a) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{3}{4}=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{cases}}\)

- \(cosx=\frac{1}{2}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

- \(cosx=\frac{-1}{2}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}=-\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\sqrt{3}}=\frac{-\sqrt{3}}{3}\)

b) Bạn làm tương tự câu a) nha. 

\(11:\)

\(\frac{-\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{2}\left(B\right)\)

\(12:B\)

\(13:\sqrt{25x}-\sqrt{9x}=8\)

\(\sqrt{25}\sqrt{x}-\sqrt{9}\sqrt{x}=8\)

\(\sqrt{x}\left(5-3\right)=8\)

\(\sqrt{x}=4< =>x=16\left(C\right)\)

\(14:\frac{4}{\sqrt{5}-1}-\sqrt{5}\)

\(\frac{4-5+\sqrt{5}}{\sqrt{5}-1}\)

\(\frac{\sqrt{5}-1}{\sqrt{5}-1}=1\left(B\right)\)

\(15:\)

\(-\sqrt{a^2\frac{b}{a}}\)

\(-\sqrt{a.b}\left(C\right)\)

\(1:4\left(B\right)\)

\(16:\sqrt{12}-\sqrt{27}+\sqrt{3}\)

\(\sqrt{3}\left(\sqrt{4}-\sqrt{9}+1\right)\)

\(\sqrt{3}\left(2-3+1\right)=0\left(B\right)\)

\(17:\sqrt{18}+\frac{2}{\sqrt{2}}-3\sqrt{8}\)

\(\sqrt{2}\left(\sqrt{9}+1-3\sqrt{4}\right)\)

\(\sqrt{2}.2=2\sqrt{2}\left(D\right)\)

\(18:\sqrt{x^2}=\left|x\right|=13\)

\(x=\pm13\left(D\right)\)

\(19:\left|x-1\right|\left(C\right)\)

\(20:\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(\left|\sqrt{5}-2\right|=\sqrt{5}-2\left(B\right)\)

hok tốt

chịu luôn

ko dc vì

tui hok lớp 4

áp dụng bđt bunhia- cốp xki với bộ số (a,b,c)(1,1,1)

\(\left(a^2+b^2+c^2\right)\left(1+1+1\right)\ge\left(a+b+c\right)^2\)

\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(< =>ĐPCM\)

Áp dụng BĐT Cauchy ta có:
\(a^2+b^2\ge2ab\) ; \(b^2+c^2\ge2bc\) ; \(c^2+a^2\ge2ca\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)=\left(a+b+c\right)^2\)

Dấu "=" xảy ra khi: a = b = c

Do: Góc ABD = Góc ACE (= 90 - A)
=> Δ ABD ∼ Δ ACE (2 Δ vuông)
=> AD.AC = AE.AB (tỉ lệ đồng dạng)
<=> AM² = AN² (Hệ thức lượng trong Δ vuông)
<=> AM = AN
Hay Δ AMN cân tại A.=>....

     #HT#

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