Tính giá trị của biểu thức
A=1.2.3........100-1.2.3.....99-1.2.3.....99^2
B=\(\frac{\left\{3.4.2^{16}\right\}}{11.2^{13}.4^{11}-4^9.2^{18}}\)
C=1.2+2.3+3.4+...+98.99
D=\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+....+\(\frac{2}{99.101}\)
1) 1.2.3...9 - 1.2.3...8 - 1.2.3...7.82
2) \(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
3)13 - 12 + 11 + 10 - 9 + 8 - 7 - 6 + 5 -4 +3 +2 -1
Câu 1 dễ mà :
1.2.3...9 - 1.2.3...8 - 1.2.3...7.82
= 1.2.3...8.9 - 1.2.3...8.1 - 1.2.3...7.8.8
= 1.2.3...8.( 9 - 1 - 8 )
= 1.2.3...8.0
= 0
còn câu 2 và 3 thì sao
K ghi lại đề câu 2 nha :
\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{35}.2}\)
\(=\frac{3^2.2^{36}}{2^{35}\left(11-2\right)}\)
\(=\frac{3^2.2^{36}}{2^{35}.3^2}\)
\(=\frac{3^2}{3^2}.\frac{2^{36}}{2^{35}}\)
\(=1.2=2\)
\(\text{Thực hiện các phép tính sau một cách hợp lý:}\)
\(a\)) \(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
\(b\)) \(1.2.3...9-1.2.3...8-1.2.3...7.8^2\)
\(c\)) \(\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(d\)) \(1152-\left(374+1152\right)+\left(-65+374\right)\)
\(e\)) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)
thực hiện các phép tính sau một cách hợp lí:
a,\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
b,\(1.2.3...9-1.2.3...8-1.2.3...7.8^2\)
c,\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
d,1152-(374+1152)+(-65+374)
e,13-12+11+10-9+8-7-6+5-4+3-2-1
a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)
=1
b)=(1.2.3....8).(9-1-8)
=(1.2.3....8).0
=0
mik chỉ giải được zậy thôi.
t mik nha.
a, \(\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\\ \)
b,\(1.2.3.....9-1.2.3....8-1.2.3.....7.8^2\)
c, \(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
d, \(1152-\left(374+1152\right)+\left(-65+374\right)\)
a) Đặt \(A=\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)
- Ta có: \(A=\left(100+121+144\right)\div\left(169+196\right)\)
\(\Leftrightarrow A=365\div365=1\)
Vậy \(A=1\)
b) Đặt \(B=1.2.3.....9-1.2.3.....8-1.2.3.....8^2\)
- Ta có: \(B=1.2.3.....8.\left(9-1\right)-1.2.3.....8^2\)
\(\Leftrightarrow B=1.2.3.....8.8-1.2.3.....8.8=0\)
Vậy \(B=0\)
c) Đặt \(C=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
- Ta có: \(C=\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(\Leftrightarrow C=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)
\(\Leftrightarrow C=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(\Leftrightarrow C=\frac{9.2^{36}}{2^{35}.9}\)
\(\Leftrightarrow C=2\)
Vậy \(C=2\)
d) Đặt \(D=1152-\left(374+1152\right)+\left(-65+374\right)\)
- Ta có: \(D=1152-374-1152-65+374\)
\(\Leftrightarrow D=\left(1152-1152\right)+\left(374-374\right)-65\)
\(\Leftrightarrow D=-65\)
Vậy \(D=-65\)
Bài 5) Tính giá trị của biểu thức
a) A=1.2+2.3+3.4+...+9.10
b) B=3.4+4.5+5.6+...+198.199+199.200
c) C=1.2.3+2.3.4+3.4.5+...+8.9.10
d) D=31.32.33+32.33.34+...+58.59.60
e) E=1.3+3.5+5.7+...+95.97+97.99
f) F=51.53+53.55+...+153.155+155.157
g) G=1.3.5+3.5.7+...+15.17.19+17.19.21
h) H=2.4+4.6+6.8+...+96.98+98.100
Bài 5:
a) Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\)
\(\Leftrightarrow3\cdot A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+9\cdot10\cdot\left(11-8\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+8\cdot9\cdot10-8\cdot9\cdot10+9\cdot10\cdot11\)
\(\Leftrightarrow3\cdot A=9\cdot10\cdot11=90\cdot11=990\)
hay A=330
Vậy: A=330
thực hiện các phép tính sau một cách hợp lí:
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
1.2.3.....................9 - 1.2.3......8 - 1.2.3........7.\(^{8^2}\)
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
1152 - (374 + 1152 ) + (-65 + 374)
13 - 12 + 11 + 10 - 9 + 8 - 7 - 6 + 5 - 4 + 3 + 2 - 1
(10*2+11*2+12*2):(13*2+14*2)=100+121+144):(169+196)=1
Tính giá trị của biểu thức:
\(\frac{\left(3.4.2^{16}\right)^{^2}}{11.2^{13}.4^{11}-16^9}\)
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.4^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)
\(=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3^2.2^{36}}{2^{35}.9}=2\)
Cho n!=1.2.3....n,đọc là n giai thừa.Chứng minh rằng:
a.\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{99}{100!}< \)\(1\)
b.\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+....+\frac{99.100-1}{100!}< 2\)
TÌm x: ( 1.2.3 + 2.3.4 + ... + 20.21.22 ) - 5 - 7 - 9 - ... - 55 - 3x = \(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)-\frac{5}{1.3}-\frac{5}{3.5}-...-\frac{5}{53.55}-\) 8.9 - 9.10 - ... - 20.21