b) bạn trục mẫu đi nha dựa vào hằng đẳng thức a^2 -b^2=(a-b)(a+b)

rồi bạn tính nói chung mẫu bằng -1

tính cái trên tử kết quả là 4

c) bạn dựa vào câu b .\(\dfrac{1}{\sqrt{3}}=\dfrac{2}{2\sqrt{3}}>\dfrac{2}{\sqrt{3}+\sqrt{4}}\)

từ đó suy ra B > 2A vậy B>8

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)

cả 2 ý bạn trục căn thức ở mấu là xong nhé:

vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy

a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)

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

=4

Thay x=4 vào B, ta được:

\(B=\dfrac{2-4}{2}=-1\)

Bài 1 :

a) \(Cos30^o=Cos\left(2.15^o\right)=2cos^215^o-1\)

\(\Rightarrow cos^215^o=\dfrac{cos30^o+1}{2}\)

\(\Rightarrow cos^215^o=\dfrac{\dfrac{\sqrt[]{3}}{2}+1}{2}\)

\(\Rightarrow cos^215^o=\dfrac{\sqrt[]{3}+2}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{\sqrt[]{3}+2}}{2}\)

\(\Rightarrow cos15^o=\dfrac{2\sqrt[]{\sqrt[]{3}+2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{4\sqrt[]{3}+8}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{6+2.2\sqrt[]{2}\sqrt[]{6}+2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{\left(\sqrt[]{6}+\sqrt[]{2}\right)^2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{6}+\sqrt[]{2}^{ }}{4}\left(dpcm\right)\)

a)

 Dựng tam giác ABC vuông tại A với \(\widehat{C}=15^o\). Trên đoạn thẳng AC lấy điểm D sao cho \(\widehat{CBD}=15^o\). Không mất tính tổng quát, ta chuẩn hóa \(AB=1\). \(\Rightarrow\left\{{}\begin{matrix}BD=\dfrac{AB}{cos60^o}=2\\AD=AB.tan60^o=\sqrt{3}\end{matrix}\right.\)

 Dễ thấy tam giác DBC cân tại D \(\Rightarrow BD=CD=2\) \(\Rightarrow AC=AD+DC=2+\sqrt{3}\)

  \(\Rightarrow tanC=\dfrac{AB}{AC}=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}\) 

\(\Rightarrow\dfrac{sinC}{cosC}=2-\sqrt{3}\)

\(\Rightarrow sinC=\left(2-\sqrt{3}\right)cosC\)

Mà \(sin^2C+cos^2C=1\)

\(\Rightarrow\left(7-4\sqrt{3}\right)cos^2C+cos^2C=1\)

\(\Leftrightarrow\left(8-4\sqrt{3}\right)cos^2C=1\)

\(\Leftrightarrow cos^2C=\dfrac{1}{8-4\sqrt{3}}=\dfrac{2+\sqrt{3}}{4}\)

\(\Leftrightarrow cosC=\sqrt{\dfrac{2+\sqrt{3}}{4}}\) \(=\dfrac{\sqrt{2+\sqrt{3}}}{2}=\dfrac{\sqrt{8+4\sqrt{3}}}{4}\) \(=\dfrac{\sqrt{6}+\sqrt{2}}{4}\) 

\(\Rightarrow cos15^o=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)

b) \(A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{32cos^415^o-10-8\sqrt[]{3}}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{32.\dfrac{1}{4^4}\left(\sqrt[]{6}+\sqrt[]{2}\right)^4-10-8\sqrt[]{3}}}\) \(\left(Cos15^o=\dfrac{\sqrt[]{6}-\sqrt[]{2}}{4}\right)\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{\dfrac{1}{8}\left(8+2\sqrt[]{12}\right)^2-10-8\sqrt[]{3}}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{\dfrac{1}{8}\left[64+32\sqrt[]{12}+48-80-64\sqrt[]{3}\right]}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\dfrac{1}{2}\sqrt[3]{32}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\dfrac{1}{2}.2\sqrt[3]{4}}=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{4}}\)

giúp mình với ạ 

b: Ta có: \(\dfrac{4}{\sqrt{3}+1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{6}{3-\sqrt{3}}\)

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

\(=2\sqrt{3}-4\)

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

Lời giải:
a.

\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)

$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.

$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$

$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.

$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$

$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$

$=1(3+\sqrt{2})=3+\sqrt{2}$

Với n\(\in N\)^* có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow\)\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\) (*)

a) Áp dụng (*) vào T

\(\Rightarrow T=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

b) Có \(VT=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)\(=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)

\(\Leftrightarrow\sqrt{n+1}=5\Leftrightarrow n=24\) (tm)

Vậy n=24.