a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

⇔\(\left(x+4\right)\left(x+4\right)=100\)

⇔\(\left(x+4\right)^2=10^2\)

⇔\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

`x-1/9 =8/3`

`=>x=8/3 +1/9`

`=> x= 24/9 +1/9`

`=>x= 25/9`

Vậy `x=25/9`

__

`x-2/20=-5/2-x`

`=>x+x=-5/2 +2/20`

`=> 2x= -50/20 +2/20`

`=> 2x= -48/20`

`=> x= -12/5:2`

`=>x=-12/5 xx1/2`

`=>x= -12/10`

`=>x= -6/5`

Vậy `x=-6/5`

`@` `\text {Ans}`

`\downarrow`

`a)`

Đề là \(\dfrac{x-1}{9}=\dfrac{8}{3}\) phải hongg bạn?

\(\dfrac{x-1}{9}=\dfrac{8}{3}\)

`=>` `(x-1)3 = 8*9`

`=> (x-1)3=72`

`=> x-1=72 \div 3`

`=> x-1=24`

`=> x=25`

`b)`

\(\dfrac{x-2}{20}=\dfrac{-5}{2-x}\)

`=>` `(x-2)(2-x)=20*(-5)`

`=> (x-2)(2-x)=-100`

`=> -[(x-2)(x-2)]=-100`

`=> -(x-2)^2 = -100`

`=> (x-2)^2 = 100`

`=> (x-2)^2 = (+-10)^2`

`=>`\(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

Bài làm

x + y = 4

=> ( x + y )² = 16

=> x² + 2xy + y² = 16

=> 10 + 2xy = 16

=> 2xy = 6

=> xy = 3

Ta có : P = x³ + y³ + 20

= ( x + y )³ - 3xy( x + y ) + 20

= 4³ - 3.3.4 + 20

= 64 - 36 + 20

= 48

Ta có:\(x+y=4\Rightarrow\left(x+y\right)^2=16\)

\(\Rightarrow x^2+2xy+y^2=16\)

\(\Rightarrow2xy+10=16\)

\(\Rightarrow2xy=6\Rightarrow xy=3\)

Ta có:\(P=x^3+y^3+20\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+20\)

\(=4\left(10-3\right)+20=48\)

-x + 20 = - (-15) - (8) + 13 

-x + 20 = 15 - 8 + 13 

-x + 20 = 7 + 13 

- x + 20 = 20

x = 20 - 20 

x = 0

-(-10) + x = -13 + (-9) + (-6) 

10 + x = -13 - 9 - 6 

10 + x = -28 

x = -28 - 10 

x = -38 

8 - (-12) + 10 = -(-14) - x 

8 + 12 + 10 = 14 - x 

30 = 14 - x 

x = 14 - 30 

x = -16

=> 1(x+2) +2(x+2) +3(x+2)+....+ 10(x+2) = 242

=> (1+2+3+...+10)X ( x+2) =242

=> 55X (x+2) =242

=> x+2 = 242 :55

=> x+2=22/5

=> x=12/ 5 

( nhớ click đúng và kết bạn với tớ nha )

(x+2)+(x+2).2+(x+2).3+...+(x+2).10=242

(x+2)(1+2+3+...+10)=242

(x+2).55=242

(x+2)=242:55=4,4

=>x=4,4-2=2,2

(x+2)+(x+2).2+(x+2).3+...+(x+2).10=242

(x+2)(1+2+3+...+10)=242

(x+2).55=242

(x+2)=242:55=4,4

=>x=4,4-2=2,2

to hoc nhu the day , chac chan 100% 

a) x+5=20-(12-7)

    x+5=15

    x    =10

b) 12-3.(x+2)=3

         3.(x+2)=9

            x+2  =3

            x       =1

c) |x-3|-16=-4

    |x-3|      =-4+16

    |x-3|      =12

=>    x-3=12 (1)

hoặc x-3=-12 (2)

Từ (1), => x=15

Từ (2) => x=-9

Vậy x={-9;15}

a, x + 5= 20-(12-7)

   x = -5+ 20- 5 = 10 

x= 10

b, 12-3.(x+2)=3

-3.(x+2)= 3-12

-3x- 6 = -9

3x= 3

x= 1

c, | x-3|-16= -4

|x-3|= 16-4

|x-3|= 12

Trường hợp 1: 

x-3= 12

x= 15

Trường hợp 2:

-(x-3)=12

-x+3= 12

x= -9

Theo t/c dãy tỉ số bằng nhau, ta có:

\(\frac{x}{2}-\frac{y}{3}=\frac{y^2-x^2}{3^2-2^2}=\frac{20}{5}=4\)

\(=>\hept{\begin{cases}\frac{x}{2}=4\\\frac{y}{3}=4\end{cases}}=>\hept{\begin{cases}x=8\\y=12\end{cases}}\)

lộn r lộn r

Ta có: \(\frac{x}{2}=\frac{y}{3}\)

\(=>\frac{x^2}{4}=\frac{y^2}{9}\)

Theo t/c dãy tỉ số bằng nhau: \(\frac{x^2}{4}=\frac{y^2}{9}=>\frac{y^2-x^2}{9-4}=\frac{20}{5}=4\)

Khi đó: \(\hept{\begin{cases}\frac{x^2}{4}=4\\\frac{y^2}{9}=4\end{cases}}=>\hept{\begin{cases}x=4\\y=6\end{cases}}\)

Vậy ...