a)\(1-2+3-4+5-6+7-8+8-9+9-10\)

=\(\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+\left(8-9\right)+\left(9-10\right)\)

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

\(=\left(-1\right).6\)

\(=-6\)

b)\(1-2+3-4+...+99-100\)

\(=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\)}\(\left[\left(100-1\right):1+1\right]:2=50\)(cặp)

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)} 50 số (-1)

\(=\left(-1\right).50\)

\(=-50\)

c)\(1-3+5-7+9-11+13-15\)

\(=\left(1-3\right)+\left(5-7\right)+\left(9-11\right)+\left(13-15\right)\)

\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+\left(-2\right)\)

\(=\left(-2\right).4\)

\(=-8\)

d)\(1-3+5-7+...-99+101\) (Đối với bài này, có vẻ đề sai, mình đã sửa lại rồi

\(=\left(1-3\right)+\left(5-7\right)+...+\left(97-99\right)+101\) } \(\left[\left(99-1\right):2+1\right]:2=25\)(cặp)

\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+...+\left(-2\right)\) } 25 số (-2)

\(=\left(-2\right).25\)

\(=-50\)

e)\(-1-2-3-4-...-99-100\)

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

\(=\left[\left(-1\right)+\left(-100\right)\right]+\left[\left(-2\right)+\left(-99\right)\right]+...+\left[\left(-51\right)+\left(-50\right)\right]\) } \(\left[\left(100-1\right):1+1\right]:2=50\)(cặp) (phần này của đề bài, không thay được như (-100) hoặc (-1))

\(=\left(-100\right)+\left(-100\right)+\left(-100\right)+...+\left(-100\right)\)} 50 số (-100)

\(=\left(-100\right).50\)

\(=-5000\)

a, -5

b, -50

c, -8

d, -50

e, -5050

1.

a, => 21-x+3 < 0

=> 24-x < 0

=> x < 24

b, => 7+x > 0

=> x > -7

c, => x-1 < 0 ; x+2 > 0 ( vì x-1 < x+2 )

=> x < 1 ; x > -2

=> -2 < x < 1

Tk mk nha

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2003\cdot2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+....\left(\frac{1}{2003}-\frac{1}{2003}\right)-\frac{1}{2004}\)

\(=1-0+0+0+....+0-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

x= -4, -3, -2, -1, 0

a, (a-b) + (c+d)

= a-b + c+d

= (a+c) - (b-d)

=> (a-b) + (c+d) = (a+c) - (b-d)

b, (a-b) - (a-d) 

= a-b - a + d

= (a+d) - (b-d)

=> (a-b) - (a-d) = (a+d) - (b-d)

\(a)\) \(\left(a-b\right)+\left(c+d\right)\)

\(=\)\(a-b+c+d\)

\(=\)\(\left(a+c\right)+\left(-b+d\right)\)

\(=\)\(\left(a+c\right)-\left(b-d\right)\)

Vậy ...

\(b)\) \(\left(a-b\right)-\left(c-d\right)\)

\(=\)\(a-b-c+d\)

\(=\)\(\left(a+d\right)+\left(-b-c\right)\)

\(=\)\(\left(a+d\right)-\left(b+c\right)\)

Vậy ... 

\(\hept{\begin{cases}\left|x-2\right|\ge0\\\left|x+5\right|\ge0\\3>0\end{cases}}\Rightarrow\hept{\begin{cases}\left|x-2\right|=x-2\\\left|x+5\right|=x+5\end{cases}}\Rightarrow x-2+x+5=3\)

\(\Rightarrow\left(x+x\right)+\left(5-2\right)=3\)

\(\Rightarrow2x+3=3\)

\(\Rightarrow2x=3-3=0\)

\(\Rightarrow x=0:2=0\)

2 câu còn lại làm tương tự

Trình bày lời giải chi tiết nha !

Vì có 1002 đường thẳng đi qua điểm A nên có 1002.(1001-1)=1003002(góc). 

k cho mình nha. 

a)\(\left|2x-3\right|-\left|3x+2\right|=0\)

\(\Rightarrow\left|2x-3\right|=\left|3x+2\right|\)

\(\Rightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x+2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x-3x=3+2\\2x+3x=3+2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=5\\5x=5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-5\\x=1\end{cases}}\)

b)\(\left|2+3x\right|=\left|4x-3\right|\)

\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-4x-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x-4x=-2-3\\3x+4x=-2-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=-5\\7x=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-\frac{5}{7}\end{cases}}\)