Xét tổng 1 + 3 +5 +... + 99:

- Số số hạng là: (99-1):2+1= 50 số

- Tổng là: (1 + 99) x 50 : 2 = 2500

Vậy ta có biểu thức:

 \(2500=\left(x-2\right)^2\\ \Leftrightarrow50^2=\left(x-2\right)^2\\\Leftrightarrow \left[{}\begin{matrix}x-2=50\\x-2=-50\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=52\\x=-48\end{matrix}\right.\)

Vậy...

\(1+3+5+...+99=\left(x-2\right)^2\)

\(\Rightarrow\left[\left(99-1\right):2+1\right]\left(1+99\right):2=\left(x-2\right)^2\)

\(\Rightarrow50.100:2=\left(x-2\right)^2\)

\(\Rightarrow\left(x-2\right)^2=2500=50^2\)

\(\Rightarrow\left[{}\begin{matrix}x-2=50\\x-2=-50\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=52\\x=-48\end{matrix}\right.\)

x=52

1+3+5+..........+97+99=(x+2)²

[(99-1)/2+1]*(99+1)/2=(x+2)²

2500=(x+2)*(x+2)

Vì 2500=50*50 nên x=50-2=48

1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\ \left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\ \left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\ \frac{98}{99}-x=\frac{7}{2}\\ x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)

2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)

1.

\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}+\frac{x-5}{95}=5\)

\(\Rightarrow\left(\frac{x-1}{99}-1\right)+\left(\frac{x-2}{98}-1\right)+\left(\frac{x-3}{97}-1\right)+\left(\frac{x-4}{96}-1\right)+\left(\frac{x-5}{95}-1\right)\)\(=5-1-1-1-1-1\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}+\frac{x-100}{95}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\ne0\)

\(\Rightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy x=100

Chúc bạn học tốt

 a, (x+1) + (x+3) + (x+5) +.......+(x+99) =0

\(\Rightarrow\)x+x+x+...+x(50 số hạng)  +  1+3+5+...+99=0

\(\Rightarrow\)50.x+4950=0

50.x=0-4950

50.x=(-4950)

x=(-4950):50

x=(-99)