a)1+2-3-4+5+6-7-8+...-1999-2000+2001+2002-2003
b)1.2.3...9-1.2.3....8-1.2.3....7.82
Làm hộ em với ạ
1.2.3...9 - 1.2.3...8 - 1.2.3...7.82
\(1.2.3.....9-1.2.3.....8-1.2.3.....7.8^2\)
\(=1.2.3.....8\left(9-1-8\right)\)
\(=1.2.3.....8\cdot0\)
\(=0\)
1.2.3.....9−1.2.3.....8−1.2.3.....7.821.2.3.....9−1.2.3.....8−1.2.3.....7.82
=1.2.3.....8(9−1−8)=1.2.3.....8(9−1−8)
=1.2.3.....8⋅0=1.2.3.....8⋅0
=0
1.2.3...9-1.2.3...8-1.2.3...7.82
mn giải giúp em bài toán với ạ !
BÀI 1 :TÍNH NHANH
A=3/4*5 +3/5*6 +3/6*7 +3/7*8 +...+3/99*100BÁI 2 :KHÔNG THỰC HIỆN PHÉP TÍNH , HÃY SO SÁNH TỔNG SAU VỚI 4
1999/2000 +2000/2001 +2001/2002 +2002/2003
Ta có :
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(A=3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(A=3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=3\left(\frac{1}{4}-\frac{1}{100}\right)\)
\(A=3.\frac{6}{25}\)
\(A=\frac{18}{25}\)
Vậy \(A=\frac{18}{25}\)
Chúc bạn học tốt ~
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(\Rightarrow A=3.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{100}\right)=\frac{3.24}{100}\)
\(=\frac{3.4.6}{25.4}\)
\(\Rightarrow A=\frac{18}{25}\)
tính : A=1+2-3-4+5+6-7-8+...-1999-2000+2001+2002-2003
A=(1+2-3)+(-4+5+6-7)+(-8+9+10-11)+......(-2000+2001+2002-2003)
A=0+0....+0
A=0
Ta thấy 2-3-4=-5
6-7-8=-9
.............
1998-1999-2000=-2001
=> 1+2-3-4+5+6-7-8+....-1999-2000+2001-2003=1-5+5-9+9-...-2001+2001+2002-2003
=> A= 1+2002-2003=0
Vậy A=0
\(=\left(1+2-3\right)+\left(-4+5+6-7\right)+...+\left(-2000+2001+2002-2003\right)\)
\(=0+0+0+...+0\)
\(=0\)
học tốt
Bài 1 Tính nhanh
1.2.3...9-1.2.3...8-1.2.3...7.8.8
1152-(374+1152)+(-65+374)
13-12+11+10-9+8-7-6+5-4+3+2-1
Tính nhanh
a) 1.2.3...9 - 1.2.3...8 - 1.2.3...7.82
b) 1-2-3+4+5-6-7+8+9-10-11+12+13
Bài 2
Chứng minh A=2+22+23+24+...+260 chia hết cho 7
Ai biết giúp mình với
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 :
\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{35}.2}\)
\(=\frac{3^2.2^{36}}{2^{35}\left(11-2\right)}\)
\(=\frac{3^2.2^{36}}{2^{35}.3^2}\)
\(=\frac{3^2}{3^2}.\frac{2^{36}}{2^{35}}\)
\(=1.2=2\)
Tính hợp lý:
a. (102+112+122) : (132+142)
b. 1.2.3...9-1.2.3...8-1.2.3...7.82
c. 13-12+11+10-9+8-7-6+5-4+3+2-1
\(b)\) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)
\(=\)\(1.2.3...8\left(9-1-8\right)\)
\(=\)\(1.2.3...8\left(9-9\right)\)
\(=\)\(1.2.3...8.0\)
\(=\)\(0\)
A=1+2-3-4+5+6-7-8+...-1999-2000+2001+2002-2003
Hãy rút gọn biểu thức A
A=1+2-3-4+5+6-7-8+...-1999-2000+2001+2002-2003
A=1+(2-3-4+5)+(6-7-8+9)+...+(1998-1999-2000+2001)+(2002-2003)
A=1+0+0+...+0+(-1)
A=1+(-1)
A=0
Tick cho mk nha
A=(1+2-3)+(-4+5+6-7)+(-8+9+10-11)+...+(-2000+2001+2002-2003)
A=0+0+0+...+0
A=0