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\(\frac{258^2-242^2}{254^2-246^2}=\frac{\left(258+242\right)\left(258-242\right)}{\left(254+246\right)\left(254-246\right)}=\frac{500.16}{500.8}=2\)
\(263^2+74.263+37^2=263^2+2.37.263+37^2=\left(263+37\right)^2=300^2=90000\)
\(136^2-92.136+46^2=136^2-46.2.136+46^2=\left(136-46\right)^2=90^2=8100\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+.....+\left(2^2-1^2\right)=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+....+\left(2+1\right)\left(2-1\right)=\left(50+49+....+1\right)=\frac{51.50}{2}=51.25=1275\)
Bài 1:
a, \(\dfrac{63^2-47^2}{215^2-105^2}=\dfrac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}=\dfrac{16.110}{110.220}=\dfrac{16}{220}=\dfrac{4}{55}\)
b, \(\dfrac{427^2-373^2}{527^2-473^2}=\dfrac{\left(427-373\right)\left(427+373\right)}{\left(527-473\right)\left(527+473\right)}=\dfrac{54.800}{54.1000}=\dfrac{4}{5}\)
Bài 2:
\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50\)
\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52\)
Vì \(2.50< 2.52\Leftrightarrow A< B\)
Vậy A < B
Bài 3: Chỉ cần nhân hết cái trong ngoặc ở VT ra rồi nó sẽ bằng VP
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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
A - B = (502+482+462+.....+42+22) - (492+472+452+.....+32+12)
= 502 + 482 + 462 +... + 42+ 22 - 492 - 472 - .... - 32 - 12
= (502 - 492) + (482 - 472) + ... + (42 - 32) + (22 - 12)
= (50+49) (50 - 49) + (48 - 47) (48+47)+....+(4-3)(4+3) + (2-1)(2+1)
= 50 + 49 + 48 + 47 + 46 + 45+...+4+3+2+1
= [(50 - 1) : 1 + 1] * (50+1) : 2 = 1275
vậy A - B = 1275
Câu b :
\(A=\left(50^2+48^2+46^2+.........+4^2+2^2\right)-\left(49^2+47^2+45^2+.........+5^2+3^2+1^2\right)\)
\(A=\left(50^2-49^2\right)+\left(48^2-47^2\right)+.........\left(4^2-3^2\right)+\left(2^2-1^2\right)\)
\(A=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+..........+\left(4+3\right)\left(4-3\right)+\left(2+1\right)\left(2-1\right)\)
\(A=50+49+48+..........+3+2+1\)
\(A=\dfrac{50.51}{2}\)
\(\Rightarrow A=1275\)
a, \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Tính giá trị biểu thức sau:
a, A= \(258^2-\dfrac{242^2}{254^2}-246^2\approx\) 6047,1
b, B= \(263^2+74.263+37^2=90000\)
c, C= \(136^2-92.136+46^2=8100\)
d, D = \(\left(50^2+48^2+46^2+...+2^2\right)-\left(49^2+47^2+45^2+...+1^2\right)\)
= 22100 - 20825= 1275