\(a,x^2+20x+100=\left(x+10\right)^2\)

\(b,16x^2+24xy+9y^2=\left(4x+3y\right)^2\)

\(c,\left(2a+3b\right)\left(4a^2-6ab+9b^2\right)=8a^3+27a^3\)

\(d,\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)=125x^3-64y^3\)

a, \(\left(8a^3-27b^3\right)-2a\left(4a^2-9b^2\right)\)

\(=\left(2a-3b\right)\left[\left(2a\right)^2+2a.3b+\left(3b\right)^2\right]-2a\left(2a-3b\right)\left(2a+3b\right)\)

\(=\left(2a-3b\right)\left[4a^2+6ab+9b^2-2a\left(2a+3b\right)\right]\)

\(=\left(2a-3b\right)\left(4a^2+6ab+9b^2-4a^2-6ab\right)\)

\(=\left(2a-3b\right).9b^2\)

b, \(\left(x^3-y^3\right)+\left(x-y\right)^2\)

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

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

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

c, \(\left(m^3+n^3\right)+\left(m+n\right)^2\)

\(=\left(m+n\right)\left(m^2-mn+n^2\right)+\left(m+n\right)^2\)

\(=\left(m+n\right)\left(m^2-mn+n^2+m+n\right)\)

Chúc bạn học tốt!!!

a) 5ay - 3bx + ax - 15by

= (5ay + ax) - (3bx + 15by)

= a (5y + x) - 3b (x + 5y)

= (5y + x) (a - 3b)

b) x^3 + x^2 - x - 1

= (x^3 + x^2) - (x + 1)

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

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

c) (2a + b)^2 - (2b + a)^2

= 4a^2 + 4ab + b^2 - 4b^2 - 4ab - a^2

= 3a^2 - 3b^2

= 3 (a^2 - b^2)

d) (8a^3 - 27b^3) - 2a (4a^2 - 9b^2)

= 8a^3 - 27b^3 - 8a^3 + 18ab^2

= 27b^3 + 18ab^2

= 9b^2 (3b + 2a)

a) 4a²b³ - 6a³b² = 2a²b²( 2b - 3a )

b) ( a - b )² - ( b - a ) = ( a - b )² + ( a - b ) = ( a - b )( a - b + 1 )

c) ( 8a³ - 27b³ ) - 2a( 4a² - 9b² ) = 8a³ - 27b³ - 8a³ + 18ab² = 18ab² - 27b³ = 9b²( 2a - 3b )

d) 10x² + 10xy + 5x + 5y = 10x( x + y ) + 5( x + y ) = ( x + y )( 10x + 5 ) = 5( x + y )( 2x + 1 )

e) 5ay - 3bx + ax - 15by = 5y( a - 3b ) + x( a - 3b ) = ( a - 3b )( 5y + x )

a) \(4a^2.b^3-6a^3.b^2=2a^2.b^2\left(2b-3a\right)\)

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

\(=\left(a-b\right).\left(a-b+1\right)\)

c) \(8a^3-27b^3-2a.\left(4a^2-9b^2\right)=8a^3-27b^3-8a^3+18ab^2\)

\(=-27b^3+18ab^2=18ab^2-27b^3=9b^2.\left(2a-3b\right)\)

d) \(10x^2+10xy+5x+5y=5.\left(2x^2+2xy+x+y\right)\)

\(=5.\left[\left(2x^2+2xy\right)+\left(x+y\right)\right]=5.\left[2x\left(x+y\right)+\left(x+y\right)\right]\)

\(=5\left(x+y\right)\left(2y+1\right)\)

e) \(5ay-3bx+ax-15by=\left(5ay-15by\right)-\left(3bx-ax\right)\)

\(=5y\left(a-3b\right)-x\left(3b-a\right)=5y\left(a-3b\right)+x\left(a-3b\right)\)

\(=\left(a-3b\right)\left(x+5y\right)\)

a) \(N=8a^3-27b^3\)

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

\(=\left(2a-3b\right)^3+18ab\left(2a-3b\right)\)

\(=5^3+18\cdot12\cdot5\)

\(=125+1080=1205\)

b) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)

\(=a^3+b^3+6a^2b^2+3a^3b+3ab^3\)

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

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

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

\(=1^3+3ab\cdot1\cdot0\)

\(=1\)

a ) \(N=8a^3-27b^3\)

\(\Leftrightarrow N=\left(2a-3b\right)\left(4x^2+6ab+9b^2\right)\)

\(\Leftrightarrow N=5\left(4x^2+9b^2+72\right)\)

Ta có : \(2a-3b=5\)

\(\Leftrightarrow4a^2+9b^2=25+6ab\)

Thay vào ta được : \(N=5\left(25+6ab+72\right)=845\)

b ) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)

\(\Leftrightarrow K=\left(a+b\right)^3-3ab\left(a+b\right)+6a^2b^2\left(a+b\right)+3ab\left(a+b\right)^2-6a^2b^2\)

\(\Leftrightarrow K=1-3ab+6a^2b^2+3ab-6a^2b^2=1\)

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

\(\Leftrightarrow P=\left(\dfrac{x}{4}+\dfrac{y}{2}\right)^3-3\left[\left(\dfrac{x}{4}\right)^2\dfrac{y}{2}+\dfrac{x}{4}\left(\dfrac{y}{2}\right)^2\right]\)

\(\Leftrightarrow P=\left(\dfrac{2\left(x+2y\right)}{8}\right)^3-3\left[\dfrac{x^2y}{32}+\dfrac{xy^2}{16}\right]\)

\(\Leftrightarrow P=8-3xy\left(\dfrac{x+2y}{32}\right)\)

\(\Leftrightarrow P=8-3.4\left(\dfrac{8}{32}\right)=5\)

a) (-2)+ (-5) = -7

Vì: -7< -5

=> (-2)+ (-5) < -7

b) (-3)+ (-8)= -11

Vì: (-10) > (-11)

=> -10> (-3)+ (-8)

a)(-2)+(-5)<(-5)

b)(-10)>(-3)+(-8)

a) \(\left(-2\right)+\left(-5\right)..........\left(-5\right)\)

\(\left(-7\right)< \left(-5\right)\)

Vậy \(\left(-2\right)+\left(-5\right)< \left(-5\right)\)

b) \(\left(-10\right)...........\left(-3\right)+\left(-8\right)\)

\(\left(-10\right)>\left(-11\right)\)

Vậy \(\left(-10\right)>\left(-3\right)+\left(-8\right)\)

a)\({\left[ {{{\left( {\frac{{ - 2}}{3}} \right)}^2}} \right]^5} = {\left( {\frac{{ - 2}}{3}} \right)^{2.5}} = {\left( {\frac{{ - 2}}{3}} \right)^{10}}\)

Vậy dấu “?” bằng 10.

b) \({\left[ {{{\left( {0,4} \right)}^3}} \right]^3} = {\left( {0,4} \right)^{3.3}} = {\left( {0,4} \right)^9}\)

Vậy dấu “?” bằng 9.

c) \({\left[ {{{\left( {7,31} \right)}^3}} \right]^0} = 1\)

Vậy dấu “?” bằng 1.

a) (-5) . (-4) + (-5) . 14 = (-5 ) . [ (-4) + 14 ] = -50

b ) 3 . ( 5 + 8 ) = 13 . ( -3 ) + 13 . 8 = 65