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

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

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

\(=\frac{\sqrt{3}.2}{\sqrt{3}}\)

\(=2\)

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

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

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

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

\(=\frac{2\sqrt{3}}{\sqrt{3}}\)

= 2

Giúp với :((

bạn có chắc chắn đề đúng

\(P=\frac{\sqrt{a}+3}{\sqrt{a}-2}-\frac{\sqrt{a}-1}{\sqrt{a}+2}+\frac{4\sqrt{a}-4}{4-a}\)

\(=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{a-4}-\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-4}-\frac{4\sqrt{a}-4}{a-4}\)

\(=\frac{a+5\sqrt{a}+6-\left(a-3\sqrt{a}+2\right)-\left(4\sqrt{a}-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4}{\sqrt{a}-2}\)

_Hình hơi xấu , thông cảm _

Kẻ \(\(DE\perp AC\)\)

Có \(\(\widehat{AOB}=\widehat{DOC}\)\)

Xét tam giác vuông \(\(DKO\)\), ta có :

\(\(AK=DO.\sin\widehat{DOK}\)\)hay \(\(AK=DO.\sin\widehat{AOB}\)\)

Do đó:

\(\(S_{\Delta ADC}=\frac{1}{2}.AC.DO.\sin\widehat{AOB}\left(1\right)\)\)

Tương tự :

\(\(S_{\Delta ACB}=\frac{1}{2}.AC.BO.\sin\widehat{AOB}\left(2\right)\)\)

Từ \(\(\left(1\right)\&\left(2\right)\Rightarrow S_{ABCD}=S_{\Delta ADC}+S_{\Delta ACB}=\frac{1}{2}.AC.\left(DO+BO\right).\sin\widehat{AOB}\)\)

\(\(\Leftrightarrow S_{ABCD}=\frac{1}{2}AC.BD.\sin\widehat{AOB}\left(dpcm\right)\)\)

_Minh ngụy_

Cách 2 :

Ta có : \(\(\sin\widehat{AOD}=\sin\widehat{AOB}=\sin\widehat{COB}=\sin\widehat{COD}\left(=\sin a\right)\)\)

Mặt khác

\(\(2S_{\Delta AOD}=AO.OD.\sin a\)\)

\(\(2S_{AOB}=AO.OB.\sin a\)\)

\(\(2S_{BOC}=BO.OC.\sin a\)\)

\(\(2S_{COD}=DO.OC.\sin a\)\)

\(\(\Rightarrow2\left(S_{AOD}+S_{AOB}+S_{BOC}+S_{COD}\right)\)\)

\(\(=AO.OD.\sin a+AO.OB.\sin a+BO.OC.\sin a+DO.OC.\sin a\)\)

\(\(=\sin a.[\left(AO\left(OD+OB\right)+OC\left(OB+OD\right)\right)]\)\)

\(\(=\sin a.\left(OD+OB\right)\left(AO+OC\right)\)\)

\(\(=\sin a.BD.AC\)\)

\(\(\Rightarrow S_{\Delta AOD}+S_{\Delta AOB}+S_{\Delta BOC}+S_{\Delta COD}=\frac{1}{2}.AC.BD.\sin a\)\)

hay \(\(S_{ABCD}=\frac{1}{2}AC.BD.\sin a\)\)mà \(\(\sin\widehat{AOB}=\sin a\)\)

\(\(\Rightarrow S_{ABCD}=\frac{1}{2}AC.BD.\sin\widehat{AOB}\)\)

_Minh ngụy_

a) \(\cos^4\alpha-\sin^4\alpha=\left(\cos^2\alpha+\sin^2\alpha\right)\left(\cos^2\alpha-\sin^2\alpha\right)=\cos^2\alpha-\sin^2\alpha\)

\(2\cos^2\alpha-\left(\sin^2\alpha+\cos^2\alpha\right)=2\cos^2\alpha-1\)

b) \(\frac{\cos\alpha}{1-\sin\alpha}=\frac{1+\sin\alpha}{\cos\alpha}\)\(\Leftrightarrow\)\(\left(1-\sin\alpha\right)\left(1+\sin\alpha\right)=\cos^2\alpha\)

\(\Leftrightarrow\)\(1-\left(\sin^2\alpha+\cos^2\alpha\right)=0\)\(\Leftrightarrow\)\(1-1=0\) ( luôn đúng ) 

c) \(\frac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha.\cos\alpha}=\frac{2\cos\alpha.2\sin\alpha}{\sin\alpha.\cos\alpha}=4\)

um, hình như câu b) chỗ 1-.... đó hơi sai nếu viết từ bước trên xuống á bạn!

mình nghĩ là: sau dấu bằng đầu tiên, sau đó là:

\(=cos^2\alpha=1-sin^2\alpha\)(luôn đúng)

CẢM ƠN bạn nhiều lắm luôn nha!!!!!