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Câu 1 :
A=1+2+3+..+100
=> số số hạng của A là : (100-1):1+1=100(số)
Giá trị của A là : ( 100+1)100:2= 5050
Câu 2 :
B=1.2+2.3+...+99.100
=> 3B = 3(1.2+2.3+...+99.100)
=> 3B = 1.2.3+2.3.3+...+99.100.3
=> 3B = 1.2.(3-0)+2.3.(4-1)+...+99.100.(101-98)
=> 3B = 1.2.3-0.1.2+2.3.4-1.2.3+....+99.100.101-98.99.100
=> 3B = 99.100.101
=> 3B = 999900
=> B = 999900:3=333300
Câu 3 :
C = 1 + 22 + 23 + ... + 299 + 2100
=>2C= 2+ 23 + 24+ ... + 2100 + 2101
=> 2C-C = ( 2+ 23 + 24+ ... + 2100 + 2101 ) - ( 1 + 22 + 23 + ... + 299 + 2100)
=> C = 2101- 1
c ) S = 1.2 + 2.3 + 3.4 + .... + 99.100
=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3
=> 3S = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 99.100.( 101 - 98 )
=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 99.100.101 - 98.99.100
=> 3S = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 98.99.100 - 98.99.100 ) + 99.100.101
=> 3S = 99.100.101 => S = \(\frac{99.100.101}{3}\)
d ) Ta có \(\frac{1}{2^2}<\frac{1}{2.1}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
..........
\(\frac{1}{100^2}<\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{100}=\frac{99}{100}<1\)
1/
a. \(x^3-2=25\)
\(x^3=25+2\)
\(x^3=27\)
\(\Rightarrow x=3\)
b.\(\left(x-3\right)^2=25\)
\(\left(x-3\right)^2=5^2\)
\(\Rightarrow x-3=5\)
\(\Rightarrow x=8\)
1,a, x^3-2=25 b, (x-3)^2=25 c, x^3-x^2=55 d,[(8.x-12):4].3^7=3^10
x^3=27 (x-3)^2=5^2 không có giá trị x (8.x-12):4=3^3
x^3=3^3 x-3=5 8.x-12=108
x=3 x=8 8.x=120
x=15
2, a, \(7^6:7^4+3^4.3^2-3^7:3\) b, 1736-(21-16).32+6.7^2 c,56.17+17.44-4^3.5+6.(3^2-2)
=\(7^2+3^6-3^6\) =1736-5.32+6.49 =17.(56+44)-320+42
=\(49\) =1736-160+294 =17.10-278
=1736+134 =170-278
=1870 =-108
d, 3.10^2-[1200-(4^2-2.3)^3]
=300-[1200-(16-6)^3]
=300-(1200-10^3)
=300-(1200-1000)
=300-200
=100
a. Áp dụng CT: n.9n+1)/2
=>S=(101.100)/2
b. SSH=(998-2) : 2+1
TBC=(998+2):2
Nhân SSH với TBC => S
c.
Đặt A= 1.2 + 2.3 + 3.4 + ...+ 99.100
3A = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3A= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3A= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3A = 99.100.101 3S = 3.33.100.101
A=33.100.101= 333300
d.
Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101
4A=98.99.100.101
=>A=98.99.100.101/4
a. S= 1+2+3+4+.....+98+99+100
S= (100 -1) : 1 + 1 =100
b. S= 2+4+6+8+.....+996+998
S= (998 - 2 ) : 2 + 1 = 499
c. S= 1.2+2.3+3.4+.....+98.99+99.100
Bài này hôm qua đã làm -.- vào thống kê của tôi mà nhìn :)
d. S= 1.2.3+2.3.4+3.4.5+......+97.98.99+98.99.100
S = (1.2.3.2.3.4.5.4.5.6+98.99.100)4
S=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+97.98.99+98.99.100
S=101 - 97
S=1.2.3.5.2.4.+2.1.2.3.4.3.4.5.5.6-2.4.5.4.5.6.7-3.4.5.6-3.4.5.6+.......100
S=1.2.3.3.4.5.5.6.7.7.8.9......+97.98.99+98.99.100
S=1.2.3.4.4.3.2.1+2.3.5-2.3.4.5+3.4.5.6.6.7.3.4.5.6+........97.98.99+98.99.100
S= 98.99.100.101
S=98.99.100.\(\frac{101}{4}\)
e. S= 12+22+32+.....982+992+1002
S= 1002 - 992 + 982 -972 +...+ 22- 12
S= (100 - 99) (100+99) (98 - 97) (98+97) +....+(2-1) (2+1)
S=(1+100) 100 :2
s=5050
+) \(A=3\left(x-4\right)^4-4\ge-4\)
Min A = -4 \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
+) \(B=5+2\left(x-2019\right)^{2020}\ge5\)
Min B = 5 \(\Leftrightarrow x-2019=0\Leftrightarrow x=2019\)
+) \(C=5+2018\left(2020-x\right)^2\)
Min C = 5 \(\Leftrightarrow2020-x=0\Leftrightarrow x=2020\)
+) \(D=\left(x-1\right)^{2020}+\left(y+x\right)-1\ge-1\)
Min D = -1 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-x\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)
+) \(E=2\left(x-1\right)^2+3\left(2x-y\right)^4-2\ge-2\)
Min E = -2 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\2x-y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\2x=y\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)
\(x=1.2+2.3+3.4+...+99.100\)
\(x=1.\left(1+1\right)+2.\left(2+1\right)+3.\left(3+1\right)+...+99.\left(99+1\right)\)
\(x=1^2+1.1+2^2+2.1+3^2+3.1+...+99^2+99.1\)
\(x=y+\left(1+2+3+...+99\right)\)
\(x=y+\frac{99.\left(99+1\right)}{2}=y+4950\)
\(x-y=4950\)