915370

là kết quả

Bài 7: 

Ta có: \(C=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)

\(=\dfrac{\sqrt{2}\left(4+\sqrt{7}\right)}{6+\sqrt{8+2\sqrt{7}}}+\dfrac{\sqrt{2}\left(4-\sqrt{7}\right)}{6-\sqrt{8-2\sqrt{7}}}\)

\(=\dfrac{\sqrt{2}\left(4+\sqrt{7}\right)}{7+\sqrt{7}}+\dfrac{\sqrt{2}\left(4-\sqrt{7}\right)}{7-\sqrt{7}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{7}-1\right)\left(4+\sqrt{7}\right)}{6\sqrt{7}}+\dfrac{\sqrt{2}\left(\sqrt{7}+1\right)\left(4-\sqrt{7}\right)}{6\sqrt{7}}\)

\(=\dfrac{\sqrt{2}\left(-3+3\sqrt{7}+3+3\sqrt{7}\right)}{6\sqrt{7}}\)

\(=\sqrt{2}\)

6.

Ta có:

\(A=\sqrt{20+\sqrt{20+...+\sqrt{20}}}>\sqrt{20+\sqrt{\dfrac{1}{16}}}=\dfrac{9}{2}\)

\(B=\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}>\sqrt[3]{24}=\sqrt[3]{\dfrac{192}{8}}>\sqrt[3]{\dfrac{125}{8}}=\dfrac{5}{2}\)

\(\Rightarrow A+B>\dfrac{9}{2}+\dfrac{5}{2}=7\)

\(A=\sqrt[]{20+\sqrt[]{20+...+\sqrt[]{20}}}< \sqrt[]{20+\sqrt[]{20+...+\sqrt[]{25}}}=5\)

\(B=\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}< \sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{27}}}=3\)

\(\Rightarrow A+B< 5+3=8\)

8.

Ta có:

\(a=\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}-\dfrac{\sqrt{2}}{8}\Rightarrow\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}=a+\dfrac{\sqrt{2}}{8}\)

\(a^2=\dfrac{1}{4}\left(\sqrt{2}+\dfrac{1}{8}\right)+\dfrac{1}{32}-\dfrac{\sqrt{2}}{4}.\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}\)

\(=\dfrac{\sqrt{2}}{4}+\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{\sqrt{2}}{4}\left(a+\dfrac{\sqrt{2}}{8}\right)\)

\(=\dfrac{\sqrt{2}}{4}+\dfrac{1}{16}-\dfrac{\sqrt{2}}{4}a-\dfrac{1}{16}\)

\(=\dfrac{\sqrt{2}}{4}\left(1-a\right)\)

\(\Rightarrow a^4=\dfrac{1}{8}\left(a^2-2a+1\right)\)

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

\(\Rightarrow R=a^2+\dfrac{\sqrt{2}}{4}\left(a+3\right)=\dfrac{\sqrt{2}}{4}\left(1-a\right)+\dfrac{\sqrt{2}}{4}\left(a+3\right)=\sqrt{2}\)

vẽ lại mạch ta có RAM//RMN//RNB

đặt theo thứ tự 3 R là a,b,c

ta có a+b+c=1 (1)

điện trở tương đương \(\dfrac{1}{R_{td}}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(\Rightarrow I=\dfrac{U}{R_{td}}=9.\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) với a,b,c>0

áp dụng bất đẳng thức cô si cho \(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\)  \(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{\sqrt[3]{abc}}\ge\dfrac{3}{\left(\dfrac{a+b+c}{3}\right)}=\dfrac{9}{a+b+c}=9\)

\(\Leftrightarrow9\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge81\Leftrightarrow I\ge81\) I min =81 ( úi dồi ôi O_o hơi to mà vẫn đúng đá nhỉ)

dấu ''='' xảy ra \(\Leftrightarrow a=b=c\left(2\right)\)

từ (1) (2) \(\Rightarrow a=b=c=\dfrac{1}{3}\left(\Omega\right)\)

vậy ... (V LUN MẤT CẢ BUỔI TỐI R BÀI KHÓ QUÁ EM ĐANG ÔN HSG À )

em ơi chụp cả cái mạch điện a xem nào sao chụp nó bị mất r

Đề bài thiếu thì phải

32, A

33,B

34,C

III. 

A. 

1. B

2. C

3. B

4. A

5. D

B. 

1. C

2. B

3. B

4. A

5. C

1 b c b a d

2 c b b a c nha

1 The meeting was canceled 3 days ago

2 She told me she was watching a film with her sister then

3 I admire the guitarist who is perfroming on the stage

4 Had it not been for Pauline's interest, the project would have been abandoned 

IV. 

1. The meeting was canceled 3 days ago.

2. She told me shewas watching a film with her sister then

3. I admire the guitarist who is performing on the stage

4. Had it not been for Pauline's interest, the project was abandoned. (chắc thế)