\(g,\left(x^2y-xy+xy^2+y^3\right).3xy^2\\ =\left(3xy^2.x^2y\right)-\left(3xy^2.xy\right)+\left(3xy^2.xy^2\right)+\left(3xy^2.y^3\right)\\ =3x^3y^3-3x^2y^3+3x^2y^4+3xy^5\)

\(h,\dfrac{2}{3}x^2y\left(15x-0,9y+6\right)\\ =\left(\dfrac{2}{3}x^2y.15x\right)-\left(\dfrac{2}{3}x^2y.0,9y\right)+\left(\dfrac{2}{3}x^2y.6\right)\\ =10x^3y-\dfrac{3}{5}x^2y^2+4x^2y\)

\(i,-\dfrac{3}{7}x^4\left(2,1y^2-0,7x+35\right)\\ =\left(-\dfrac{3}{7}x^4.2,1y^2\right)-\left(-\dfrac{3}{7}x^4.0,7x\right)+\left(-\dfrac{3}{7}x^4.35\right)\\ =-\dfrac{9}{10}x^4y^2+\dfrac{3}{10}x^5-15x^4\)

g: =x^2y*3xy^2-xy*3xy^2+xy^2*3xy^2+y^3*3xy^2

=3x^3y^3-3x^2y^3+3x^2y^4+3xy^5

h: =2/3x^2y*15x-2/3x^2y*0,9y+2/3x^2y*6

=10x^3y-0,6x^2y^2+4x^2y

i: =-3/7x^4*2,1y^2+3/7x^4*0,7x-3/7x^4*35

=-0,9x^4y^2+3/10x^5-15x^4

g)(x²y-xy+xy²+y³).3xy²

 =x²y.3xy²  -   xy.3xy²+   xy².3xy²   +   y³.3xy²

 =    3x³y³   -   3x²y³   +    3x²y⁴         +3xy⁵

h)2/3x²y(15x-0,9y+6)

=2/3x2y.15x     -        2/3x²y.0,9y   +     2/3x²y.6

=     10x³y        -         3/5x²y²          +    4x²y

i)-3/7x⁴(2,1y²-0,7x+35)

=   -3/7x⁴.2,1y²        +     3/7x⁴.0,7x      -      3/7x⁴.35

=        -9/10x⁴y²           +         3/10x⁵          -       15x⁴

\(a.x^3+12x^2+48x+64=x^3+3.4x^2+3.4^2x+4^3=\left(x+4\right)^3\)

Thay \(x=6\) vào \(\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

\(b,B=x^3-6x^2+12x-8=\left(x-2\right)^3\)

Thay \(x=22\) vào \(\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

\(c,C=x^3+9x^2+27x+27=x^3+3.3x^2+3.3^2x+3^3=\left(x+3\right)^3\)

Thay \(x=-103\) vào \(\left(x+3\right)^3=\left(-103+3\right)^3=\left(-100\right)^3=-1000000\)

\(d,D=x^3-15x^2+75x-125=x^3-3.5x^2+5^2.3x-5^3=(x-5)^3\)

Thay \(x=25\) vào \(\left(x-5\right)^3=\left(25-5\right)^3=20^3=8000\)

a) \(A=x^3+12x^2+48x+64\)

\(=x^3+3\cdot4\cdot x^2+3\cdot4^2\cdot x+4^3\)

\(=\left(x+4\right)^3\)

Thay \(x=6\) vào biểu thức A ta có:

\(\left(6+4\right)^3=10^3=1000\)

Vậy: ...

b) \(B=x^3-6x^2+12x-8\)

\(=x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-2^3\)

\(=\left(x-2\right)^3\)

Thay \(x=22\) vào biểu thức B ta có:

\(\left(22-2\right)^3=20^3=8000\)

Vậy: ...

c) \(C=x^3+9x^2+27x+27\)

\(=x^3+3\cdot3\cdot x^2+3\cdot3^2\cdot x+3^3\)

\(=\left(x+3\right)^3\)

Thay \(x=-103\) vào biểu thức C ta được:

\(\left(-103+3\right)^3=\left(-100\right)^3=-1000000\)

Vậy: ...

d) \(D=x^3-15x^2+75x-125\)

\(=x^3-3\cdot5\cdot x^2+3\cdot5^2\cdot x-5^3\)

\(=\left(x-5\right)^3\)

Thay \(x=25\) vào biểu thức D ta được:

\(\left(25-5\right)^3=20^3=8000\)

Vậy: ...

a: A=(x+4)^3

Thay x=6 vào A, ta được:

A=(6+4)^3=10^3=1000

b: B=(x-2)^3

Khi x=22 thì B=(22-2)^3=8000

c: C=(x+3)^3

Khi x=-103 thì C=(-103+3)^3=(-100)^3=-1000000

d: D=(x-5)^3

Khi x=25 thì

D=(25-5)^3=20^3=8000

câu 4 hình nhỏ quá 

Câu 5 : A

Câu 4: D - Vì 0% = 30%?

