Câu 1:

(x-18)-42=(23-43)-(70+x)

x-18-42=-20-70-x

x-18-42+20+70+x=0

2x+30=0

2x=-30

x=-15

Câu 2 : Tính tổng

a,1+(-2)+3+(-4)+...+19+(-20)

Từ 1 đến -20 có 20 số hạng 

=> Có 10 nhóm

=>(1-2)+(3-4)+...+(19-20)

=-1-1-1-....-1

=-1.10

=-10

b,c,d,e làm tương tự ta được : 

b) -50

c) -24

d) -99

e) -100

Câu 3 : Tìm x

a)\(x\left(x+7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-7\end{cases}}}\)

Vậy : x={0;-7}

b)\(\left(x+12\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-12\\x=3\end{cases}}}\)

Vậy:....

c)\(\left(-x+5\right)\left(3-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=3\end{cases}}}\)

Vậy:......

d)\(x\left(2+x\right)\left(7-x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\2+x=0\\7-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-2\\x=7\end{cases}}}\)

Vậy:.....

e) \(\left(x-1\right)\left(x+2\right)\left(-x-3\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+2=0\\-x-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=-2\\x=-3\end{cases}}}\)

Vậy:........

Câu 4 : 

a) ab+ac

=a(b+c)

b) ab-ac+ad

=a(b-c+d)

c) ax-bx-cx+dx

=x(a-b-c+d)

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

=(b+c)(a-d)

e) ac-ad+bc-bd

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

=(c-d)(a+b)

f) ax+by+bx+ay

=x(a+b)+y(a+b)

=(a+b)(x+y)

#H

bạn giải chi tiết giúp mik ý d,e câu 2 đc k.Mik k hiểu chỗ đó

\(A=\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{100\cdot104}\)

\(A=\frac{7-4}{4\cdot7}+\frac{11-7}{7\cdot11}+\frac{15-11}{11\cdot15}+...+\frac{104-100}{100\cdot104}\)

\(A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)

\(A=\frac{1}{4}-\frac{1}{104}\)

\(A=\frac{25}{104}\)

\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)

\(B\cdot2=\left(\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\right)\cdot2\)

\(B\cdot2=\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\)

\(B\cdot2=\frac{27-25}{25\cdot27}+\frac{29-27}{27\cdot29}+\frac{31-29}{29\cdot31}+...+\frac{75-73}{73\cdot75}\)

\(B\cdot2=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)

\(B\cdot2=\frac{1}{25}-\frac{1}{75}\)

\(B\cdot2=\frac{2}{75}\)

\(B=\frac{2}{75}\frac{\cdot}{\cdot}2\)

\(B=\frac{1}{75}\)

\(C=\frac{6}{15\cdot18}+\frac{6}{18\cdot21}+\frac{6}{21\cdot24}+...+\frac{6}{87\cdot90}\)

\(\frac{C}{2}=\frac{3}{15\cdot18}+\frac{3}{18\cdot21}+\frac{3}{21\cdot24}+...+\frac{3}{87\cdot90}\)

\(\frac{C}{2}=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\)

\(\frac{C}{2}=\frac{1}{15}-\frac{1}{90}\)

\(\frac{C}{2}=\frac{1}{18}\)

\(C=\frac{1}{18}\cdot2\)

\(C=\frac{1}{9}\)

A) ab.101=ab

B) abc. 7 .11.13=abcabc

k mình nha

a, ab.101=abab

b, abc.7.11.13=abc.1001=abcabc

ab . 101 = abab

abc . 7 . 11 . 13  = abc . 1001 = abcabc

1) (a+b).(a+b)=(a+b)²=a²+2ab+b²

2) (a-b)2=a²-2ab+b²

3) (a+b).(a-b)=a²-b²

4) (a+b)³=a³+3a²b+3ab²+b³

5) (a-b)³=a³-3a²b+3ab²-b³

6) (a+b).(a²-ab+b²)=a³⁺b³

7) (a-b).(a²+ab+b²)=a³-b³

mấy cái ày là hằng đẳng thức đáng nhớ mà

lấy a+a b+b

lấy b^2-a

lấy a.b b.a

a^3 +b

b^3-a

hai câu cuối thì mình k biết

 a / ab.101=ab(100+1) 
= ab.100+ab.1 
=ab00+ab 
=abab 
b/ abc.7.11.13=abc.1001 
=abc(1000+1) 
=abc.1000+abc.1 
=abc000+abc=abcabc

