Bài 1: 

a: Tổng là:

(-19+19)+(-18+18)+...+20=20

b: Tổng là:

-18+(-17+17)+...+0=-18

1, có từ 1đến 100 có 100 số hạng .Chia thành 50 nhóm .Mỗi nhóm co 2 số hạng

Suy ra A= [1+(-2)]+[3+(-4)]+......+[99+(-100)]

A= (-1)+(-1)+.... +(-1)

A= (-1).50=(-50)

2,A=(1-2)+(3-4)+.....+(2015-2016)

A=(-1)+(-1)+....+(-1)

A có 2016 số hạng .Chia thành 1008 nhóm .Mỗi nhóm co 2 số hạng và có tổng =(-1)

A=(-1).1008=(-1008)

\(A=\left(1+3+...+99\right)-\left(2+4+...+100\right)\)

\(A=\left(\left(1+99\right)\cdot\frac{50}{2}\right)-\left(\left(2+100\right)\cdot\frac{50}{2}\right)\)

\(A=2500-2550=-50\)

Đúng ko ta lâu rồi ko làm.

\(A=\left(\left(1+99\right)\cdot\frac{50}{2}\right)-\left(\left(2+100\right)\cdot\frac{50}{2}\right)\)

\(A=1\cdot2^2+2\cdot3^2+3\cdot4^2+\cdots+98\cdot99^2\)

\(=2^2\left(2-1\right)+3^2\left(3-1\right)+4^2\left(4-1\right)+\cdots+99^2\left(99-1\right)\)

\(=\left(2^3+3^3+\cdots+99^3\right)-\left(2^2+3^2+\cdots+99^2\right)\)

\(=\left(1^3+2^3+\cdots+99^3\right)-\left(1^2+2^2+3^2+\cdots+99^2\right)\)

\(=\left(1+2+3+\cdots+99\right)^2-\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}\)

\(=\left(99\cdot\frac{100}{2}\right)^2-\frac{99\cdot100\cdot199}{6}\)

\(=\left(99\cdot50\right)^2-33\cdot50\cdot199=\left(3\cdot33\cdot50\right)^2-33\cdot50\cdot199\)

\(=9\cdot33^2\cdot50^2-33\cdot50\cdot199=33\cdot50\left(9\cdot33\cdot50-199\right)\)

\(=33\cdot50\cdot14651=24174150\)

\(Câu\text{ }4:\\ Ta\text{ }có:\text{(x^2 – 3x + 2) + (4x^3– x^2+ x – 1)}\\ =x^2-3x+2+4x^3-x^2+x-1\\ =\text{4x}^3+\left(x^2-x^2\right)+\left(-3x+x\right)+\left(2-1\right)\\ =4x^3-2x+1\)

\(Câu\text{ }5:Đặt\text{ }tính\text{ }trừ\text{ }như\text{ }sau:\)

Câu 6:

A+B

\(=6x^4-4x^3+x-\frac13+\left(-3x^4-2x^3-5x^2+x+\frac23\right)\)

\(=6x^4-4x^3+x-\frac13-3x^4-2x^3-5x^2+x+\frac23\)

\(=3x^4-6x^3-5x^2+2x+\frac13\)

A-B

\(=6x^4-4x^3+x-\frac13-\left(-3x^4-2x^3-5x^2+x+\frac23\right)\)

\(=6x^4-4x^3+x-\frac13+3x^4+2x^3+5x^2-x-\frac23\)

\(=9x^4-2x^3+5x^2-1\)

Câu 4:

\(\left(x^2-3x+2\right)+\left(4x^3-x^2+x-1\right)\)

\(=4x^3+\left(x^2-x^2\right)+\left(-3x+x\right)+\left(2-1\right)\)

\(=4x^3-2x+1\)

Câu 3:

\(A+B=2x^5+5x^3-2\)

=>\(B+x^4-3x^2-2x+1=2x^5+5x^3-2\)

=>\(B=2x^5+5x^3-2-x^4+3x^2+2x-1=2x^5-x^4+5x^3+3x^2+2x-3\)

\(A-C=x^3\)

=>\(x^4-3x^2-2x+1-C=x^3\)

=>\(C=x^4-3x^2-2x+1-x^3\)

Câu 2:

2 câu c,d làm tương tự  

5:

a: \(3^{2n}=\left(3^2\right)^n=9^n\)

\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)

=>\(3^{2n}>2^{3n}\)

b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)

\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)

mà \(1568239201< 8036054027\)

nên \(199^{20}< 2003^{15}\)

4: \(100< 5^{2x-1}< 5^6\)

mà \(25< 100< 125\)

nên \(125< 5^{2x-1}< 5^6\)

=>3<2x-1<6

=>4<2x<7

=>2<x<7/2

mà x nguyên

nên x=3

2^2022 mà bạn chuche ơi

\(\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{100}\left(1+2+...+100\right)\)

\(=\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+...+100}{100}\)

\(=\frac{\left(1+2\right).2:2}{2}+\frac{\left(1+2+3\right).3:2}{3}+...+\frac{\left(1+2+...+100\right).100:2}{100}\)

\(=\left(1+2\right):2+\left(1+2+3\right):2+....\left(1+2+...+100\right):2\)

\(=\left[\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+...+100\right)\right]:2\)

\(=\left(100.1+99.2+....+1.100\right):2=171700:2=85850\)

Nếu không hiểu cái trong ngoặc tính sao thì báo tớ ;) 

Ta có: \(\frac{3}{1\times2}+\frac{3}{2\times3}+\cdots+\frac{3}{2016\times2017}\)

\(=3\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{2016\times2017}\right)\)

\(=3\times\left(1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(=3\times\left(1-\frac{1}{2017}\right)=3\times\frac{2016}{2017}=\frac{6048}{2017}\)