Sách Giáo Khoa

a) (37 - 17).(-5) + 23.(-13 - 17)
 = 20.(-5) + 23.(-30)
 = (-100) + (-690)
 = -790

b) (-57).(67 - 34) - 67.(34 - 57)
 = (-57).33 - 67.(-23)
 = -1881 + 1541
 = -340
   
hoặc:
   (-57).(67 - 34) - 67.(34 - 57)
 = (-57).67 – (-57).34 – 67.34 + 67.57
 = [67.(-57) + 67.57] – [(-57).34 + 67.34]
 = 67(-57 + 57) - 34(-57 + 67)
 = 67.0 - 34.10
 = 0 - 

a) (37 - 17).(-5) + 23.(-13 - 17)

= 20.(-5) + 23.(-30)

= (-100) + (-690) = -790

b) (-57).(67 - 34) - 67.(34 - 57)

= (-57).33 - 67.(-23)

= -1881 + 1541

= -340

hoặc: (-57).(67 - 34) - 67.(34 - 57)

= (-57).67 – (-57).34 – 67.34 + 67.57

= [67.(-57) + 67.57] – [(-57).34 + 67.34]

= 67(-57 + 57) - 34(-57 + 67)

= 67.0 - 34.10

= 0 - 340

= -340

\(35+98=\left(35-2\right)+\left(98+2\right)=33+100=133\)

\(46+29=\left(46+4\right)+\left(29-4\right)=50+25=75\)

35 + 98 = (35 - 2) + (98 + 2) = 33 + 100 = 133.

46 + 29 = (46 - 1) + (29 + 1) = 45 + 30 = 75.

Lời giải:

a. $(-2018)+2018=2018-2018=0$

b) $57+(-93)=-(93-57)=-36$

c) $(-38)+46=46-38=8$

Tính: 

a) 

b) 

c) 

\(=5+5+...+5\)

Tổng trên có \(\left[\left(57-2\right):5+1\right]:2=6\left(\text{số 5}\right)\)

Vậy tổng là \(6\cdot5=30\)

a)

\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

b) S =\(\frac{1}{x}-\frac{1}{x+5}+\frac{1}{x+5}=\frac{1}{x}\)

Dùng hằng đẳng thức đáng nhớ thôi b

Ta có y² - x² = (y - x)(y + x)

Mà theo đêc bài thì mẫu có (y + x) rồi nên chỉ cần nhân cho (y - x) nữa là được

Mình ko hiểu bạn muốn hỏi gì? Câu hỏi mập mờ quá!

a.\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{x+1}{x\left(x+1\right)}\)-\(\dfrac{x}{x\left(x+1\right)}\)=\(\dfrac{x+1-x}{x\left(x+1\right)}\)=\(\dfrac{1}{x\left(x+1\right)}\)

b. Ta có:

\(\dfrac{1}{x\left(x+1\right)}\)= \(\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}\)=\(\dfrac{x+1}{x\left(x+1\right)}\)-\(\dfrac{x}{x\left(x+1\right)}\)=\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)

Ta lại có:

\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)=\(\dfrac{1}{x+1}\)-\(\dfrac{1}{x+2}\);

\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)=\(\dfrac{1}{x+2}\)-\(\dfrac{1}{x+3}\);

\(\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)=\(\dfrac{1}{x+3}\)-\(\dfrac{1}{x+4}\);

\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)=\(\dfrac{1}{x+4}\)-\(\dfrac{1}{x+5}\);

Do đó:

\(\dfrac{1}{x\left(x+1\right)}\)+\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)+\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)+\(\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)+\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)+\(\dfrac{1}{x+5}\) = \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)+\(\dfrac{1}{x+1}\)-\(\dfrac{1}{x+2}\)+\(\dfrac{1}{x+2}\)-...... -\(\dfrac{1}{x+5}\)+\(\dfrac{1}{x+5}\)=\(\dfrac{1}{x}\)

Vậy tổng trên bằng \(\dfrac{1}{x}\)

Ta có : \(D=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)

\(=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)

\(=57+12\sqrt{57}+36-\left(54+12\sqrt{57}+38\right)\)

\(=93-92=1\)

Vậy : \(D=1\)

\(D=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-2\sqrt{6}-\sqrt{38}+6\right)\)

\(=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)

\(=\left(93+12\sqrt{57}\right)-\left(92+12\sqrt{57}\right)\)

\(=1\)