A = -2 +( -2) +........+ (-2)

A = 50 x (-2)

A = -100

Ko bít đúng hay ko

A=(-2). 50 =(-100)

B=(-2).100 =(-200)

\(a,1-2+3-4+5-6+......+199-200\)

\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+.....+\left(199-200\right)\)( 100 cặp )

\(=-1+\left(-1\right)+\left(-1\right)+........+\left(-1\right)\)( 100 số hạng )

\(=-1.100\)

\(=-100\)

\(a.1-2+3-4+5-6+...+199-200\)

\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(199-200\right)\)  (có tất cả \(200:2=100\)cặp)

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)

\(=\left(-1\right).200=-200\)

\(b.1+2-3-4+5+6-7-8+...+97+98-99-100\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)  (có \(100:4=25\)cặp)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(=\left(-4\right).25=-100\)

\(c.1+\left(-6\right)+11+\left(-16\right)+...+21+\left(-26\right)\)

\(=\left[1+\left(-6\right)\right]+\left[11+\left(-16\right)\right]+...+\left[21+\left(-26\right)\right]\) (có tất cả \(26:2=13\)cặp)

\(=\left(-5\right)+\left(-5\right)+...+\left(-5\right)\)

\(=-5.13=-65\)

Câu a tớ làm sai, bạn tham khảo của bạn kia nhé!

S=(1-2)+(3-4)+(5-6)+...+(199-200)

S=(-1)+(-1)+...+(-1)

S=(-1).100=-100

S=1+(2-3)+(-4+5)+...+(98-99)+(-100+101)

S=1+(-1)+1+..+(-1)+1

S=1+25.(-1)+25.1

S=1+(-25)+25

S=1+0

=1

Nhanh nha mk cần gấp đó!

S= 1+(1-3)+3+(1-5)+5+(1-7)+........+199+(1-199)=200

viết có chắc chữ giải mà cũng đúng thật vô lý

giúp với

\(M=\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{8}+...+\frac{197}{198}-\frac{199}{200}\)
\(=\left(1-\frac{1}{2}\right)-\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{6}\right)-\left(1-\frac{1}{8}\right)+...+\left(1-\frac{1}{198}\right)-\left(1-\frac{1}{200}\right)\)=\(=-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}+\frac{1}{8}-...-\frac{1}{198}+\frac{1}{200}\)
\(=-\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=-\frac{1}{2}\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\right]\)
\(=-\frac{1}{2}\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\)
\(=-\frac{1}{2}\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)\)
\(=-\frac{1}{2}.N\)
\(Tacó:\)
\(M:N=-\frac{1}{2}.N:N=-\frac{1}{2}\)