\(a,\overline{ab}\cdot101=\overline{abab}\)

\(b,\overline{abc}\cdot7\cdot11\cdot13=\overline{abc\cdot1001}=\overline{abcabc}\)

thank nha

1) \(\left(a+b\right).\left(a+b\right)=a.\left(a+b\right)+b.\left(a+b\right)=a^2+ab+b^2+ab\)

2) \(\left(a-b\right)^2=\left(a-b\right).\left(a-b\right)=a.\left(a-b\right)-b.\left(a-b\right)=a^2-ab-ab+b^2\)

\(=a^2+\left(-ab\right)+\left(-ab\right)+b^2\)

3) \(\left(a+b\right).\left(a-b\right)=a.\left(a-b\right)+b.\left(a-b\right)=a^2-ab+ab-b^2=a^2-b^2\)

\(=a^2+-\left(b^2\right)\)

4) \(\left(a+b\right)^3=\left(a+b\right).\left(a+b\right).\left(a+b\right)=a.\left(a+b\right).\left(a+b\right)+b.\left(a+b\right).\left(a+b\right)\)

\(=\left[a.\left(a+b\right)\right].\left(a+b\right)+\left[b.\left(a+b\right)\right].\left(a+b\right)=\left(a^2+ab\right).\left(a+b\right)+\left(ab+b^2\right).\left(a+b\right)\)

\(=a^2.\left(a+b\right)+ab.\left(a+b\right)+ab.\left(a+b\right)+b^2.\left(a+b\right)\)

\(=a^3+a^2b+a^2b+ab^2+a^2b+ab^2+b^2a+b^3\)

5) \(\left(a-b\right)^3=\left(a-b\right).\left(a-b\right).\left(a-b\right)=a.\left(a-b\right).\left(a-b\right)-b.\left(a-b\right).\left(a-b\right)\)

\(=\left(a^2-ab\right).\left(a-b\right)-\left(ba-b^2\right).\left(a-b\right)\)

\(=a^2.\left(a-b\right)-ab.\left(a-b\right)-ba.\left(a-b\right)+b^2.\left(a-b\right)\)

\(=a^3-a^2b-a^2b+ab^2-ba^2+b^2a-ba^2+b^2a-b^3\)

6) \(\left(a+b\right).\left(a^2-ab+b^2\right)=a.\left(a^2-ab+b^2\right)+b.\left(a^2-ab+b^2\right)\)

\(=a^3-a^2b+ab^2+ba^2-ab^2+b^3\)

\(=a^3+b^3\)

7) \(\left(a-b\right).\left(a^2+ab+b^2\right)=a.\left(a^2+ab+b^2\right)-b.\left(a^2+ab+b^2\right)\)

\(=a^3+a^2b+ab^2-ba^2-ab^2-b^3\)

\(=a^3-b^3\)

1 a^2+2ab+b^2

2 a^2-2ab+b^2

3 a^2-b^2

4 a^3+3a^2b+3ab^2+b^3

5 a^3-3a^2b+3ab^2-b^3

6 a^3+b^3

7 a^3-b^3

\(1)\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2=a^2+2ab+b^2\)

\(2)\left(a-b\right)^2=a^2-2ab+b^2\)

\(3)\left(a+b\right)\left(a-b\right)=a^2-b^2\)

\(4)\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)

\(5)\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)

\(6)\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\)

\(7)\left(a-b\right)\left(a^2+ab+b^2\right)=a^3-b^3\)

pé pun

a﴿ ab x 101 = ab x ﴾100 + 1﴿

= ab x 100 + ab x 1

= ab00 + ab

= abab

b﴿ abc x 7 x 11 x 13

= abc x 1001

= abc x ﴾1000 + 1﴿

= abc x 1000 + abc x 1

= abc000 + abc

= abcabc

\(\frac{37}{10}+\frac{21}{100}+\frac{249}{10}+\frac{5684}{1000}+\frac{317}{1000}=3,7+0,21+24,9+5,684+0,317=34,811\)

\(\frac{93}{100}=93\%\); \(\frac{72}{10}=\frac{720}{100}=720\%\); \(\frac{853}{1000}=\frac{85,3}{100}=85,3\%\); \(\frac{7}{8}=\frac{56}{1000}=\frac{5,6}{100}=5,6\%\)