Bài 1.

a) ( 3x + 4y )² = ( 3x )² + 2.3x.4y + ( 4y )² = 9x² + 24xy + 16y²

b) ( x² + 1 )² = ( x² )² + 2.x².1 + 1² = x⁴ + 2x² + 1

c) ( 3 - 2y )² = 3² - 2.3.2y + ( 2y )² = 9 - 12y + 4y²

d) ( xy² - 2 )² = ( xy² )² - 2.xy².2 + 2² = x²y⁴ - 4xy² + 4

Bài 2.

a) x² - 9 = x² - 3² = ( x - 3 )( x + 3 )

b) 25 - 4y² = 5² - ( 2y )² = ( 5 - 2y )( 5 + 2y )

c) 9x⁴ - 4y² = ( 3x² )² - ( 2y )² = ( 3x² - 2y )( 3x² + 2y )

d) ( x + 1 )² - y² = ( x - y + 1 )( x + y + 1 )

B1:

a) \(\left(3x+4y\right)^2=\left(3x\right)^2+2.3x.4y+\left(4y\right)^2=9x^2+24xy+16y^2\)

b) \(\left(x^2+1\right)^2=\left(x^2\right)^2+2.x^2.1+1^2=x^4+2x^2+1\)

c) \(\left(3-2y\right)^2=3^2-2.3.2y+\left(2y\right)^2=9-12y+4y^2\)

d) \(\left(xy^2-2\right)^2=\left(xy^2\right)^2-2.xy^2.2+2^2=xy^4-4xy^2+4\)

B2:

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

b) \(25-4y^2=5^2-\left(2y\right)^2=\left(5-2y\right)\left(5+2y\right)\)

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

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

ko bt tự làm đê!!!!
 

27(1-x)(x²+x+1)+81x(x-1)

=(27-27x)(x²+x+1)+81x²-81x

=27x²-27x³+27x-27x²+27-27x+81x²-81x

=-27x³+81x²-81x+27

\(27\left(1-x\right)\left(x^2+x+1\right)+81x\left(x-1\right)\)

\(=-27\left(x-1\right)\left(x^2+x+1\right)+81x\left(x-1\right)\)

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

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

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

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

Có nguyên tố khối à ? Hay bạn viết ngược vậy ?

\(\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{9}{20}\)

Ta có : \(\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{9}{20}\)(Vì a + b = 9 ; ab = 20)

Vậy \(\frac{1}{a}+\frac{1}{b}=\frac{9}{20}\)

\(\left(x^3-x+1\right)\left(x^3+x+1\right)=\left(x^3+1\right)-x^2=x^6+2x^3-x^2+1.\text{Bậc 3 là 2; Bậc 2 là 1}\)

( x³ + x + 1 )( x³ - x + 1 )

= [ ( x³ + 1 ) + x ][ ( x³ + 1 ) - x ]

= ( x³ + 1 )² - x² ( HĐT số 3 )

= x⁶ + 2x³ - x² + 1

Hệ số của lũy thừa bậc 3 : 2

                                      2 : -1

                                      1 : 0 

\(A=x^3+y^3+3xy=\left(x+y\right)^3-3xy\left(x+y\right)+3xy=1+0=1\)

\(B=\left(x-y\right)^3+3xy\left(x-y\right)-3xy=1\)

\(c,M=a^2-ab+b^2+3ab\left(a^2+b^2\right)+6a^2b^2=3ab\left(a^2+2ab+b^2\right)+a^2-ab+b^2\)

\(=3ab+a^2-ab+b^2=\left(a+b\right)^2=1\)

\(x+y=2;x^2+y^2=10\text{ do đó:}xy=-3\text{ nên }\left(x-y\right)^2=16\text{ do đó: }x-y=4\text{ hoặc }x-y=-4\)

\(\text{giải ra được:}x=3;y=-1\text{ hoặc ngược lại nên }x^3+y^3=-26\text{ hoặc }26\)

A = x³ + y³ + 3xy

= x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy² + 3xy

= ( x³ + 3x² + 3xy² + y³ ) - ( 3x²y + 3xy - 3xy )

= ( x + y )³ - 3xy( x + y - 1 )

= 1³ - 3xy( 1 - 1 )

= 1³ - 3xy.0

= 1 - 0 = 1

Vậy A = 1

b) B = x³ - y³ - 3xy

= x³ - 3x²y + 3xy² - y³ + 3x²y - 3xy² - 3xy

= ( x³ - 3x²y + 3xy² - y³ ) + ( 3x²y - 3xy² - 3xy )

= ( x - y )³ + 3xy( x - y - 1 )

= 1³ + 3xy( 1 - 1 )

= 1 + 3xy.0

= 1 + 0 = 1

Vậy B = 1

M = a³ + b³ + 3ab( a² + b² ) + 6a²b²( a + b )

= ( a + b )( a² - ab + b² ) + 3ab[ ( a + b )² - 2ab ] + 6a²b²( a + b )

= ( a + b )[ ( a + b )² - 3ab ] + 3ab[ ( a + b )² - 2ab ] + 6a²b²( a + b )

= 1.( 1 - 3ab ) + 3ab( 1 - 2ab ) + 6a²b².1

= 1 - 3ab + 3ab - 6a²b² + 6a²b²

= 1

Vậy M = 1

d) x + y = 2

⇔ ( x + y )² = 4

⇔ x² + 2xy + y² = 4

⇔ 10 + 2xy = 4 ( gt x² + y² = 10 )

⇔ 2xy = -6

⇔ xy = -3

x³ + y³ = x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy²

            = ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² )

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

            = 2³ - 3.(-3).(2)

            = 8 + 18 = 26