a, y x 8 = y + 10

=> 8y=y+10

=> 8y-y=10

=> 7y=10

=> y=10/7

b, y x 10 + 9 = y x 3 + 3

=> 10y+9=3y+3

=>10y+9-3y=3

=> 10y-3y+9=3

=> 10y-3y=3-9

=> 7y=-6

=> y=-6/7

:)

y= 6/7 nhaaaaaaa!

a: =>7/9:x=1/18-2/9=1/18-4/18=-3/18=-1/6

=>x=-7/9:1/6=-7/9*6=-42/9=-14/3

b: =>x*7/5=2/15+2/5=8/15

=>x=8/15:7/5=8/21

c: =>x-1/2=3/14:4/7=3/8

=>x=3/8+4/8=7/8

d: =>0,4x+0,3x-0,2x=0,7

=>0,5x=0,7

=>x=1,4

7/10

MG42E3RSDáidsdPDIOHSA8DR0RYE9FHEUCnd09f7eqg03

ai giải hộ mình mình k liền

Bài 1:

a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)

\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{6}{7}\)

b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)

\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)

\(\Rightarrow x=\dfrac{1}{6}\)

help me

Bài 2:

a) \(\dfrac{3}{10}+\dfrac{2}{5}:\dfrac{4}{15}=\dfrac{3}{10}+\dfrac{3}{2}=\dfrac{9}{5}\)

b) \(\dfrac{4}{7}.3+\dfrac{2}{3}.2=\dfrac{12}{7}+\dfrac{4}{3}=\dfrac{64}{21}\)

 11 x 10 - 10 x 9 + 9 x 8 - 8 x 7 + 7 x 6 - 6 x 5 + 5 x 4 - 4 x 3 + 3 x 2 - 2 x 1 = 60

\(\dfrac{x+8}{12}+\dfrac{x+9}{11}+\dfrac{x+10}{10}+3=0\\ \Leftrightarrow\dfrac{x+8}{12}+1+\dfrac{x+9}{11}+1+\dfrac{x+10}{10}+1=0\\ \Leftrightarrow\dfrac{x+20}{12}+\dfrac{x+20}{11}+\dfrac{x+20}{10}=0\\ \Leftrightarrow\left(x+20\right)\left(\dfrac{1}{12}+\dfrac{1}{11}+\dfrac{1}{10}\right)=0\\ \Leftrightarrow x+20=0\Leftrightarrow x=-20\\ KL:...\)

`<=>((x+8)/12+1)+((x+9)/11+1)+((x+10)/10+1)=0`

`<=>(x+20)/12+(x+20)/11+(x+20)/10=0`

`<=>(x+20)(1/12+1/11+1/10)=0`

Vì `1/12+1/11+1/10 ≠ 0`

`=>x+20=0`

`=>x=0-20`

`=>x=-20`

Mình xin trình bày 2 cách, một là phân tích bình thường, 2 là xài L'Hospital. Bởi c3 ko ai cho xài L'Hospital để hack tự luận cả

C1: Normal

\(\left(2-x\right)+\left(2-x\right)^2+...+\left(2-x\right)^9-9\)

\(=\left[\left(2-x\right)-1\right]+\left[\left(2-x\right)^2-1\right]+...+\left[\left(2-x\right)^9-1\right]\)

\(=\left(2-x-1\right)+\left(2-x-1\right)\left(2-x+1\right)+\left(2-x-1\right)\left[\left(2-x\right)^2+\left(2-x\right)+1\right]+...+\left(2-x-1\right)\left[\left(2-x\right)^8+\left(2-x\right)^7+...+1\right]\)

\(=-\left(x-1\right)\left(1+2-x+1+\left(2-x\right)^2+\left(2-x\right)+1+....+\left(2-x\right)^8+\left(2-x\right)^7+...+1\right)\)

Lai co:

\(x+x^2+...+x^{10}-10=\left(x-1\right)+\left(x^2-1\right)+...+\left(x^{10}-1\right)\)

\(=\left(x-1\right)+\left(x-1\right)\left(x+1\right)+....+\left(x-1\right)\left(x^9+x^8+...+1\right)\)

\(=\left(x-1\right)\left[1+x+1+x^2+x+1+....+x^9+x^8+...+1\right]\)

\(\Rightarrow\lim\limits_{x\rightarrow1}....=\lim\limits_{x\rightarrow1}\dfrac{-[1+2-x+1+\left(2-x\right)^2+\left(2-x\right)+1+...+\left(2-x\right)^8+\left(2-x\right)^7+...+1]}{1+x+1+x^2+x+1+...+x^9+x^8+...+1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{-[9.1+8.\left(2-x\right)+7\left(2-x\right)^2+6\left(2-x\right)^3+5\left(2-x\right)^4+4\left(2-x\right)^5+3\left(2-x\right)^6+2\left(2-x\right)^7+\left(2-x\right)^8]}{10.1+9x^2+8x^3+7x^4+6x^5+5x^6+4x^7+3x^8+2x^9+x^{10}}\)

\(=\dfrac{-[1+2+3+...+9]}{1+2+3+...+10}=\dfrac{-45}{55}\)

