a.x^3-1^3

b.x^3-5^3

c)(2x)^3+3^3

d)x^3+1/2^3

1a)  x + 2/5 = 1/2 =5/10-4/10=1/10

b=2/7+2/5=10/35+14/35=24/35

c)Xx3/4=9/10

         x=9/10:3/4

        x36/30=6/5

\(\frac{2}{3}\left(x-1\right)-x-\frac{3}{4}=1\)

<=> \(\frac{2}{3}x-\frac{2}{3}-x-\frac{3}{4}=1\)

<=> \(-\frac{1}{3}x-\frac{17}{12}=1\)

<=> \(-\frac{1}{3}x=\frac{29}{12}\)

<=> \(x=-\frac{29}{4}\)

\(\frac{5}{6}\left(x+2\right)-x-\frac{1}{2}=\frac{1}{3}\)

<=> \(\frac{5}{6}x+\frac{5}{3}-x-\frac{1}{2}=\frac{1}{3}\)

<=> \(-\frac{1}{6}x+\frac{7}{6}=\frac{1}{3}\)

<=> \(-\frac{1}{6}x=-\frac{5}{6}\)

<=> \(x=5\)

học tốt

sai đề bạn ạ

\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)

\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)

\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)

a) 3x-5 ⋮ x+2

+ (x+2) ⋮ (x+2)

⇒ 3(x+2) ⋮ (x+2)

⇒3x+6   ⋮ x+2

mà 3x-5 ⋮ x+2

⇒ 3x-5-(3x+6) ⋮ x+2

⇒ 3x-5-3x-6    ⋮ x+2

⇒ 3x-3x-5-6    ⋮ x+2

⇒-1    ⋮ x+2

⇒ x+2=-1

     x    =-1+2

      x    =1

 vậy x=1

*câu b bnj cho đề bài rõ ràng hơn nhé

nếu đúng thì tích đúng cho mình nha

a) 3x-5 ⋮ x+2

+ (x+2) ⋮ (x+2)

⇒ 3(x+2) ⋮ (x+2)

⇒3x+6   ⋮ x+2

mà 3x-5 ⋮ x+2

⇒ 3x-5-(3x+6) ⋮ x+2

⇒ 3x-5-3x-6    ⋮ x+2

⇒ 3x-3x-5-6    ⋮ x+2

⇒-1    ⋮ x+2

⇒ x+2=-1

     x    =-1+2

      x    =1

 vậy x=1

a) 2/5 < x < 6/5

=> x = 1 ( =5/5 )    (vì x thuộc Z)

Vậy x = 1

b) 3/5 < 3/x < 3/2

=> 5 > x > 2

=> x thuộc { 4 ; 3 }   (vì x thuộc Z)

Vậy ...

c) 3/8 + -11/8 < x < 22/9 + 5/18

=> -8/8 < x < 49/18

=>-1 < x < 2+13/18

=> x thuộc {0; 1; 2}  ( vì x thuộc Z )

Vậy...

c)\(\dfrac{3}{8}\times\dfrac{5}{8}+y=\dfrac{5}{4}\) 

   \(\dfrac{15}{64}+y=\dfrac{5}{4}\) 

           \(y=\dfrac{5}{4}-\dfrac{15}{64}\) 

           \(y=\dfrac{65}{64}\)

d, \(\dfrac{3}{8}+\dfrac{5}{8}\times y=\dfrac{5}{4}\) 

          \(\dfrac{5}{8}\times y=\dfrac{5}{4}-\dfrac{3}{8}\) 

          \(\dfrac{5}{8}\times y=\dfrac{7}{8}\) 

                 \(y=\dfrac{7}{8}:\dfrac{5}{8}\) 

                \(y=\dfrac{7}{5}\)

a) 3/4 x y = 9/10

y = 9/10 : 3/4

y = 6/5

b) 3/5 : y = 7/20

y = 3/5 : 7/20 

y = 12/7

e)\(\dfrac{1}{3}+\dfrac{2}{3}:x=\dfrac{5}{2}\) 

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

       \(\dfrac{2}{3}:x=\dfrac{13}{6}\) 

             \(x=\dfrac{2}{3}:\dfrac{13}{6}\) 

             \(x=\dfrac{4}{13}\)