a) (x² + 2)²

= (x²)² + 2.x².2 + 2²

= x⁴ + 4x² + 4

b) (x + y + z)²

= [(x + y) + z]²

= (x + y)² + 2(x + y).z + z²

= x² + 2xy + y² + 2xz + 2yz + z²

= x² + y² + z² + 2xy + 2xz + 2yz

sos

a,

   (\(x^2\) + 2)² 

= (\(x^2\))² + 2.\(x^2\).2 + 2² 

= \(x^4\) + 4\(x^2\) + 4

b, (\(x\) + \(y\) + z)²

=  (\(x\) + y)² + 2(\(x+y\))z + z²

= \(x^2\) + 2\(xy\) + y² + 2\(xz\) + 2yz + z²

xy⁵  -5xy⁴+10xy³-10xy²+5xy-1

a: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

\(=6x^2y+2y^3\)

\(\left(x+y\right)^3-\left(x-y\right)^3\)

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

\(=8y^3+6y\left(x^2-y^2\right)\)

\(=8y^3+6x^2y-6y^3\)

\(=2y^3+6x^2y\)

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

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

\(\left(2y-3\right)^3=8y^3-36y^2+54y-27\)

a: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

\(=6x^2y+2y^3\)

a) (x - 1/2x²y)²

= x² - 2x . 1/2 x²y + (1/2x²y)²

= x² - x³y + 1/4 x⁴y²

b) (2xy² - 1)(1 + 2xy²)

= (2xy²)² - 1²

= 4x²y⁴ - 1

c) (x - y + 2)²

= (x - y)² + 2(x - y).2 + 2²

= x² - 2xy + y² + 4x - 4y + 4

= x² + y² - 2xy + 4x - 4y + 4

d) (x + 1/2)(1/2 - x)

= (1/2)² - x²

= 1/4 - x²

e) (x² - 1/3)²

= (x²)² - 2x².1/3 + (1/3)²

= x⁴ - 2/3 x² + 1/9

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

Câu 7:

a: Ta có: \(A=x^2-2x+7\)

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

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

Dấu '=' xảy ra khi x=1

b: Ta có: \(B=5x^2-20x\)

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

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

Dấu '=' xảy ra khi x=2

a) \({\left( {3x + y} \right)^4} = {\left( {3x} \right)^4} + 4.{\left( {3x} \right)^3}y + 6.{\left( {3x} \right)^2}{y^2} + 4.\left( {3x} \right){y^3} + {y^4}\)

\( = 81{x^4} + 108{x^3}y + 54{x^2}{y^2} + 12x{y^3} + {y^4}\)

b) \(\begin{array}{l}{\left( {x - \sqrt 2 } \right)^5} = \left( {x + (-\sqrt 2) } \right)^5 ={x^5} + 5.{x^4}.\left( { - \sqrt 2 } \right) + 10.{x^3}.{\left( { - \sqrt 2 } \right)^2} + 10.{x^2}.{\left( { - \sqrt 2 } \right)^3} + 5.x.{\left( { - \sqrt 2 } \right)^4} + 1.{\left( { - \sqrt 2 } \right)^5}\\ = {x^5} - 5\sqrt 2 .{x^4} + 20{x^3} - 20\sqrt 2 .{x^2} + 20x - 4\sqrt 2 \end{array}\)

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

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

\(=\left(2x^2\right)^2-y^2\)    Ta sử dụng \(\left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right]\)

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

\(=2^2x^4-y^2\)

\(=4x^4-y^2\)

Vậy khai triển của biểu thức \(2.\left(2x^2+\dfrac{1}{2}y\right)\left(2x^2-y\right)\) là: \(4x^4-y^2\)

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

\(=2.2.\left(x^2+\dfrac{1}{2}\right)\left(x^2-\dfrac{1}{2}\right)\)

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

Đề bài

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

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

= 2^2x^4 - y^2

= 4x^4 - y^2