=> \(\left(a-2\right)\left(a+1\right)\left(a+3\right)=0\)

=> TH1 : a-2=0 => a=2

TH2: a+1=0 => a=-1

TH3: a+3=0 => a=-3 

Vậy a={-3;-1;2}

Còn cách phân tích là : 

=> \(a^3+a^2+a^2+a-6a-6=0\)

=> \(a^2\left(a+1\right)+a\left(a+1\right)-6\left(a+1\right)=0\)

=> \(\left(a^2+a-6\right)\left(a+1\right)=0\)

=> \(\left(a^2-2a+3a-6\right)\left(a+1\right)=0\)

=> \(\left(a+1\right)\left(a-2\right)\left(a+3\right)=0\)

Hai phân số \(\frac{abab}{cdcd}\) và \(\frac{ababab}{cdcdcd}\) có bằng nhau . 

Vì :

\(\frac{ababab}{cdcdcd}\) \(=\) \(\frac{ababab:10001}{cdcdcd:10001}\) \(=\) \(\frac{ab}{cd}\)

\(\frac{abab}{cdcd}\) \(=\) \(\frac{abab:101}{cdcd:101}\) \(=\) \(\frac{ab}{cd}\)

\(\Rightarrow\) Hai phân số này bằng nhau .

:)

abab/cdcd = abab : 101 / cdcd : 101 = ab / cd

ababab / cdcdcd = ababab : 10101 / cdcdcd : 10101 = ab/cd

Vậy 2 phân số acac/cdcd = ababab/cdcdcd

a gặp nhiều rồi , nhưng dài lắm chú ak

Dài cũng được

a. Ta có :

B = 308/1 + 307/2 +306/3+....+1/308

B = (1+1+....+1) ₊ 307/2 + ....+ 1/308

B = (1 + 307/2) + (1+306/3) + ...+ (1+ 1/308) + 1

B = 309/2 + 309/3 + ....+ 309/308 + 309/309

B = 309.(1/2 + 1/3 + ....+1/309)

Vậy A/B: 1/2 + 1/3 + ... + 1/309 / 308/1 + 307/2 +....+ 2/307+1/308

       A/B = 1/2 + 1/3 +... + 1/309 /  309.(1/2 + 1/3 + ....+1/309)

       A/B = 1/309

b.7/10.11 + 7/11.12 + .... +7 /69.70

= 7. (1/10.11+1/11.12 + ...+ 1/69.70)

= 7.(1/10-1/11+1/11-1/12+....+1/69-1/70)

= 7.(1/10 - 1/70)

= 7. 3/35

= 3/5

    11-(5-x)=8                                                                                                                                                                                                                       

          5-x=11-8

          5-x=3

             x=5-3

             x=2.        

5-x=11-8

5-x=3

   x=5-3

   x=2

giúp mình với

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20 + ( 4 - x ) = -7

         4 - x   = -7 - 20

         4 - x   = -27

              x   = 4 - ( -27 ) 

              x   = 31 

:)

20 + ( 4 - x ) = -7

4 - x = -7 - 20

4 - x = -27

x = 4 - ( - 27 )

x = 31

Vậy x = 31

k mk nha !!>>

Vì \(On\in\widehat{xOy}\)

\(\Leftrightarrow\widehat{xOn}+\widehat{nOy}=\widehat{xOy}\)

\(\Leftrightarrow\widehat{xOn}=\widehat{xOy}-\widehat{nOy}\)

                \(=150-110\)

                \(=40^o\) 

Vì \(\widehat{xOn}< \widehat{xOm}\)

\(\Leftrightarrow\widehat{xOn}+\widehat{nOm}=\widehat{xOm}\)

\(\Leftrightarrow\widehat{nOm}=\widehat{xOm}-\widehat{xOn}\)

                 \(=90-40\)

                 \(=50^o\)