A) \(x^3+27\)

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

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

B) \(x^3-\dfrac{1}{8}\)

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

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

C) \(8x^3+y^3\)

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

\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

D) \(8x^3-27y^3\)

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

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

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

b)\(\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)

c)\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

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

c) \(x^6-3x^4y+3x^2y^2-y^3\)

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

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

d) \(\left(x-y\right)^3+\left(x-y\right)^2+\dfrac{1}{2}\left(x-y\right)+\dfrac{1}{27}\)

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

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

a) \(7x^2-63y^2\)

\(=7\cdot\left(x^2-9y^2\right)\)

\(=7\left[x^2-\left(3y\right)^2\right]\)

\(=7\left(x-3y\right)\left(x+3y\right)\)

b) \(36-4a^2+20ab-25b^2\)

\(=36-\left(4a^2-200ab+25b^2\right)\)

\(=6^2-\left(2a-5b\right)^2\)

\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)

c) \(2x-2y-x^2+2xy-y^2\)

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

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

\(=\left(x-y\right)\left(2-x+y\right)\)

a: =7(x^2-9y^2)

=7(x-3y)(x+3y)

b: =36-(4a^2-20ab+25b^2)

=36-(2a-5b)^2

=(6-2a+5b)(6+2a-5b)

c: =2(x-y)-(x-y)^2

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

\(x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}\)

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

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

=x^3+3*x^2*1/3+3*x*(1/3)^2+(1/3)^3

=(x+1/3)^3

1) \(f\left(x\right)=6x^2-15x+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}-\dfrac{25}{36}\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)+4-\dfrac{25}{6}\)

\(\Rightarrow f\left(x\right)=6\left(x-\dfrac{5}{6}\right)^2-\dfrac{1}{6}\ge-\dfrac{1}{6}\left(6\left(x-\dfrac{5}{6}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{1}{6}\left(tạix=\dfrac{5}{6}\right)\)

2) \(f\left(x\right)=4x^2-13x+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}-\dfrac{169}{64}\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)+5-\dfrac{169}{16}\)

\(\Rightarrow f\left(x\right)=4\left(x-\dfrac{13}{8}\right)^2-\dfrac{89}{16}\ge-\dfrac{89}{16}\left(4\left(x-\dfrac{13}{8}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{89}{16}\left(tạix=\dfrac{13}{8}\right)\)

r) \(100x^2-\left(x^2-25\right)^2\)

\(=\left(10x\right)^2-\left(x^2+25\right)^2\)

\(=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)

\(=\left(-x^2+10x-25\right)\left(x^2+10x+25\right)\)

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

v) \(\left(x+y\right)^2-2\left(x+y\right)+1\)

\(=\left(x+y\right)^2-2\left(x+y\right)\cdot1+1^2\)

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

y) \(12y-36-y^2\)

\(=-y^2+12x-36\)

\(=-\left(y^2-12x+36\right)\)

\(=-\left(y-6\right)^2\)

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

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

=-(x^2-10x+25)(x+5)^2

=-(x-5)^2(x+5)^2

t: =(2x-1)^2-(x+1)^2

=(2x-1-x-1)(2x-1+x+1)

=3x*(x-2)

v: =(x+y)^2-2(x+y)*1+1^2

=(x+y-1)^2

u: =(x-y+5)^2-2(x-y+5)*1+1^2

=(x-y+5-1)^2

=(x-y+4)^2

x: =-(x^2+2xy+y^2)

=-(x+y)^2

y: =-(y^2-12y+36)

=-(y-6)^2

a:

ABCD là hbh

=>AC cắt BD tại trung điểm của mỗi đường

=>O là trung điểm chung của AC và BD

OM=OD/2

ON=OB/2

mà OD=OB

nên OM=ON

=>O là trung điểm của MN

Xét tứ giác AMCN có

O là trung điểm chung của AC và MN

=>AMCN là hbh

b: Xét tứ giác AFCE có

AF//CE

AE//CF
=>AFCE là hbh

=>AF=CE

AF+FB=AB

CE+ED=CD

mà AF=CE và AB=CD

nên FB=ED

\(72^2+144\cdot16+16^2-12^2\)

\(=\left(72^2+144\cdot16+16^2\right)-12^2\)

\(=\left(72^2+2\cdot72\cdot16+16^2\right)-12^2\)

\(=\left(72+16\right)^2-12^2\)

\(=88^2-12^2\)

\(=\left(88+12\right)\left(88-12\right)\)

\(=100\cdot76\)

\(=7600\)

\(48^2-42^2+64-52^2\)

\(=\left(48^2-52^2\right)-\left(42^2-64\right)\)

\(=\left(48^2-52^2\right)-\left(42^2-8^2\right)\)

\(=\left(48-52\right)\left(48+52\right)-\left(42+8\right)\left(42-8\right)\)

\(=-4\cdot100-50\cdot34\)

\(=-8\cdot50-50\cdot34\)

\(=-50\cdot\left(8+34\right)\)

\(=-50\cdot42\)

\(=-2100\)

a: \(\dfrac{5}{6xy^2}=\dfrac{5\cdot3\cdot x}{18x^2y^2}=\dfrac{15x}{18x^2y^2}=\dfrac{15x^2}{18x^3y^2}\)

\(\dfrac{2}{9x^3y}=\dfrac{2\cdot2y}{18x^3y^2}=\dfrac{4y}{18x^3y^2}\)

b: \(\dfrac{5}{14x^2y}=\dfrac{5\cdot y^4\cdot3}{42x^2y^5}=\dfrac{15y^4}{42x^2y^5}\)

\(\dfrac{4}{21xy^5}=\dfrac{4\cdot2\cdot x}{42x^2y^5}=\dfrac{8x}{42x^2y^5}\)

c: \(\dfrac{2x}{\left(x+2\right)^3}=\dfrac{4x^2}{2x\left(x+2\right)^3}\)

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

d: \(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-1}{2x\left(x+3\right)}=\dfrac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)

\(\dfrac{5-3x}{x^2-9}=\dfrac{2x\left(5-3x\right)}{2x\left(x-3\right)\left(x+3\right)}\)

f: \(\dfrac{x-1}{2x+2}=\dfrac{\left(x-1\right)^2}{2\left(x+1\right)\left(x-1\right)}\)

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

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