Ta có:

\(99^{20}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)

\(9999^{10}=\left(99.101\right)^{10}\)

Vì \(99.99< 99.101\) \(\Rightarrow\left(99.99\right)^{10}< \left(99.101\right)^{10}\Leftrightarrow99^{20}< 9999^{10}\)

a. \(\dfrac{\left(x+1\right)}{10}+\dfrac{\left(x+1\right)}{11}+\dfrac{\left(x+1\right)}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)

\(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)

\(\Rightarrow x+1=0\)

\(x=-1\)

b, \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\\ \left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\\ \dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\\ x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)\)

vì \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\\ \Rightarrow x+2004=0\\ x=-2004\)

nucuoicuapi! @phynit @Bùi Thị Vân là thầy cô giáo đấy!

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\) (k\(\in\) N*)

\(\Rightarrow x=2k;y=3k;z=5k\)

\(xyz=2k.3k.5k=30k^3=81\Rightarrow k^3=2,7\Leftrightarrow k=\)

đến đây thì mk chịu! Máy tính bỏ tui nhà mk thì đang hỏng. bn lấy máy tính bỏ túi mà tính.

b, \(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{7}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=-\dfrac{1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)

Vậy .....

Đậy nè!

\(\dfrac{10.4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{5.2.2^{12}.3^{10}+3^9.2^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\dfrac{5.2^{13}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}=\dfrac{2^{12}.3^{10}.5\left(2+1\right)}{2^{11}.3^{11}.5}\)

\(=\dfrac{2^{11}.2.3^{10}.5.3}{2^{11}.3^{10}.3.5}=2\)

Bn có chép sai đề bài ko zợ!