e: =>(3x-1-x-5)(3x-1+x+5)=0

=>(2x-6)(4x+4)=0

=>x=3 hoặc x=-1

f: =>(2x-1-x+3)(2x-1+x-3)=0

=>(x+2)(3x-4)=0

=>x=4/3 hoặc x=-2

g: =>4x^2-4x+1-4x^2+1=0

=>-4x+2=0

=>x=1/2

g: =>(x^2+4)(x^2-1)=0

=>x^2-1=0

=>x=1 hoặc x=-1

m: =>(x+3)(2-x)=0

=>x=2 hoặc x=-3

l: =>(x-2)(x^2+2x+4-x+12)=0

=>(x-2)(x^2+x+16)=0

=>x-2=0

=>x=2

k: =>(2x-5)(2x+5-2x-7)=0

=>2x-5=0

=>x=5/2

a: Xét ΔABE vuông tai A và ΔHBE vuông tại H có

BE chung

gócABE=gócHBE

=>ΔABE=ΔHBE

b: ΔBAE=ΔBHE

=>BA=BH; EA=EH

=>BE là trung trực của AH

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

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

\(a,3\left(2a-1\right)+5\left(3-a\right)\)

\(=6a-3+15-5a\)

\(=a-12\)

Thay \(a=\dfrac{-3}{2}\) vào biểu thức trên 

\(a-12\)

\(=\dfrac{-3}{2}-12\)

\(=\dfrac{-27}{2}\)

\(b,25x-4\left(3x-1\right)+7\left(5-2x\right)\)

\(=25x-12x+4+35-14x\)

\(=-1x+39\)

Thay \(x=2,1\) vào biểu thức trên

\(-1x+39\)

\(=-1.2,1+39\)

\(=-2,1+39\)

\(=36,9\)

\(c,4a-2\left(10a-1\right)+8a-2\)

\(=4a-20a+2+8a-2\)

\(=-8a\)

Thay \(a=-0,2\) vào biểu thức trên

\(-8a\)

\(=-8.\left(-0,2\right)\)

\(=1,6\)

\(d,12\left(2-3b\right)+35b-9\left(b+1\right)\)

\(=24-36b+35b-9b-9\)

\(=-10b-15\)

Thay \(b=\dfrac{1}{2}\) vào biểu thức trên 

\(-10b-15\)

\(=-10.\dfrac{1}{2}-15\)

\(=-20\)

a: =6y^3-3y^2-y^2+2y-y+y^2-y^3

=5y^3-3y^2+y

b: =2x^2a-a-2x^2a-a-x^2-ax

=-x^2-ax-2a

c: =2p^3-p^3+1+2p^3+6p^2-3p^5

=3p^3+6p^2-3p^5+1

d: =-3a^3+5a^2+4a^3-4a^2=a^3+a^2