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\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)
\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)
\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)
\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)
\(50^2-49^2+48^2-47^2+...+2^2-1\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48-47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=99+95+...+3\)
\(=\frac{\left(99+3\right)\left[\left(99-3\right):4+1\right]}{2}\)
\(=\frac{102.\left(24+1\right)}{2}=\frac{102.25}{2}=1275\)
\(\left(50^2+48^2+...+2^2\right)-\left(49^2+47^2+...+1^2\right)\)
\(=50^2+48^2+...+2^2-49^2-47^2-...-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=50+49+48+47+...+2+1\)
\(=\dfrac{50\left(50+1\right)}{2}=\dfrac{50\cdot51}{2}=1275\)
Ta có : ( 502 + 482 + ... + 22 ) - ( 492 +472 + ... + 12 )
= 502 + 482 +...+ 22 - 492 -472 - 12
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
= \(\left(50-49\right)\left(50+49\right)+\left(48-47\right)+..+\left(2-1\right).\left(2+1\right)\)
= \(50+49+48+47+...+2+1\)
= \(1257\)
\(50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)\(=50+49+48+47+...+2+1\)
\(=\dfrac{50\left(50+1\right)}{2}\)
\(=\dfrac{50.51}{2}\)
\(=1275\)
Ta có : \(\frac{x}{50}+\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-150}{25}=0\)
=> \(\frac{x}{50}-1+\frac{x-1}{49}-1+\frac{x-2}{48}-1+\frac{x-3}{47}-1+\frac{x-150}{25}+4=0\)
=> \(\frac{x-50}{50}+\frac{x-50}{49}+\frac{x-50}{48}+\frac{x-50}{47}+\frac{x-50}{25}=0\)
=> \(\left(x-50\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{25}\right)=0\)
=> \(x-50=0\)
=> \(x=50\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{50\right\}\)
\(50^2-49^2+48^2-47^2+46^2-45^2+...+4^2-3^2+2^2\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+\left(46-45\right)\left(46+45\right)...+\left(4-3\right)\left(4+3\right)+4\)
\(=99+95+91+...+7+3+1\)
\(=\left(3+99\right).\left[\left(99-3\right):4+1\right]:2+1\)
\(=102x25:2+1=1276\)
Giải:
Đặt \(A=50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(\Leftrightarrow A=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(\Leftrightarrow A=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(\Leftrightarrow A=50+49+48+47+...+2+1\)
\(\Leftrightarrow A=\dfrac{\left(50+1\right).50}{2}\)
\(\Leftrightarrow A=1275\)
Vậy giá trị của biểu thức \(50^2-49^2+48^2-47^2+...+2^2-1^2\) là 1275.
Chúc bạn học tốt!!!
\(50^2-49^2+48^2-47^2+...........+2^2-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+.......+\left(2^2-1^2\right)\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48-47\right)+.......+\left(2-1\right)\left(2+1\right)\)
\(=50+49+..........+1\)
\(=\dfrac{\left(50+1\right)50}{2}=1275\)
=(502-492)+...+(22-12)
=(50-49)(50+49)+(48-47)(48+47)+...+(2-1)(2+1)
=1.99+1.95+1.91+...+1.3
=99+95+91+...+3
=(99+3)+(95+7)+...+
\(50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=99.1+95.1+...+3.1\)
\(=99+95+...+3\)
\(=3+...+95+99\)
Từ 3 đến 99 có: \(\left(99-3\right):4+1=25\left(\text{số hạng}\right)\)
Tổng là: \(\frac{\left(99+3\right)\times25}{2}=1275\)
học tập 2 rồi hả