giải chi tiết giúp mik ạ

a: =11-11=0

c: =23x100=2300

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

a) \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(x-2\right)=1\)

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

\(\Leftrightarrow x^3-27-x\left(x^2-2^2\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

\(\Leftrightarrow4x-27=1\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

b) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow\left[\left(x+1\right)-\left(x-1\right)\right]\left[\left(x+1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x-1\right)^2\right]-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left[\left(x+1\right)^2+\left(x^2-1\right)+\left(x-1\right)^2\right]-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left[\left(x^2+2x+1\right)+\left(x^2-1\right)+\left(x^2-2x+1\right)\right]-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left(3x^2+1\right)-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow6x^2+2-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=-\dfrac{6}{12}=-\dfrac{1}{2}\)

\(a,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\\ \Leftrightarrow x^3-3x^2+3x^2-9x+9x-27+\left(x^2+2x\right)\left(2-x\right)-1=0\\ \Leftrightarrow x^3-3x^2+3x^2-9x+9x-27+2x^2-x^3+4x-2x^2-1=0\\ \Leftrightarrow x^3-x^3-3x^2+3x^2+2x^2-2x^2-9x+9x+4x=1+27\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)

\(b,\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\\ \Leftrightarrow\left[\left(x+1\right)-\left(x-1\right)\right].\left[\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]-6\left(x^2-2x+1\right)=-10\\ \Leftrightarrow2.\left[\left(x^2+2x+1\right)+\left(x^2-1\right)+\left(x^2-2x+1\right)\right]-6x^2+12x-6=-10\\ \Leftrightarrow2.\left(3x^2+1\right)-6x^2+12x-6=-10\\ \Leftrightarrow6x^2-6x^2+12x=-10+6-2\\ \Leftrightarrow12x=-6\\ \Leftrightarrow x=-\dfrac{1}{2}\)

b: 4x^2+4x+1=(2x+1)^2

c: =(6x+1)^2

d: =(3x-4y)^2

e: =(1/2x+2y)^2

f: =-(x^2-10x+25)

=-(x-5)^2

g: =-(16a^4b^6+24a^5b^5+9a^6b^4)

=-a^4b^4(16b^2+24ab+9a^2)

=-a^4b^2(4b+3a)^2

h: =(5x)^2-2*5x*2y+(2y)^2

=(5x-2y)^2

i: =(5x^2)^2-2*5x^2*y+y^2

=(5x^2-y)^2

d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)

\(=\left[\left(3x+1\right)-2\left(x-2\right)\right]\left[\left(3x+1\right)+2\left(x-2\right)\right]\)

\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)

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

c) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)

\(=\left[3\left(2x+3\right)-2\left(x+1\right)\right]\left[3\left(2x+3\right)+2\left(x+1\right)\right]\)

\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)

\(=\left(4x-7\right)\left(8x+11\right)\)

f) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

\(=\left[2bc-\left(b^2+c^2-a^2\right)\right]\left[2bc+\left(b^2+c^2-a^2\right)\right]\)

\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)

g: =(ax+by-ay-bx)(ax+by+ay+bx)

=[a(x-y)-b(x-y)]*[a(x+y)+b(x+y)]

=(x-y)(x+y)(a-b)(a+b)

h: =(a^2+b^2-5-2ab-4)(a^2+b^2-5+2ab+4)

=[(a-b)^2-9][(a+b)^2-1]

=(a-b-3)(a-b+3)(a+b-1)(a+b+1)

i: =(4x^2-3x-18-4x^2-3x)(4x^2-3x-18+4x^2+3x)

=(-6x-18)(8x^2-18)

=-12(x+3)(4x^2-9)

=-12(x+3)(2x-3)(2x+3)

k: =(3x+3y-3)^2-(4x+6y+2)^2

=(3x+3y-3-4x-6y-2)(3x+3y-3+4x+6y+2)

=(-x-3y-5)(7x+9y-1)

i: =25-(2x-3y)^2

=(5-2x+3y)(5+2x-3y)

m: =(x-y)^2-(2m-n)^2

=(x-y-2m+n)(x-y+2m-n)

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

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

\(=a-12\)

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

a) \(x^3+6x^2+12x+8\)

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

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

b) \(x^3-3x^2+3x-1\)

\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)

\(=\left(x-1\right)^3\)

c) \(1-9x+27x^2-27x^3\)

\(=1^3-3\cdot3x\cdot1^2+3\cdot\left(3x\right)^2\cdot1-\left(3x\right)^3\)

\(=\left(1-3x\right)^3\)

d) \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)

\(=x^3\cdot3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3\)

\(=\left(x+\dfrac{1}{2}\right)^3\)

e) \(27x^3-54x^2y+36xy^2-8y^3\)

\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)

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