ta có

(121-1²).(121-2²).....(121-11²)

= (11²-1²).(11²-2²).....(11²-11²)

= (121-1²).(11²-2²).....0

= 0

\(B=\left(11^2-1^2\right)×...×\left(11^2-11^2\right)×...×\left(11^2-29^2\right)\)

\(B=0\)

`(3*x+2)^2=121`

\(=>\left[{}\begin{matrix}3x+2=11\\3x+2=-11\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x=11-2\\3x=-11-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x=9\\3x=-13\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{13}{3}\left(loại\right)\end{matrix}\right.\)

`(3x+2)^2=121`

`=>( 3x+2)^2= +-11^2`

\(\Rightarrow\left[{}\begin{matrix}3x+2=11\\3x+2=-11\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=9\\3x=-13\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{13}{3}\end{matrix}\right.\)

(121-1^2)*(121-2^2)*...*(121-11^2)*...*(121-29^2)=(121-1^2)*(121-2^2)*...*(121-121)*...*(121-29^2)=(121-1^2)*(121-2^2)*...*0*...*(121-29^2)=0

Lời giải:

$(3x+2)^2=121=11^2=(-11)^2$

$\Rightarrow 3x+2=11$ hoặc $3x+2=-11$

$\Rightarrow x=3$ hoặc $x=\frac{-13}{3}$

Vì $x$ là số tự nhiên nên $x=3$

\(\left(\frac{3}{5}+\frac{17}{3}\right):\frac{121}{19}+\left(\frac{-17}{3}+\frac{2}{5}\right):\frac{121}{19}\)

\(=\left(\frac{3}{5}+\frac{17}{3}\right).\frac{19}{121}+\left(\frac{-17}{3}+\frac{2}{5}\right).\frac{19}{121}\)

\(=\frac{19}{121}.\left[\left(\frac{3}{5}+\frac{17}{3}\right)+\left(\frac{-17}{3}+\frac{2}{5}\right)\right]\)

\(=\frac{19}{121}.\left[\frac{3}{5}+\frac{17}{3}+\left(\frac{-17}{3}\right)+\frac{2}{5}\right]\)

\(=\frac{19}{121}.\left[\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{-17}{3}+\frac{17}{3}\right)\right]\)

\(=\frac{19}{121}.\left[1+0\right]\)

\(=\frac{19}{121}.1\)

\(=\frac{19}{121}\)

\(\frac{121}{19}\)

A=(121-2²)(121-3²)...(121-11²)...(121-2011²)

A=(121-2²)(121-3²)...0...(121-2011²)

A=0 

Cảm ơn nhìu

bn ghi như rứa mk ko hiểu

Tìm số tự nhiên xx thỏa mãn: \left(3.x+2\right)^{2}=121(3.x+2)2=121

Tổng quát:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow S=\dfrac{10}{11}\)

Ta có công thức tổng quát như sau:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\right]\left[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}\right]}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)

\(=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)

\(=\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}\)

Áp dụng vào tổng S ta có:

\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{121\sqrt{120}+120\sqrt{121}}\)

\(S=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{120}}+\dfrac{1}{\sqrt{121}}\)

\(S=1-\dfrac{1}{\sqrt{121}}=1-\dfrac{1}{11}=\dfrac{10}{11}\)