Question

[Quỳnh Anh]

giải hộ mình  bải 10 với ạ. Mình cám ơn nhiều ạ.

 

Answer 1

[Toru]

1) $-xyz^2-3xz.yz=-xyz^2-3xyz^2=-4xyz^2$

2) $-8x^2y-x.(xy)=-8x^2y-x^2y=-9x^2y$

3) $4xy^2.x-(-12x^2y^2)=4x^2y^2+12x^2y^2=16x^2y^2$

4) $\frac12 x^2y^3-\frac13 x^2y.y^2=\frac12 x^2y^3-\frac13 x^2y^3=\frac16 x^2y^3$

5) $3xy.(x^2y)-\frac56 x^3y^2=3x^3y^2-\frac56 x^3y^2=\frac{13}{6}x^3y^2$

6) $\frac34 x^4y-\frac16 xy.x^3=\frac34 x^4y-\frac16 x^4y=\frac{7}{12}x^4y$

7) $\frac45y^2x^5-x^3.x^2y^2=\frac45 x^5y^2-x^5y^2=-\frac15 x^5y^2$

8) $-xy^3-\frac27 y^2.xy=-xy^3-\frac27 xy^3==\frac97 xy^3$

9) $\frac56 xy^2z-\frac14 xyz.y=\frac56 xy^2z-\frac14 xy^2z=\frac{7}{12} xy^2z$

10) $15x^4+7x^4-20x^2.x^2$

$=22x^4-20x^4=2x^4$

11) $\frac12 x^5y-\frac34 x^5y+xy.x^4$

$=-\frac14 x^5y+x^5y=\frac34 x^5y$

12) $13x^2y^5-2x^2y^5+x^6$

$=11x^2y^5+x^6$

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 10:

1: \(-xyz^2-3xz\cdot yz=-xyz^2-3xyz^2=-4xyz^2\)

2: \(-8x^2y-x\cdot xy=-8x^2y-x^2y=-9x^2y\)

3: \(4xy^2\cdot x-\left(-12x^2y^2\right)=4x^2y^2+12x^2y^2=16x^2y^2\)

4: \(\dfrac{1}{2}x^2y^3-\dfrac{1}{3}x^2y\cdot y^2=\dfrac{1}{2}x^2y^3-\dfrac{1}{3}x^2y^3=\dfrac{1}{6}x^2y^3\)

5: \(3xy\cdot\left(x^2y\right)-\dfrac{5}{6}x^3y^2=3x^3y^2-\dfrac{5}{6}x^3y^2=\dfrac{13}{6}x^3y^2\)

6: \(\dfrac{3}{4}x^4y-\dfrac{1}{6}xy\cdot x^3=\dfrac{3}{4}x^4y-\dfrac{1}{6}x^4y=x^4y\left(\dfrac{3}{4}-\dfrac{1}{6}\right)=\dfrac{7}{12}x^4y\)

7: \(\dfrac{4}{5}x^5y^2-x^3\cdot x^2y^2=\dfrac{4}{5}x^5y^2-x^5y^2=-\dfrac{1}{5}x^5y^2\)

8: \(-xy^3-\dfrac{2}{7}\cdot y^2\cdot xy=-xy^3-\dfrac{2}{7}xy^3=-\dfrac{9}{7}xy^3\)

9: \(\dfrac{5}{6}xy^2z-\dfrac{1}{4}xyz\cdot y=\dfrac{5}{6}xy^2z-\dfrac{1}{4}xy^2z=xyz^2\left(\dfrac{5}{6}-\dfrac{1}{4}\right)=\dfrac{7}{12}xyz^2\)

10:

\(15x^4+7x^4-20x^2\cdot x^2=22x^4-20x^4=2x^4\)

11:

\(\dfrac{1}{2}x^5y-\dfrac{3}{4}x^5y+xy\cdot x^4\)

\(=\dfrac{1}{2}x^5y-\dfrac{3}{4}x^5y+x^5y\)

\(=x^5y\left(\dfrac{1}{2}-\dfrac{3}{4}+1\right)=\dfrac{3}{4}x^5y\)

12: \(13x^2y^5-2x^2y^5+x^6=x^2y^5\left(13-2\right)+x^6=x^6+11x^2y^5\)