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a: \(=\dfrac{\left(x^4-y^4\right)^2}{x^2+y^2}=\left(x^2-y^2\right)^2\cdot\left(x^2+y^2\right)\)
b: \(=\dfrac{\left(4x+3\right)\left(16x^2-12x+9\right)}{16x^2-12x+9}=4x+3\)
5) \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)^2-2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left[\left(x-y\right)-\left(x+y\right)\right]^2\)
\(=\left(x-y-x-y\right)^2\)
\(=\left(-2y^2\right)\)
\(=4y^2\)
6) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)
\(=\left(x-5\right)^2-2\left(x-5\right)\left(x+5\right)+\left(x+5\right)^2\)
\(=\left[\left(x-5\right)-\left(x+5\right)\right]^2\)
\(=\left(x-5-x-5\right)^2\)
\(=\left(-10\right)^2=100\)
7) \(\left(x-2\right)^2+\left(x+1\right)^2+2\left(x-2\right)\left(-1-x\right)\)
\(=\left(x-2\right)^2-2\left(x-2\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left[\left(x-2\right)-\left(x+1\right)\right]^2\)
\(=\left(-3\right)^2=9\)
8) \(-\left(2x+3y\right)^2+\left(2x-3y\right)^2-2\left(4x^2-9y^2\right)\)
\(=\left(2x-3y\right)^2+2\left(2x+3y\right)\left(2x-3y\right)+\left(2x+3y\right)^2\)
\(=\left[\left(2x+3y\right)+\left(2x-3y\right)\right]^2\)
\(=\left(4x\right)^2=16x^2\)
a) ( x2 - 5 )( x + 3 ) = x3 + 3x2 - 5x - 15
b) ( x + 4 )( x - x2 ) = x2 - x3 + 4x - 4x2 = -x3 - 3x2 + 4x
c) ( x2 - 6 )( x + 2 ) + ( x + 3 )( x - x2 ) = x3 + 2x2 - 6x - 12 + x2 - x3 + 3x - 3x2 = -3x - 12 = -3( x + 4 )
d) x( x - y ) - y( x - y ) = ( x - y )( x - y ) = ( x - y )2
e) x2( x + y ) - x( x2 - y ) = x3 + x2y - x3 + xy = x2y + xy = xy( x + 1 )
f) 3x( 12x - 4 ) - 9x( 4x - 3 ) = 36x2 - 12x - 36x2 + 27x = 15x
Bài làm
a) ( x2 - 5 )( x + 3 )
= x3 + 3x2 - 5x - 15
b) ( x + 4 )( x - x2 )
= ( x + 4 ) . x( 1 - x )
= x( x + 4 )( 1 - x )
= x( x - x2 + 4 - 4x )
= x( 4 - x2 - 3x )
= 4x - x3 - 3x2
c) ( x2 - 6 )( x + 2 ) + ( x + 3 )( x - x2 )
= ( x - 3 )( x + 3 )( x + 2 ) + ( x + 3 )( x - x2 )
= ( x + 3 )[ ( x - 3 )( x + 2 ) + ( x - x2 )]
= ( x + 3 ) [ x2 + 2x - 3x - 6 + x2 - x2 ]
= ( x + 3 ) ( x2 - x - 6 )
= x3 - x2 - 6x + 3x2 - 3x - 18
= x3 + 2x2 - 9x - 18
d) x( x - y ) - y( x - y )
= ( x - y )( x - y )
= ( x - y )2
= x2 - 2xy + y
e) x2( x + y ) - x( x2 - y )
= x3 + x2y - x3 + xy
= x2y + xy
f) 3x( 12x - 4 ) - 9x( 4x - 3 )
= 3x . 3( 4x - 1 ) - 9x( 4x - 3 )
= 9x( 4x - 1 ) - 9x( 4x - 3 )
= 9x( 4x - 1 - 4x + 3 )
= 9x . 2
= 18x
1. x2 - 2xy + y2 - ( y + 1 )2 = ( x - y )2 - ( y + 1)2
= \(\left[\left(x-y\right)-\left(y+1\right)\right]\left[\left(x-y\right)+\left(y+1\right)\right]\)
= (x-2y-1) ( x +1 )
5. x6 - y6 = (x3)2 - (y3)2
= ( x3 - y3 ) ( x3 + y3 )
=\(\left[\left(x-y\right)\left(x^2+xy+y^2\right)\right]\left[\left(x+y\right)\left(x^2-xy+y^2\right)\right]\)
Vì bài dài nên mình sẽ tách ra nhé.
