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\(\left(50^2-49^2\right)+\left(48^2-47^2\right)+\left(46^2-45^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...\)
\(+\left(4-3\right)\left(4+3\right)+\left(2-1\right)\left(2+1\right)\)
(ta thấy trong mỗi tích đều có 1 thừa số bằng 1, VD: 50-49=1)
\(A=99+95+91+...+7+3\) số hạng cách nhau 4 đơn vị
Số số hạng của A là \(\left(99-3\right):4+1=25\)
=> \(A=\left(99+3\right).25:2=1275\)
Theo bài ra ta có:
\(50^2-49^2+48^2-47^2+....+2^2-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=1\times\left(50+49\right)+1\times\left(48+47\right)+...+1\times\left(2+1\right)\)
\(=50+49+48+47+...+2+1\)
\(=\left(50+49\right)\times50\div2=2475\)
Vậy giá trị biểu thức = 2475
C = 50 - 49 + 48 - 47 + ... + 2 - 1 ( có 50 số hạng )
=> C = ( 50 - 49 ) + ( 48 - 47 ) + ... + ( 2 - 1 ) ( có đủ 25 nhóm )
=> C = 1 + 1 + ... + 1 ( 25 số hạng 1 )
=> C = 1 . 25 = 25
a) \(87^2+26\cdot87+13^2=87^2+2\cdot87\cdot13+13^2=\left(87+13\right)^2=100^2=10000\)
Sủa đề : tính \(D=\left(50^2+48^2+46^2+....+2^2\right)-\left(49^2+47^2+45^2+...+1^2\right)\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+\left(46^2-45^2\right)+.....+\left(2^2-1^2\right)\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+....+\left(2-1\right)\left(2+1\right)\)
\(=50+49+48+.....+2+1\)
\(=\frac{50\left(50+1\right)}{2}=1275\)
D=(502-492)+(482-472)+...+(22-12)
= ( (50-49)(50+49)+(48-47)(48+47)+...+(2-1)(2+1)
= 50+49+48+47+...+2+1
=\(\frac{\left(50+1\right).50}{2}\)
=1275
\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)
\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)
\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)
\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)
a: \(A=\dfrac{\left(258-242\right)\left(258+242\right)}{\left(254-246\right)\left(254+246\right)}=\dfrac{16}{8}=2\)
b: \(=\left(263+37\right)^2=300^2=90000\)
c: \(=\left(136-46\right)^2=90^2=8100\)
d: \(=50^2-49^2+48^2-47^2+...+2^2-1^2\)
=50+49+...+2+1
=51x50:2=1275
Đặt \(THANG=50^2-49^2+48^2-47^2+....+2^2-1\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+....+\left(2^2-1\right)\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+....+\left(2-1\right)\left(2+1\right)\)
\(=50+49+48+47+....+2+1\)
\(=\dfrac{50\cdot\left(50+1\right)}{2}=\dfrac{50\cdot51}{2}=1275\)
\(C=50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=50+49+48+47+...+2+1\)
\(=\frac{50\left(50+1\right)}{2}=1275\)