Câu 5: A

4. A (40%+30%+20%+10100%)

5. A

\(\text{≌₰⇴⩸⨙⩸※◡⨦}\)

a: 

b: Thời điểm 11h là nhiều khách nhất, 13h là ít khách nhất

c: Tổng số lượt khách là: 40+50+20+35+45=190(lượt)

Số lượt khách lúc 11h tăng: 50/40-100%=10/40=25%

Số lượt khách lúc 13h giảm: (50-20)/50=30/50=60%

d.

$(\frac{x}{2}-y)^3=(\frac{x}{2})^3-3(\frac{x}{2})^2y+3.\frac{x}{2}y^2-y^3$

$=\frac{x^3}{8}-\frac{3x^2y}{4}+\frac{3xy^2}{2}-y^3$

e.

$(\frac{x}{2}+\frac{y}{3})^3=(\frac{x}{2})^3+3(\frac{x}{2})^2\frac{y}{3}+3.\frac{x}{2}(\frac{y}{3})^2+(\frac{y}{3})^3$

$=\frac{x^3}{8}+\frac{x^2y}{4}+\frac{xy^2}{6}+\frac{y^3}{27}$

f.

$(\frac{2x}{3}-2y)^3=(\frac{2x}{3})^3-3(\frac{2x}{3})^2.2y+3.\frac{2x}{3}(2y)^2-(2y)^3$

$=\frac{8x^3}{27}-\frac{8x^2y}{3}+8xy^2-8y^3$

g.

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

$=2x^3+6xy^2$

Lời giải:

a.
$(3-y)^3=3^3-3.3^2y+3.3y^2-Y63=27-27y+9y^2-y^3$
b.

$(3x+2y^2)^3=(3x)^3+3.(3x)^2(2y^2)+3.3x(2y^2)^2+(2y^2)^3$
$=8y^6+24xy^4+24x^2y^2+8x^3$

c.

$(x-3y^2)^3=x^3-3x^2(3y^2)+3x(3y^2)^2-(3y^2)^3$
$=x^3-9x^2y^2+27xy^4-27y^6$

\(a,-x^3+3x^2-3x+1=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)

\(b,8-12x+6x^2-x^3=2^3-3.2^2x+3.2x^2-x^3=\left(2-x\right)^3\)

\(c,x^3-6x^2y+12xy^2-8y^3=x^3-3.2y.x^2+2.\left(2y\right)^2x-\left(2y\right)^3=\left(x-2y\right)^3\)

\(d,8x^3+12x^2+6x+1\\ =\left(2x\right)^3+3.1\left(2x\right)^2+3.2x.1^2+1^3=\left(2x+1\right)^3\)

\(c,x^3+x^2+\dfrac{x}{3}+\dfrac{1}{27}=x^3+3.x^2.\dfrac{1}{3}+3.x.\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3=\left(x+\dfrac{1}{3}\right)^3\\ f,x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=x^3+3.x^2.\dfrac{1}{2}=3.x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3=\left(x+\dfrac{1}{2}\right)^3\)

a) (x - 2)³ - x(x + 1)(x - 1) + 6x(x + 3)

= x³ - 6x² + 12x - 8 - x(x² - 1) + 6x² + 18x

= x³ + 30x - 8 - x³ + 8

= 30x

b) (x - 2)(x² - 2x + 4)(x + 2)(x² + 2x + 4)

= [(x - 2)(x² + 2x + 4)][(x + 2)(x² - 2x + 4)]

= (x³ - 2³)(x³ + 2³)

= x⁶ - 2⁶

= x⁶ - 64

c) (2x + y)(4x² - 2xy + y²) - (2x - y)(4x² + 2xy + y²)

= [(2x)³ + y³] - [(2x)³ - y³]

= 8x³ + y³ - 8x³ + y³

= 2y³

d) (x + y)³ - (x - y)³ - 2y³

= x³ + 3x²y + 3xy² + y³ - x³ + 3x²y - 3xy² + y³ - 2y³

= 6x²y

e) (x + y + z)² - 2(x + y + z)(x + y) + (x + y)

= x² + y² + z² + 2xy + 2xz + 2yz - 2x² - 2xy - 2xy - 2y² - 2xz - 2yz + x + y

= -x² - y² + z² + x + y

a: =x^3-6x^2+12x-8+6x^2-18x-x(x^2-1)

=x^3-6x-8-x^3+x

=-5x-8

b: =(x-2)(x^2+2x+4)(x+2)(x^2-2x+4)

=(x^3-8)(x^3+8)=x^6-64

c: =8x^3+y^3-8x^3+y^3=2y^3

d: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

e: =(x+y+z)(x+y+z-2x-2y)+(x+y)

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

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

=z^2-x^2-2xy-y^2+x+y

Câu e hình như đề bị thiếu (x + y)² mới đúng ấy em nhé. Hỏi lại giáo viên ra đề