\(x=\dfrac{7}{25}+\dfrac{-1}{5}=\dfrac{7}{25}-\dfrac{1}{5}=\dfrac{2}{25}.\\ x=\dfrac{5}{11}+\dfrac{4}{-9}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{1}{99}.\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\Leftrightarrow\dfrac{5}{9}+x=-\dfrac{1}{3}.\Leftrightarrow x=-\dfrac{8}{9}.\)

\(x=\dfrac{7}{25}+-\dfrac{1}{5}=>\dfrac{7}{25}+-\dfrac{5}{25}=>x=\dfrac{2}{25}\)

\(x=\dfrac{5}{11}+\dfrac{4}{-9}=>\dfrac{-45}{-99}+\dfrac{44}{-99}=>x=\dfrac{-1}{-99}=\dfrac{1}{99}\)

\(\dfrac{5}{9}-\dfrac{x}{-1}=-\dfrac{1}{3}=>-\dfrac{1}{3}-\dfrac{5}{9}=>\dfrac{x}{-1}=-\dfrac{8}{9}=>x=-\dfrac{8}{9}\)

\(x=\dfrac{7}{25}+\dfrac{-1}{5}\\ \Rightarrow x=\dfrac{7}{25}+\dfrac{-5}{25}\\ \Rightarrow x=\dfrac{2}{25}\\ x=\dfrac{5}{11}+\dfrac{4}{-9}\\ \Rightarrow x=\dfrac{-45}{-99}+\dfrac{44}{-99}\\ \Rightarrow x=\dfrac{45}{99}+\dfrac{44}{99}\\ \Rightarrow x=\dfrac{95}{99}\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{5}{9}-\dfrac{-1}{3}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{5}{9}-\dfrac{-3}{9}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{8}{9}\Rightarrow x\cdot9=-1\cdot8=-8\\ \Rightarrow x\cdot9=-8\\ \Rightarrow x=\dfrac{-8}{9}\)

` x : 1 / 2 + x : 1 / 3 + x : 1 / 4 + x = 2010`

`x . 2 + x . 3 + x . 4 + x = 2010`

`x ( 2 + 3 + 4 + 1 ) = 2010`

`x . 10 = 2010`

`x = 2010 : 10`

`x = 201`

Vậy ` x= 201`

\(=>x=2010:\left(2+3+4\right)=2010:9=\dfrac{670}{3}\)

\(\Leftrightarrow xy=63\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;63\right);\left(3;21\right);\left(7;9\right);\left(-63;-1\right);\left(-21;-3\right);\left(-9;-7\right)\right\}\)

\(x-\dfrac{1}{8}=\dfrac{4}{x-2}\) hay \(x-\dfrac{1}{8}=\dfrac{4}{x}-2\) vậy bạn?

\(\dfrac{x-1}{8}=\dfrac{4}{x-2}\)

=>\(\dfrac{\left(x-1\right)\left(x-2\right)}{8.\left(x-2\right)}=\dfrac{4.8}{\left(x-2\right).8}\)

=>khử mẫu

=>(x-1)(x-2)=32

.......

Là \(\dfrac{x+1}{8}=\dfrac{4}{x-2}\) nhoa

`a) 3-5+(-x+3)=6`

`=>5+(-x+3)=3-6`

`=>5+(-x+3)=-3`

`=>-x+3=-3-5`

`=>-x+3=-8`

`=>-x=-8-3`

`=>-x=-11`

`=>x=11`

__

`b)(-4-x)+(4-15)=-15`

`=>(-4-x)+-11=-15`

`=>-4-x=-15-(-11)`

`=>-4-x=-15+11`

`=>-4-x=-4`

`=>x=-4-(-4)`

`=>x=-4+4`

`=>x=0`

`c)(11+x)-(-11-9)=32`

`=>(11+x)-(-20)=32`

`=>(11+x)+20=32`

`=>11+x=32-20`

`=>11+x=12`

`=>x=12-11`

`=>x=1`

`a)3-5+(-x+3)=6`

`5+(-x+3)=3-6`

`5+(-x+3)=-3`

`-x+3=-3-5`

`-x+3=-8`

`-x=-8-3`

`-x=-11`

`x=11`

`b,(-4-x)+(4-15)=-15`

`(-4-x)+(-11)=-15`

`-4-x=-15-(-11)`

`-4-x=-15+11`

`-4-x=-4`

`x=-4-(-4)`

`x=-4+4`

`x=0`

`c)(11+x)-(-11-9)=32`

`(11+x)-(-20)=32`

`(11+x)+20=32`

`11+x=32-20`

`11+x=12`

`x=12-11`

`x=1`

a ) 3 − 5 + ( − x + 3 ) = 6 ⇒ 5 + ( − x + 3 ) = 3 − 6 ⇒ 5 + ( − x + 3 ) = − 3 ⇒ − x + 3 = − 3 − 5 ⇒ − x + 3 = − 8 ⇒ − x = − 8 − 3 ⇒ − x = − 11 ⇒ x = 11 __ b ) ( − 4 − x ) + ( 4 − 15 ) = − 15 ⇒ ( − 4 − x ) ± 11 = − 15 ⇒ − 4 − x = − 15 − ( − 11 ) ⇒ − 4 − x = − 15 + 11 ⇒ − 4 − x = − 4 ⇒ x = − 4 − ( − 4 ) ⇒ x = − 4 + 4 ⇒ x = 0 c ) ( 11 + x ) − ( − 11 − 9 ) = 32 ⇒ ( 11 + x ) − ( − 20 ) = 32 ⇒ ( 11 + x ) + 20 = 32 ⇒ 11 + x = 32 − 20 ⇒ 11 + x = 12 ⇒ x = 12 − 11 ⇒ x = 1

Bài 2: 

a: =>-5x=-200

hay x=40

b: =>x*2/3=1244

hay x=1866

Ta có: 

\(x^4=y^4\)

\(\Rightarrow x^4-y^4=0\)

\(\Rightarrow\left(x^2\right)^2-\left(y^2\right)^2=0\)

\(\Rightarrow\left(x^2-y^2\right)\left(x^2+y^2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-y^2=0\\x^2+y^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-y=0\\x+y=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

_______________

Ta có: 

\(x^5=y^5\)

\(\Rightarrow x^5-y^5=0\)

\(\Rightarrow x-y=0\)

\(\Rightarrow x=y\)

ta có : `x/2 = y/3 = z/4=> (2x)/4 =(3y)/9 = z/4`

`=> (2x)/4 =(3y)/9 = z/4` và `2x + 3y - z = 27`

Áp dụng t/c dãy tỉ số bằng nhau ta có:

`(2x)/4 =(3y)/9 = z/4 =(2x + 3y - z)/(4+9-4)=27/9=3`

`=>x/2=3=>x=3.2=6`

`=>y/3=3=>x=3.3=9`

`=>z/4=3=>z=3.4=12`

\(\dfrac{x^2}{20}=\dfrac{4}{5}\)

\(\Leftrightarrow x^2=16\)

hay \(x\in\left\{4;-4\right\}\)