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
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\(1.2.3.....9-1.2.3.....8-1.2.3.....7.8^2\)
\(=1.2.3.....8\left(9-1-8\right)\)
\(=1.2.3.....8\cdot0\)
\(=0\)
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
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 ~
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(\Rightarrow A=3.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{100}\right)=\frac{3.24}{100}\)
\(=\frac{3.4.6}{25.4}\)
\(\Rightarrow A=\frac{18}{25}\)
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
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
\(b)\) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)
\(=\)\(1.2.3...8\left(9-1-8\right)\)
\(=\)\(1.2.3...8\left(9-9\right)\)
\(=\)\(1.2.3...8.0\)
\(=\)\(0\)
A=1+2-3-4+5+6-7-8+...-1999-2000+2001+2002-2003
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