Sửa đề: 1/1.4 (số hạng thứ 2) ➜ 1/4.7

Giải:

1/1.4+1/4.7+1/7.10+...+1/x.(x+3)=6/19

1/3.(3/1.4+3/4.7+3/7.10+...+3/x.(x+3))=6/19

1/3.(1/1-1/4+1/4-1/7+1/7-1/10+...+1/x-1/x+3)=6/19

1/3.(1/1-1/x+3)                            =6/19

       1/1-1/x+3                             =6/19:1/3

       1/1-1/x+3                             =18/19

              1/x+3                            =1/1-18/19

               1/x+3                           =1/19

⇒x+3=19

      x=19-3

      x=16

Chúc bạn học tốt!

toi k0 biet

cố tình đăng lên rồi tự trả lời o biết

lớp 5 day con biết thế nào là đúng sai

a,\(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

\(\Rightarrow\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}\)

\(\Rightarrow\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}=\dfrac{x+3+63}{63}+\dfrac{x+4+62}{62}\)

\(\Rightarrow\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

\(\Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

\(\Rightarrow x+66=0\) ( vì \(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}>0\) )

\(\Rightarrow x=-66\)

\(b,\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)

\(\Rightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1=\dfrac{x-74}{15}-1+\dfrac{x-73}{16}-1\)

\(\Rightarrow\dfrac{x-12-77}{77}+\dfrac{x-11-78}{78}=\dfrac{x-74-15}{15}+\dfrac{x-73-16}{16}\)

\(\Rightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)

\(\Rightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)

\(\Rightarrow x-89=0\)

\(\Rightarrow x=89\)

b.

\(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\\ \Rightarrow\left(\dfrac{x-12}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)=\left(\dfrac{x-74}{15}-1\right)+\left(\dfrac{x-73}{16}-1\right)\\ \Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}=\dfrac{x-89}{15}+\dfrac{x-89}{16}\\ \Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\\ \\ \Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\\ \Leftrightarrow x-89=0\\ \Leftrightarrow x=89\)

=>\(x\times4+x\times8+x\times2=1,4\)

\(\Rightarrow x\times\left(4+8+2\right)=1,4\)

\(\Rightarrow x\times14=1,4\)

\(\Rightarrow x=1,4\div14\)

\(\Rightarrow x=0,1\)

x : 0,25 + x : 0,125 + x : 0,5 = 1,4 

x : ( 0,25 + 0,125 + 0,5 ) = 1,4 

x : 0,875 = 1,4 

x = 1,4 x 0,875 

x = 1,225

Chúc bn hc tốt <3

X : 0,25 + X : 0,125 + X : 0,5 = 1,4

=> X x 4 + X x 8 + X x 2 = 1,4

=> X x ( 4 + 8 + 2 ) = 1,4

=> X x 14 = 1,4

=> X = 1,4 : 14

=> 0,1

\(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)

Vì GTTĐ luôn lớn hơn hoặc bằng 0 với mọi x 

\(\Rightarrow\left|x+1,1\right|+...+\left|x+1,4\right|\ge0\forall x\)

\(\Rightarrow5x\ge0\forall x\)

\(\Rightarrow x\ge0\)

\(\Leftrightarrow x+1,1+x+1,2+x+1,3+x+1,4=5x\)

\(\Leftrightarrow4x+5=5x\)

\(\Leftrightarrow5x-4x=5\)

\(\Rightarrow x=5\)

Vậy x = 5

vì các giá trị tuyệt đối lớn hơn hoặc bằng 0=>5x  lớn hơn hoặc bằng 0

=>x lớn hơn hoặc bằng 0

=>ta có thể phá dấu giá trj tuyệt đối

ta có x+1.1+x+1.2+x+1.3+x+1.4=5x

5=5x-4x=x

=>x=5

(x+1)+(x+2)+(x+3)+(x+4)=5x

x.(1+2+3+4)=5x

x.5=5x

=>x chỉ có thể là 0 hoặc 1

Nha!!!!!!!!!!!

A = \(\dfrac{15}{1.4}\) + \(\dfrac{15}{4.7}\) + \(\dfrac{15}{7.10}\) + ... + \(\dfrac{15}{61.64}\)

A = \(5.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}...+\dfrac{3}{61.64}\right)\)

A = 5.( \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{61}\) - \(\dfrac{1}{64}\))

A = 5.( \(\dfrac{1}{1}\) - \(\dfrac{1}{64}\))

A = 5. \(\dfrac{63}{64}\)

A = \(\dfrac{315}{64}\)

\(4-\left(7-x\right)=x-\left(13-4\right)\)

\(\Leftrightarrow4-7+x=x-9\)

\(\Leftrightarrow x-3=x-9\)

\(\Leftrightarrow-3=-9\) (vô lí)

\(1.4-\left(7-x\right)=3-\left(13-4\right)\)

\(\Rightarrow4-\left(7-x\right)=x-9\)

\(\Rightarrow4-7+x=x-9\)

\(\Rightarrow\left(-3\right)+x=x-9\)

\(\Rightarrow-3=-9\) ( không có giá trị x, vô lí )

Vậy không có giá trị x thỏa mãn đề bài

5x-2x=6-7x

( 1.1 x 1.2 x 1.3 x 1.4 x 1.5 x 1.6 ) x ( 1.25 - 0.25 x 5 ) 

=( 1,1 x 1,2 x1,3 x1,4 x1,5 x1,6 ) x ( 1,25 - 1,25 )

=( 1.1x1,2 x1,3 x1,4 x1,5 x1,6 ) x 0

=0 

khum bít

:)