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-25x6 - y8 + 10x3y4
Đặt x3 = a ; y4 = b
Đa thức đã cho trở thành
-25a2 - b2 + 10ab
= -( 25a2 - 10ab + b2 )
= -( 5a - b )2
= -( 5x3 - y4 )2
\(B=25x^2+3y^2-10y+11\)
\(=25x^2+3\left(y^2-\frac{10}{3}y+\frac{11}{3}\right)\)
\(=25x^2+3\left(y^2-2.y.\frac{5}{3}+\frac{25}{9}+\frac{8}{9}\right)\)
\(=25x^2+3\left(y-\frac{5}{3}\right)^2+\frac{8}{3}\ge\frac{8}{3}\)
Đẳng thức xảy ra khi x = 0; y = 5/3
Vậy...
b) \(-y^8+10y^4x^3-25x^6\)
\(=-\left(y^8-10y^4x^3+25x^6\right)\)
\(=-\left[\left(y^4\right)^2-2.y^4.5x^3+\left(5x^3\right)^2\right]\)
\(=-\left(y^4-5x^3\right)^2\)
c) \(8x^3+36x^2y+54xy^2+27y^3\)
\(=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)
\(=\left(2x+3y\right)^3\)
d) \(-y^3+12y^2x-48yx^2+64x^3\)
\(=-\left(y^3-12y^2x+48yx^2-64x^3\right)\)
\(=-\left[y^3-3.y^2.4x+3.y.\left(4x\right)^2-\left(4x\right)^3\right]\)
\(=-\left(y-4x\right)^3\)
e) \(64x^6y^4-81x^2y^2\)
\(=\left(8x^3y^2\right)^2-\left(9xy\right)^2\)
\(=\left(8x^3y^2-9xy\right)\left(8x^3y^2+9xy\right)\)
f) \(64x^6-27y^6\)
\(=\left(4x^2\right)^3-\left(3y^2\right)^3\)
\(=\left(4x^2-3y^2\right)\left[\left(4x^2\right)^2+4x^2.3y^2+\left(3y^2\right)^2\right]\)
\(=\left(4x^2-3y^2\right)\left(16x^4+12x^2y^2+9x^4\right)\)
Câu 1:
a: \(=\left(5x+10y\right)^2-\left(8x-4y\right)^2\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(-3x+14y\right)\left(13x+6y\right)\)
b: \(=\left(0.5x-y\right)^2-\left(2x+2y\right)^2\)
\(=\left(0.5x-y-2x-2y\right)\left(0.5x-y+2x+2y\right)\)
\(=\left(-1.5y-3y\right)\left(2.5x+y\right)\)
c: \(=\left(\dfrac{2}{3}x-2y\right)^2-\left(0.2x+0.2y\right)^2\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x+2y+\dfrac{1}{5}y+\dfrac{1}{5}x\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x+\dfrac{11}{5}y\right)\)
d: \(=-\left(5x-3\right)^2\)
1) -25x6 - y8 + 10x3y4 = -( 25x6 - 10x3y4 + y8 ) = -[ ( 5x3 )2 - 2.5x3.y4 + ( y4 ) ] = -( 5x3 - y4 )2
2) 2x( 3x - 5 ) + 10 - 6x = 2x( 3x - 5 ) - 2( 3x - 5 ) = ( 3x - 5 )( 2x - 2 ) = 2( 3x - 5 )( x - 1 )
3) x2 - 9 - x2( x2 - 9 ) = ( x2 - 9 ) - x2( x2 - 9 ) = ( x2 - 9 )( 1 - x2 ) = ( x - 3 )( x + 3 )( 1 - x )( 1 + x )
4) 4x2 - 9 - ( 3x + 1 )( 2x - 3 ) = ( 2x - 3 )( 2x + 3 ) - ( 3x + 1 )( 2x - 3 )
= ( 2x - 3 )[ ( 2x + 3 ) - ( 3x + 1 ) ]
= ( 2x - 3 )( 2x + 3 - 3x - 1 )
= ( 2x - 3 )( 2 - x )
5) 8x3 - y3 - 4x + 2y = ( 8x3 - y3 ) - ( 4x - 2y )
= [ ( 2x )3 - y3 ) - 2( 2x - y )
= ( 2x - y )( 4x2 + 2xy + y2 ) - 2( 2x - y )
= ( 2x - y )( 4x2 + 2xy + y2 - 2 )
Q=3x+9y+15z+x+x4+y+y9+z+z25
\ge 108+2.2+2.3+2.5=128≥108+2.2+2.3+2.5=128
Dấu "=" xảy ra khi x+3y+5z=36, x=\dfrac{4}x, y=\dfrac{9}y, z=\dfrac{25}z\Rightarrow x=2,y=3,z=5x+3y+5z=36,x=x4,y=y9,z=z25⇒x=2,y=3,z=5
bạn tham khảo nhé
Ta có:
\(-25x^6-y^8+10x^3y^4\)
\(=-\left[\left(5x^3\right)^2-10x^3y^4+\left(y^4\right)^2\right]\)
\(=-\left(5x^3-y^4\right)^2\)