(X+2)-(x-3)=7x-2(x+1)

2x-1=5x-2

2x-1+1=5x-2+1

2x=5x-1

2x-5x=5x-1-5x

-3x=1

-3x/-3=-1/-3

X=1/3

Vậy x=1/3 

`Answer:`

`(2/3+x)-(-1/2)=3/5`

`<=>2/3+x=3/5-1/2`

`<=>2/3+x=\frac{1}{10}`

`<=>x=\frac{1}{10}-2/3`

`<=>x=\frac{-17}{30}`

`-3/4-(x-7/2)=1/2+(-2/3)`

`<=>-3/4-(x-7/2)=-1/6`

`<=>x-7/2=-3/4-(-1/6)`

`<=>x-7/2=-3/4+1/6`

`<=>x-7/2=-\frac{7}{12}`

`<=>x=\frac{-7}{12}+7/2`

`<=>x=\frac{35}{12}`

giúp mik zứi😭😭😭

a; 60 \(\times\) (\(\dfrac{7}{12}\) + \(\dfrac{4}{15}\)) 

 =  60 \(\times\) \(\dfrac{7}{12}\) + 60 \(\times\) \(\dfrac{4}{15}\)

= 35 + 16

= 51

b; \(\dfrac{1}{2}\) \(\times\)\(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\) \(\times\) \(\dfrac{5}{6}\) \(\times\) \(\dfrac{7}{8}\) \(\times\) \(\dfrac{8}{9}\)

= \(\dfrac{2\times3\times4\times5\times6\times7\times8}{2\times3\times4\times5\times6\times7\times8}\) \(\times\) \(\dfrac{1}{9}\)

= 1 \(\times\) \(\dfrac{1}{9}\)

= \(\dfrac{1}{9}\)

x+y+x=0

=) x+y=-z

(=) (x+y)^3 = (-z)^3

(=) x^3+3x^2y+3xy^2+y = -z^3

(=) x^3+y^3+z^3 = -3x^2y- 3xy^2

= x^3+y^3+z^3= -3xy(x+y)

(=) x^3+y^3+z^3 = -3xy(-z)

=) x^3+y^3+z^3 = 3xyz 

Cần chứng minh :

x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

Có :

x³ + y³ + z³ - 3xyz

= (x + y)³ - 3xy(x + y) + z³ - 3xyz

= (x + y)³ + z³ - 3xy.(x + y + z)

= (x + y + z).[(x + y)² - (x + y).z + z²) - 3xy(x + y + z)

= (x + y + z).[x² + 2xy + y² - zx - yz + z²) - 3xy(x + y + z)

= (x + y + z).(x² + y² + z² + 2xy - 3xy - yz - zx)

= (x + y + z).(x² + y² + z²  xy - yz - zx)   (Điều cần chứng minh)

=> (x + y + z).(x² + y² + z²  xy - yz - zx)  = 0   (vì x + y + z = 0)

=> x³ + y³ + z³ - 3xyz = 0

=> x³ + y³ + z³ = 3xyz 

Bn lm sai rồi

17/12

LÀ 7/12 NHÉ

\(\frac{17}{14}:\frac{x}{y}=1-\frac{1}{7}\)

\(\frac{17}{14}:\frac{x}{y}=\frac{6}{7}\)

\(\frac{x}{y}=\frac{17}{14}:\frac{6}{7}\)

\(\frac{x}{y}=\frac{17}{12}\)

a) \(\left(x-1\right)\left(y+2\right)=5\)

Th1 : \(\hept{\begin{cases}x-1=-5\\y+2=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=-3\end{cases}}}\)

Th2 : \(\hept{\begin{cases}x-1=-1\\y+2=-5\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=-7\end{cases}}}\)

TH3 : \(\hept{\begin{cases}x-1=5\\y+2=1\end{cases}\Rightarrow\hept{\begin{cases}x=6\\y=-1\end{cases}}}\)

TH4 : \(\hept{\begin{cases}x-1=1\\y+2=5\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}}\)

tớ ko hiểu mà còn ý b nữa bn

a) \(\left(x-1\right)\times\left(y+2\right)=5\)

Th1 \(\hept{\begin{cases}x-1=-5\\y+2=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=-3\end{cases}}}\)

Th2 \(\hept{\begin{cases}x-1=-1\\y+2=-5\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=-7\end{cases}}}\)

Th3\(\hept{\begin{cases}x-1=5\\y+2=1\end{cases}\Rightarrow\hept{\begin{cases}x=6\\y=-1\end{cases}}}\)

Th4\(\hept{\begin{cases}x-1=1\\y+2=5\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}}\)

Có (x-30).(y+2)=9

nên (x-30) va y+2 thuộc ước của 9

TH1:x-30=1;y+2=9

x=31;y=7

TH2:x-30=3;y+2=3

x=33;y=1

TH3:x-30=-1;y+2=-9

x=29;y=-11

TH3:x-30=-3;y+2=-3

x=-27;y=-5