1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ( 1 + 2 + 3 + 4 ) + ……+ ( 1 + 2 + 3 +…+ 99 ) = x

Ta thấy :  số 1 xuất hiện trong 99 tổng ,  số 2 xuất hiện trong 98 lần , số 3 xuất hiện trong 97 tổng , ... , 99 xuất hiện trong 1 tổng 

Nên tổng trên bằng ; 1 x 99 + 2 x 98 + 3 x 97 + ... + 97 x 3 + 98 x 2 + 99 x 1 = x 

                               [( 1 x99 ) +  ( 99 x1 )] + [( 2 x 98 ) + ( 98 x 2 ) ] + ... + [( 49 x 51 ) + ( 51 x 49 )] = x 

 ( Tự làm tiếp )

làm đi hộ mk mk k 5 k cho

\(=\frac{2-1}{\sqrt{2}+1}+\frac{3-2}{\sqrt{3}+\sqrt{2}}+\frac{4-3}{\sqrt{4}+\sqrt{3}}+...+\frac{100-99}{\sqrt{100}+\sqrt{99}}.\)

\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\sqrt{2}+1}+\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}+\frac{\left(\sqrt{4}+\sqrt{3}\right)\left(\sqrt{4}-\sqrt{3}\right)}{\sqrt{4}+\sqrt{3}}+...\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)

\(=\sqrt{100}-1=10-1=9.\)

= 1

tick đi mink giải thích cho .  hihihihihihihihiihihiiiiiiiiiiiiiiii

Ta có: \(1+2^2+3^2+4^2+...+99^2+100^2\) (đề đúng)

\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)+100\left(101-1\right)\)

\(=\left(1.2+2.3+3.4+...+99.100+100.101\right)-\left(1+2+3+...+100\right)\)

\(=\frac{1.2.3+2.3.3+...+100.101.3}{3}-\frac{\left(100+1\right)\left[\left(100-1\right)\div1+1\right]}{2}\)

\(=\frac{1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+100.101.\left(102-99\right)}{3}-5050\)

\(=\frac{1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-99.100.101+100.101.102}{3}-5050\)

\(=\frac{100.101.102}{3}-5050\)

\(=343400-5050\)

\(=338350\)

\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

\(-\frac{2}{5};-\frac{7}{20};-\frac{9}{20}\)