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\(a,=15\left(64+36\right)+100\cdot25+100\cdot60\\ =100\left(15+25+60\right)=100\cdot100=10000\\ b,Sửa:47^2+48^2-25^2+94\cdot48=\left(47+48\right)^2-25^2\\ =95^2-25^2=\left(95-25\right)\left(95+25\right)=70\cdot120=8400\)
a) Ta có 15.64 + 25.100 + 36.15 + 60.100
= (15.64 + 36.15) + (25.100 + 60.100)
= 100.(15 + 85) = 10000.
b) Ta có 47 2 + 48 2 - 25 + 94.48
= ( 47 2 +2.47.48+ 48 2 ) - 5 2 = ( 47 + 48 ) 2 - 5 2 =9000.
c) Ta có 93 -92.(-l)-9.11 + (-l).ll
= (93 +92)-(9.11 + 1.11)
= 92(9 +1) -ll.(9 + l) = 700.
a) \(26^2+52.24+24^2=26^2+2.26.24+24^2\)
= \(\left(26+24\right)^2=50^2=2500\)
b) \(52^2+47^2+94.52\) ( câu này sai đề sửa luôn)
= \(52^2+2.47.52+47^2=\left(52+47\right)^2=99^2\)
= \(9801\)
c) \(50^2-49^2+48^2-47^2+...+2^2-1^2\)
= \(\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
= \(99+95+...+3\)
Dãy số này có : \(\dfrac{99-3}{4}+1=\dfrac{96}{4}+1=25\) số hạng
\(\Rightarrow\) \(99+95+...+3\) = \(\left(99+3\right).25:2=1275\)
d) \(87^2+26.87+13^2=87^2+2.13.87+13^2\)
\(=\left(87+13\right)^2=100^2=10000\)
e) \(3003^2-3^2=\left(3003-3\right)\left(3003+3\right)\)
= \(3000.3006=9018000\)
\(a,26^2+52\cdot24+24^2\\ =26^2+2\cdot26\cdot24+24^2\\ =\left(26+24\right)^2\\ =50^2\\ =2500\)
\(b,53^2+47^2+94\cdot53\\ =53^2+2\cdot47\cdot53+47^2\\ =\left(53+47\right)^2\\ =100^2\\ =10000\)
\(c,50^2-49^2+48^2-47^2+...+2^2-1^2\\ =\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...+\left(2+1\right)\left(2-1\right)\\ =99\cdot1+97\cdot1+...+3\cdot1\\ =99+97+...+3\\ \)
\(99+97+...+3\) có số số hạng là \(\dfrac{99-3}{2}+1=49\)(số)
\(\Rightarrow99+97+...+3=\dfrac{\left(99+3\right)\cdot49}{2}=2499\)
\(d,87^2+26\cdot87+13^2\\ =87^2+2\cdot13\cdot87+13^2\\ =\left(87+13\right)^2\\ =100^2\\ =10000\)
\(e,3003^2-3^2\\ =\left(3003+3\right)\left(3003-3\right)\\ =3006\cdot3000\\ =9018000\)
\(f,85\cdot12,7+5\cdot3\cdot12,7\\ =85\cdot12,7+15\cdot12,7\\ =12,7\cdot\left(85+15\right)\\ =12,7\cdot100\\ =1270\)
\(\text{Chúc bạn học tốt}\)
(14,78-a)/(2,87+a)=4/1
14,78+2,87=17,65
Tổng số phần bằng nhau là 4+1=5
Mỗi phần có giá trị bằng 17,65/5=3,53
=>2,87+a=3,53
=>a=0,66.
a) A= 54 . 34- (152-1).(152+1)
=(5.3)4-154-1
=154-154-1
=-1
\(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\left(5^{128}-1\right)=2.5^{128}-2\)
c: Ta có: \(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{64}-1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{128}-1\right)\)
\(=2\cdot5^{128}-2\)
1.
$=153^2+2.47.153+47^2=(153+47)^2=200^2=40000$
2.
$=1,24^2-2.1,24.0,24+0,24^2=(1,24-0,24)^2=1^2=1$
3. Không phù hợp để tính nhanh
4.
$=15^8-(15^8-1)=1$
5.
$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+...+(2019^2-2020^2)$
$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+(2019-2020)(2019+2020)$
$=(-1)(1+2)+(-1)(3+4)+(-1)(5+6)+....+(-1)(2019+2020)$
$=(-1)(1+2+3+4+....+2019+2020)=(-1).2020(2020+1):2=-2041210$
6:
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^8-1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^{2020}-1\right)\left(2^{2020}+1\right)+1\\ =2^{4040}-1+1=2^{4040}\)
2809+1209+4982=9000