\(1\dfrac{1}{2}x1\dfrac{1}{3}x1\dfrac{1}{4}x1\dfrac{1}{5}x1\dfrac{1}{6}x1\dfrac{1}{7}x1\dfrac{1}{8}x1\dfrac{1}{9}\)

\(=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}x\dfrac{8}{7}x\dfrac{9}{8}x\dfrac{10}{9}\)

\(=x^7.\dfrac{3.4.5.6.7.8.9.10}{2.3.4.5.6.7.8.9}\)

\(=x^7.\dfrac{10}{2}\)

\(=5x^7\)

\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{9}{8}\times\dfrac{10}{9}=\dfrac{10}{2}=5\)

` a/`

` 2 - 1 5/6 + 2 2/3 = 2 - 11/6 - 8/3 = 1/6+ 8/3 = 1/6 + 16/6 = 17/6 `

`b/`

`5/9 xx ( 2 5/6 - 1 2/3 ) = 5/9 xx ( 17/6 - 5/3 ) = 5/9 xx 7/6 = 35/54 `

`c/`

` 1 1/3 : ( 2 + 1 1/6 : 2 5/6 ) `

`= 4/3 :  ( 2 + 7/6 : 17/6 ) `

`= 4/3 : ( 2 + 7/6 xx 6/17 )`

`= 4/3 : ( 2 + 7/17 ) `

`= 4/3 : ( 34/17 + 7/17 ) `

`= 4/3 : 41/17 `

`= 4/3 xx 17/41 `

`= 68/123`

` d/` 

` 2 3/5 : 3/4 xx 1 4/5 = 13/5 xx 4/3 xx 9/5 =52/15 xx 9/5 = 156/25`

Ta có: \(\Delta=4m^2+4m-11\)

Để phương trình có 2 nghiệm phân biệt \(\Leftrightarrow4m^2+4m-11>0\)

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m+3\\x_1x_2=2m+5\end{matrix}\right.\)

Để phương trình có 2 nghiệm dương phân biệt

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2+4m-11>0\\2m+3>0\\2m+5>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< \dfrac{-1-2\sqrt{3}}{2}\\m>\dfrac{-1+2\sqrt{3}}{2}\end{matrix}\right.\\m>-\dfrac{3}{2}\\m>-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow m>\dfrac{-1+2\sqrt{3}}{2}\)

 Mặt khác: \(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{4}{3}\)

\(\Rightarrow\dfrac{x_1+x_2+2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{16}{9}\) \(\Rightarrow\dfrac{2m+3+2\sqrt{2m+5}}{2m+5}=\dfrac{16}{9}\)

\(\Rightarrow18m+27+18\sqrt{2m+5}=32m+80\)

\(\Leftrightarrow14m-53=18\sqrt{2m+5}\)

\(\Rightarrow\) ...

giúp mình với ạ ! Mình cảm ơn ạ 

\(...=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}....x\dfrac{1000}{999}\)

\(=\dfrac{1}{2}x\dfrac{1000}{1}=500\)

=3/2x4/3x5/4x....x1000/999

=1/2x1000=500

mình chưa chắc là đúng đâu nhé

\(=>C=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}.....\cdot\dfrac{101}{100}\)

\(C=\dfrac{3\cdot4\cdot5.......\cdot101}{2\cdot3\cdot4.........\cdot100}\)

\(C=\dfrac{101}{2}\)

\(C=1\dfrac{1}{2}\cdot1\dfrac{1}{3}\cdot1\dfrac{1}{4}\cdot...\cdot1\dfrac{1}{100}\)

\(C=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}\)

\(C=\dfrac{101}{2}\)

a: \(1\dfrac{1}{2}\cdot1\dfrac{1}{3}\cdot1\dfrac{1}{4}\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\)

\(=\dfrac{5}{2}\)

b: \(1\dfrac{1}{2}:1\dfrac{1}{3}:1\dfrac{1}{4}\)

\(=\dfrac{3}{2}:\dfrac{4}{3}:\dfrac{5}{4}\)

\(=\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}=\dfrac{9}{10}\)

\(1\dfrac{1}{3}\times1\dfrac{1}{8}\times1\dfrac{1}{15}\times1\dfrac{1}{24}\times1\dfrac{1}{35}\)

= \(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}\times\dfrac{36}{35}\)

= \(\dfrac{4\times9\times16\times25\times36}{3\times8\times15\times24\times35}\)

= \(\dfrac{1\times2\times2\times25\times36}{1\times2\times15\times24\times35}\)