1a. Ta có:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$
$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$
$=-3(-z)(-x)(-y)=3xyz$
$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$
------------------------
$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$
$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$
$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$
$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$
$=-z^5+5xyz^3-5x^2y^2z$
$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$
$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$
Từ $(1);(2)$ ta có đpcm.
1b.
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$
$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$
Do đó:
$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$
$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$
$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$
$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$
$=7xyz(x^2y^2-2xyz^2+z^4)$
$=7xyz(xy-z^2)$
$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$
$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$
$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)
1. \(f\left(x\right)=25x^2-20x+\dfrac{9}{2}\)
=>\(f\left(x\right)=25x^2-20x+4+\dfrac{1}{2}\)
=> \(f\left(x\right)=(25x^2-20x+4)+\dfrac{1}{2}\)
=> \(f\left(x\right)=(5x-2)^2+\dfrac{1}{2}\)
Ta thấy: \((5x-2)^2\ge0\)
=>\(f\left(x\right)=(5x-2)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\)(đpcm)
2. \(f\left(x\right)=4x^2-28x+50\)
=> \(f\left(x\right)=(4x^2-28x+49)+1\)
=> \(f\left(x\right)=(2x-7)^2+1\)
Ta thấy: \((2x-7)^2\ge0\)
=> \(f\left(x\right)=(2x-7)^2+1\ge1>0\) (đpcm)
3. \(f\left(x\right)=-16x^2+72x-82\)
=> \(f\left(x\right)=-(16x^2-72x+82)\)
=> \(f\left(x\right)=-(16x^2-72x+81+1)\)
=> \(f\left(x\right)=-[(4x-9)^2+1]\)
Ta thấy: \((4x-9)^2\ge0\)
=> \((4x-9)^2+1\ge1>0\)
=> \(f\left(x\right)=-[(4x-9)^2+1]< 0\)
5. \(f\left(x;y\right)=4x^2+9y^2-12x+6y+11\)
=> \(f\left(x;y\right)=4x^2+9y^2-12x+6y+9+1+1\)
=> \(f\left(x;y\right)=(4x^2-12x+9)+(9y^2+6y+1)+1\)
=> \(f\left(x;y\right)=(2x-3)^2+(3y+1)^2+1\)
Ta thấy: \((2x-3)^2\ge0\)
\((3y+1)^2\ge0\)
=> \(f\left(x;y\right)=(2x-3)^2+(3y+1)^2+1\) \(\ge1>0\) (đpcm)
a,Ta có: \(x^3-4x^2-12x+27=x^3+3x^2-7x^2-21x+9x+27=x^2(x+3)-7x(x+3)+9(x+3)=(x+3)(x^2-7x+9)\)b,
\(25(x-y)^2-16(x+y)^2=(5x-5y+4x+4y)(5x-5y-4x-4y)=(9x-y)(x-9y)\)c,\(x^4+x^3+x+1=x^3(x+1)+(x+1)=(x^3+1)(x+1)=(x+1)^2(x^2-x+1)\)d, \(x(x+1)^2+x(x-5)-5(x+1)^2=(x+1)^2(x-5)+x(x-5)=(x-5)(x^2+3x+1)\)e,\(x^2-x-6=x^2-3x+2x-6=x(x-3)+2(x-3)=(x-3)(x+2)\)f,\(x^3-19x-30=x^3-5x^2+5x^2-25x+6x-30=(x-5)(x^2+5x+6)=(x-5)(x^2+2x+3x+6)=(x-5)(x+2)(x+3)\)
nãy bài 1 mk gửi thiếu 1 ý
\(x^2y+xy^2-x+y\)
có ai giúp mk ý này k
bài 2 thì k cần lm cũng đc nhé vì mk biết làm rùi còn mỗi ý này thui hu hu
sai r bạn ạ
đáp số : (x+6)2-(y-10)2-117