt{6\cdot2021}=\sqrt{12126}\)

Dấu "=" xảy ra khi: \(a=b=c=\frac{2021}{3}\)

Vậy \(Max\left(P\right)=\sqrt{12126}\Leftrightarrow a=b=c=\frac{2021}{3}\)

a 

\(\sqrt{3}\cdot\sqrt{75}=\sqrt{3\cdot75}=\sqrt{225}=15\)   

b 

\(\sqrt{72}\cdot\sqrt{18}=6\sqrt{2}\cdot3\sqrt{2}=18\cdot2=36\)   

c 

\(\sqrt{2,5}\cdot\sqrt{30}\cdot\sqrt{48}=\sqrt{2,5\cdot30}\cdot\sqrt{48}=\sqrt{75}\cdot\sqrt{48}=5\sqrt{3}\cdot4\sqrt{3}=20\cdot3=60\)   

d 

\(\sqrt{\frac{5}{49}}\cdot\sqrt{\frac{16}{125}}=\sqrt{\frac{5}{49}\cdot\frac{16}{125}}=\sqrt{\frac{16}{49\cdot25}}=\frac{4}{7\cdot5}=\frac{4}{35}\)

a) Xét tam giác \(BDC\): 

\(\widehat{DBC}=180^o-\widehat{BDC}-\widehat{DCB}=180^o-30^o-60^o=90^o\)

Do đó tam giác \(BDC\)vuông tại \(B\).

Có \(\widehat{BDC}=30^o\)nên \(BC=\frac{1}{2}DC\Rightarrow AB=AC=\frac{1}{2}DC\Rightarrow DC=12\left(cm\right)\).

\(BC^2+BD^2=CD^2\)(định lí Pythagore) 

\(\Leftrightarrow BD^2=CD^2-BC^2=12^2-6^2=108\)

\(\Leftrightarrow BD=6\sqrt{3}\left(cm\right)\)

b) \(S_{ABD}=S_{DBC}-S_{ABC}=\frac{1}{2}.6.6\sqrt{3}-\frac{6^2\sqrt{3}}{4}=9\sqrt{3}\left(cm^2\right)\)