= \(\dfrac{4\times25\times36}{30\times24\times35}\)

= \(\dfrac{1\times25\times36}{30\times6\times35}=\dfrac{1}{7}\)

\(1\dfrac{1}{3}\times1\dfrac{1}{8}\times1\dfrac{1}{15}\times1\dfrac{1}{24}\times1\dfrac{1}{35}\)

\(=\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}\times\dfrac{36}{35}\)

\(=\dfrac{2\times2}{1\times3}\times\dfrac{3\times3}{2\times4}\times\dfrac{4\times4}{3\times5}\times\dfrac{5\times5}{4\times6}\times\dfrac{6\times6}{5\times7}\)

\(=\left(\dfrac{2}{1}\times\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\right)\times\left(\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\times\dfrac{5}{6}\times\dfrac{6}{7}\right)\)

\(=6\times\dfrac{2}{7}=\dfrac{12}{7}\)

\(log_7\left(4x^2-4x+1\right)-log_72x+4x^2+1=6x\)

\(\Leftrightarrow log_7\left(4x^2-4x+1\right)+4x^2-4x+1=log_72x+2x\)

\(\Rightarrow4x^2-4x+1=2x\)

\(\Rightarrow...\)

Câu 1:

Để ý rằng \((2-\sqrt{3})(2+\sqrt{3})=1\) nên nếu đặt

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

PT đã cho tương đương với:

\(ma^x+\frac{1}{a^x}=4\)

\(\Leftrightarrow ma^{2x}-4a^x+1=0\) (*)

Để pt có hai nghiệm phân biệt \(x_1,x_2\) thì pt trên phải có dạng pt bậc 2, tức m khác 0

\(\Delta'=4-m>0\Leftrightarrow m< 4\)

Áp dụng hệ thức Viete, với $x_1,x_2$ là hai nghiệm của pt (*)

\(\left\{\begin{matrix} a^{x_1}+a^{x_2}=\frac{4}{m}\\ a^{x_1}.a^{x_2}=\frac{1}{m}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a^{x_2}(a^{x_1-x_2}+1)=\frac{4}{m}\\ a^{x_1+x_2}=\frac{1}{m}(1)\end{matrix}\right.\)

Thay \(x_1-x_2=\log_{2+\sqrt{3}}3=\log_{a^2}3\) :

\(\Rightarrow a^{x_2}(a^{\log_{a^2}3}+1)=\frac{4}{m}\)

\(\Leftrightarrow a^{x_2}(\sqrt{3}+1)=\frac{4}{m}\Rightarrow a^{x_2}=\frac{4}{m(\sqrt{3}+1)}\) (2)

\(a^{x_1}=a^{\log_{a^2}3+x_2}=a^{x_2}.a^{\log_{a^2}3}=a^{x_2}.\sqrt{3}\)

\(\Rightarrow a^{x_1}=\frac{4\sqrt{3}}{m(\sqrt{3}+1)}\) (3)

Từ \((1),(2),(3)\Rightarrow \frac{4}{m(\sqrt{3}+1)}.\frac{4\sqrt{3}}{m(\sqrt{3}+1)}=\frac{1}{m}\)

\(\Leftrightarrow \frac{16\sqrt{3}}{m^2(\sqrt{3}+1)^2}=\frac{1}{m}\)

\(\Leftrightarrow m=\frac{16\sqrt{3}}{(\sqrt{3}+1)^2}=-24+16\sqrt{3}\) (thỏa mãn)

Câu 2:

Nếu \(1> x>0\)

\(2017^{x^3}>2017^0\Leftrightarrow 2017^{x^3}>1\)

\(0< x< 1\Rightarrow \frac{1}{x^5}>1\)

\(\Rightarrow 2017^{\frac{1}{x^5}}> 2017^1\Leftrightarrow 2017^{\frac{1}{x^5}}>2017\)

\(\Rightarrow 2017^{x^3}+2017^{\frac{1}{x^5}}> 1+2017=2018\) (đpcm)

Nếu \(x>1\)

\(2017^{x^3}> 2017^{1}\Leftrightarrow 2017^{x^3}>2017 \)

\(\frac{1}{x^5}>0\Rightarrow 2017^{\frac{1}{x^5}}>2017^0\Leftrightarrow 2017^{\frac{1}{5}}>1\)

\(\Rightarrow 2017^{x^3}+2017^{\frac{1}{x^5}}>2018\) (đpcm)

Câu 3: Bạn xem lại đề bài hộ mình xem có đúng không nhe.