a: \(4x^2-12x+9=\left(2x-3\right)^2\)

b: \(4x^2+4x+1=\left(2x+1\right)^2\)

c: \(36x^2+12x+1=\left(6x+1\right)^2\)

d: \(9x^2-24xy+16y^2=\left(3x-4y\right)^2\)

a)\(=\left(2x\right)^2-2.2x.3+3^2\)

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

b)\(=\left(2x\right)^2+2.2x.1+1^2\)

\(=\left(2x+1\right)^2\)

a: Ta có: \(\left(4x-6\right)^2-16\left(x-3\right)^2\)

\(=16x^2-48x+36-16x^2+96x-144\)

\(=48x-108\)

\(=48\cdot\left(-7\right)-108\)

\(=-444\)

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

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

\(=2x^3+26\)

\(=-2+26=24\)

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)

c: \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)

d: \(3x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(3x+5\right)\)

e: \(2x^2\left(x+1\right)+4\left(x+1\right)=2\left(x+1\right)\left(x^2+2\right)\)

f: \(-3x-6xy+9xz=-3x\left(1+2y-3z\right)\)

\(a,=\left(6x+1-6x+1\right)^2=4\\ b,=3x^2-6x-5x+5x^2-8x^2-24=-11x-24\\ c,=14x^2+x-3-5x^2-18x+8-9x^2+17x=5\\ d,=6x^2+43x-40-6x^2-7x+3-36x+27=-10\)

a) \(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2=\left(6x+1-6x+1\right)^2=2^2=4\)

b) \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)

c) \(\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x=\left(7x-3\right).2x+\left(7x-3\right)-\left[\left(5x-2\right).x+4\left(5x-2\right)\right]-9x^2+17x=14x^2-6x+7x-3-\left(5x^2-2x+20x-8\right)-9x^2+17x=5x^2+18x-3-\left(5x^2+18x-8\right)=5x^2+18x-3-5x^2-18x+8=5\)

d) \(\left(6x-5\right)\left(x+8\right)-\left(3x-1\right)\left(2x+3\right)-9\left(4x-3\right)=\left(6x-5\right).x+8\left(6x-5\right)-\left[\left(3x-1\right).2x+3\left(3x-1\right)\right]-36x+27=6x^2-5x+48x-40-\left(6x^2-2x+9x-3\right)-36x+27=6x^2+7x-13-\left(6x^2+7x-3\right)=6x^2+7x-13-6x^2-7x+3=-10\)

a: Ta có: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

c: Ta có: \(4-x=2\left(x-4\right)^2\)

\(\Leftrightarrow2\left(x-4\right)^2+x-4=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

Bài 1:

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

f) \(=\left(x^2-4x+4\right)-9=\left(x-2\right)^2-3^2=\left(x-5\right)\left(x+1\right)\)

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

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

m) \(=\left(x^4+4x^2+4\right)-9=\left(x^2+2\right)^2-9=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

Bài 2:

a) \(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)

e) \(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)

f) giống câu a

g) \(=x^2-2xy=x\left(x-2y\right)\)

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

k) \(=\left(x+1\right)^3-27z^3=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

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

Xét tứ giác MNKP có

E là trung điểm của NP

E là trung điểm của MK

Do đó: MNKP là hình bình hành

mà \(\widehat{PMN}=90^0\)

nên MNKP là hình chữ nhật