C2: L'Hospital

\(=\lim\limits_{x\rightarrow1}\dfrac{-1-2\left(2-x\right)-3\left(2-x\right)^2-...-9\left(2-x\right)^8}{1+2x+3x^2+...+10x^9}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{-1-2-3-...-9}{1+2+3+...+10}=-\dfrac{45}{55}\)

https://www.mathvn.com/2020/07/qui-tac-lhopital-va-ung-dung-trong-tinh.html

Tìm hiểu thêm về quy tắc này nhá

(x+1)+(x+2)+(x+3)+...+(x+9)+(x+10)=56

⇒ x.10+(1+2+3+...+9+10)=56

⇒ x.10+[(10+1).10:2] = 56

⇒ x.10+55=56

⇒ x.10 = 56-55 = 1

⇒ x = 1:10=0,1

(\(x+1\)) + (\(x+2\)) + (\(x\) + 3)+...+ (\(x\) + 10) = 56

xét dãy số \(x+1;x+2;x+3;...;x+10\)

Dãy số trên có khoảng cách là: \(x+2-x-1\) = 1

Dãy số trên có số số hạng là: (\(x+10-x-1\)) : 1 + 1 = 10

Vậy: 

(\(x+1\))+(\(x+2\))+(\(x+3\))+...+(\(x+10\)) = (\(x+10\) + \(x+1\)).10 : 2 

⇒(2\(x\) + 11).5 = 56

   2\(x\) + 11 = 56 : 5

   2\(x\)  + 11 = 11,2 

   2\(x\)          = 11,2 - 11

   2\(x\)         = 0,2

      \(x\)        = 0,2 : 2

      \(x\)        = 0,1 

a/ (x+1)+(x+2)+(x+3)+...+(x+9)+(x+10)

= x+1 + x+2 + x+3 + ... + x+9 + x+10

= (x+x+x +...+ x+x ) + 1+2+3+...+9+10

= 10.x +55

`@` `\text {Ans}`

`\downarrow`

`a)`

`x + 10 = 20`

`=> x = 20 -10`

`=> x = 10`

Vậy, `x = 10`

`b)`

`2 * x + 15 = 35`

`=> 2x = 35 - 15`

`=> 2x = 20`

`=> x = 20 \div 2`

`=> x = 10`

Vậy, `x = 10`

`c)`

`3 * ( x + 2 ) = 15`

`=> x + 2 = 15 \div 3`

`=> x + 2 = 5`

`=> x = 5 - 2`

`=> x = 3`

Vậy, `x = 3`

`d)`

`10 * x + 15 * 11 = 20 * 10`

`=> 10x + 165 = 200`

`=> 10x = 200 - 165`

`=> 10x = 35`

`=> x = 35 \div 10`

`=> x = 3,5`

Vậy,` x = 3,5`

`e)`

`4 * ( x + 2 ) = 3 * 4`

`=> x + 2 = 12 \div 4`

`=> x + 2 = 3`

`=> x = 3 - 2`

`=> x = 1`

Vậy,` x = 1`

`f)`

`33 x + 135 = 26 * 9`

`=> 33x + 135 = 234`

`=> 33x = 234 - 135`

`=> 33x = 99`

`=> x = 99 \div 33`

`=> x = 3`

Vậy, `x = 3`

`g)`

`2 * x + 15 + 16 + 17 = 100`

`=> 2x + 48 = 100`

`=> 2x = 100 - 48`

`=> 2x = 52`

`=> x = 52 \div 2`

`=> x =26`

`h)`

`2 * (x + 9 + 10 + 11) = 4 . 12 . 25`

`=> 2 * (x + 9 + 10 + 11) = 4*25*12`

`=> 2 * (x + 9 + 10 + 11) = 100*12`

`=> x + 9 + 10 + 11 = 100*12 \div 2`

`=> x + 30 = 600`

`=> x = 600 - 30`

`=> x = 570`

Vậy, `x = 570.`

a) \(x+10=20\Leftrightarrow x=10\)

b) \(2x+15=35\Leftrightarrow2x=20\Leftrightarrow x=10\)

c) \(3.\left(x+2\right)=15\Leftrightarrow x+2=5\Leftrightarrow x=3\)

d) \(10x+15.11=20.10\Leftrightarrow10x+165=200\Leftrightarrow10x=35\Leftrightarrow x=\dfrac{35}{10}=\dfrac{7}{2}\)

e) \(4.\left(x+2\right)=3.4\Leftrightarrow x+2=3\Leftrightarrow x=1\)

f) \(35x+135=26.9\Leftrightarrow35x=234-135\Leftrightarrow35x=99\Leftrightarrow x=\dfrac{99}{35}\)

g) \(2x+15+16+17=100\Leftrightarrow2x+48=100\Leftrightarrow2x=52\Leftrightarrow x=26\)

h) \(2.\left(x+9+10+11\right)=4.12.25\)

\(\Leftrightarrow x+30=2.12.25\)

\(\Leftrightarrow x=600-30\)

\(\Leftrightarrow x=570\)