a, \(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
b, \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
c, \(x^2-10x+25=x^2-2.5x+5^2=\left(x-5\right)^2\)

1. \(25+10x+x^2\\ \Leftrightarrow5^2+2\cdot5\cdot x+x^2\\ \Leftrightarrow\left(5+x\right)^2\\ \Leftrightarrow\left(5+x\right)\left(5+x\right)\)

2. \(8x^3-\dfrac{1}{8}\\ \Leftrightarrow\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[4x^2+x+\dfrac{1}{4}\right]\)

3. \(x^2-10x+25\\ \Leftrightarrow x^2-2\cdot5\cdot x+5^2\\ \Leftrightarrow\left(x-5\right)^2\\ \Leftrightarrow\left(x-5\right)\left(x-5\right)\)

Câu 7:

a: \(A=x^2-2x+7\)

\(=x^2-2x+1+6\)

\(=\left(x-1\right)^2+6\ge6\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(B=5x^2-20x\)

\(=5\left(x^2-4x\right)\)

\(=5\left(x^2-4x+4-4\right)=5\left(x-2\right)^2-20\ge-20\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

Câu 4:

a: \(A=\left(x-y\right)^2+\left(x+y\right)^2\)

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

\(=2x^2+2y^2\)

b: \(B=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+5-8x^2+24x-18=-4x^2+20x-13\)

Câu 2:

a: \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)

b: \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=\left(x+5\right)^2\)

c: \(\frac{x^2}{4}-x+1=\left(\frac12x\right)^2-2\cdot\frac12x\cdot1+1^2=\left(\frac12x-1\right)^2\)

d: \(9\left(x+1\right)^2-6\left(x+1\right)+1\)

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

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

e: \(\left(x-2y\right)^2-8\left(x^2-2xy\right)+16x^2\)

\(=\left(x-2y\right)^2-2\cdot\left(x-2y\right)\cdot4x+\left(4x\right)^2\)

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

a. (x + y)² = x² + 2xy + y²

b. (x - 2y)² = x² - 4xy - 4x²

c. (xy² + 1)(xy² - 1) = x²y⁴ - 1

d. (x + y)²(x - y)² = (x² + 2xy + y²)(x² - 2xy + y²) = x⁴ - (2xy + y²)² = x⁴ - (4x²y² + y⁴) = x⁴ - 4x²y² - y⁴

^{Chucs hocj toots}

Câu 2: 

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(x^2+10x+25=\left(x+5\right)^2\)

d: \(9\left(x+1\right)^2-6\left(x+1\right)+1=\left(3x+2\right)^2\)

e: \(\left(x-2y\right)^2-8\left(x-2xy\right)+16x^2=\left(x-2y+4x\right)^2=\left(5x-2y\right)^2\)

a=(x+3)

b=(x+1/2)

c=(xy^2+1)

Good luck!

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

\(b,\left(x+\frac{1}{2}\right)^2\)

\(c,\left(xy^2+1\right)^2\)

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

1/2=50/100

2/5=40/100

3/4=75/100

6/25=24/100

1/2=0,5

2/5=0,4

3/4=0,75

6/25=0,24

Bài 1:

|x|<=4

=>-4<=x<=4

=>A=[-4;4]

-5<x-1<=8

=>-5+1<x<=8+1

=>-4<x<=9

=>B=(-4;9]

A\(\cup\) B=[-4;4]\(\cup\) (-4;9]

=[-4;9]

R\(A\(\cup\) B)

=R\[-4;9]

=(-∞;-4)\(\cup\) (9;+∞)

A\(\cap\) B=[-4;4]\(\cap\) (-4;9]

=(-4;4]

A\B=[-4;4]\(-4;9]

={-4}

B\A=(-4;9]\[-4;4]

=(4;9]

Bài 2:

\(x^2\le4\)

=>-2<=x<=2

=>A=[-2;2]

-2<x-1<3

=>-2+1<x<3+1

=>-1<x<4

=>B=(-1;4)

A\(\cup\) B=[-2;2]\(\cup\) (-1;4)

=[-2;4)

R\(A\(\cup\) B)

=R\[-2;4)

=(-∞;-2)\(\cup\) [4;+∞)

A\(\cap\) B=[-2;2]\(\cap\) (-1;4)

=(-1;2]

A\B=[-2;2]\(-1;4)

=[-2;-1)

B\A=(-1;4)\[-2;2]

=(2;4)

( 1/2 + x )² = ( 1/2 )² + 2.1/2.x + x² = x² + x + 1/4

( 2x + 1 )² = ( 2x )² + 2.2x.1 + 1² = 4x² + 4x + 1

lỗi

:v