​\(|x-99|^{100}+|x-100|^{101}=1\)​

* Nếu \(x=99\)\(\Rightarrow\) \(|99-99|^{100}+|99-100|^{101}=0+1=1\)(  đúng )

\(\Rightarrow x=99\)là một nghiệm của phương trình

* Nếu \(x=100\)\(\Rightarrow|100-99|^{100}+|100-100|^{101}=1+0=1\)( đúng )

\(\Rightarrow x=100\)là một nghiệm của phương trình

* Nếu \(x< 99\)\(\Rightarrow x-100< 99-100\)\(\Rightarrow x-100< -1\)

\(\Rightarrow|x-100|^{101}>1\)\(\Leftrightarrow|x-99|^{100}+|x-100|^{101}>1\)\(\Rightarrow\)Phương trình vô nghiệm

* Nếu \(x>100\)\(\Rightarrow x-99>100-99\)\(\Rightarrow x-99>1\)

\(\Rightarrow|x-99|^{100}>1\)\(\Rightarrow|x-99|^{100}+|x-100|^{101}>1\)\(\Rightarrow\)Phương trình vô nghiệm

* Nếu \(99< x< 100\)\(\Rightarrow99-99< x-99< 100-99\)\(\Rightarrow0< x-99< 1\)

\(\Rightarrow|x-99|=x-99\)\(\left(1\right)\)

Cũng có : \(99< x< 100\)\(\Rightarrow99-100< x-100< 100-100\)\(\Rightarrow-1< x-100< 0\)

\(\Rightarrow|x-100|=-x+100\)\(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow|x-99|+|x-100|=x-99-x+100\)

\(\Rightarrow|x-99|+|x-100|=1\)

Ta lại có : \(|x-99|^{100}< |x-99|\)Do(  \(0< |x-99|< 1\))

\(|x-100|^{101}< |x-100|\)Do ( \(0< |x-100|< 1\)

\(\Rightarrow|x-99|^{100}+|x-100|^{101}< |x-99|+|x-100|\)

\(\Rightarrow|x-99|^{100}+|x-100|^{101}< 1\)

\(\Leftrightarrow\)Phương trình vô nghiệm

Vậy phương trình có hai nghiệm duy nhất là \(x\in\left\{99;100\right\}\)

Bạn ơi bạn chia trường hợp kiểu gì vậy , với cả trường hợp cuối mình không hiểu gì đâu bạn ơi

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

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

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

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

\(=2\)

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

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

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

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

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

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

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

_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_