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b \(2x^4-y^4+x^2y^2+3y^2=\left(x^4-y^4\right)+\left(x^4+x^2y^2\right)+3y^2=\left(x^2-y^2\right)\left(x^2+y^2\right)+x^2\left(x^2+y^2\right)+3y^2\)
\(=\left(x^2-y^2\right)\cdot1+x^2\cdot1+3y^2=x^2-y^2+x^2+3y^2=2x^2+2y^2=2\left(x^2+y^2\right)=2\cdot1=2\)
a \(2\left(x^6+y^6\right)-3\left(x^4+y^4\right)=2\left(\left(x^2\right)^3+\left(y^2\right)^3\right)-3x^4-3y^4=2\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\)
\(-3x^4-3y^4=2\cdot1\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4=2x^4-2x^2y^2+2y^4-3x^4-3y^4\)
\(=-x^4-2x^2y^2-y^4=-\left(x^4+2x^2y^2+y^4\right)=-\left(x^2+y^2\right)^2=-1^2=-1\)
a) \(\dfrac{6x^2y^3-2x^2y+6xy}{6xy}\)
\(=\dfrac{6x^2y^3}{6xy}-\dfrac{2x^2y}{6xy}+\dfrac{6xy}{6xy}\)
\(=xy^2-\dfrac{x}{3}+1\)
b) \(\dfrac{4\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{2\left(x+y\right).2\left(x+y\right)^2}{2\left(x+y\right)}\)
\(=2\left(x+y\right)^2\)
c) \(\dfrac{8x^3+27y^3}{2x+3y}\)
\(=\dfrac{\left(2x\right)^3+\left(3y\right)^3}{2x+3y}\)
\(=\dfrac{\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]}{2x+3y}\)
\(=4x^2-6xy+9y^2\)
d) \(\dfrac{48x^4y^3-12x^2y^5+6x^2y^2}{3x^2y^2}\)
\(=\dfrac{48x^4y^3}{3x^2y^2}-\dfrac{12x^2y^5}{3x^2y^2}+\dfrac{6x^2y^2}{3x^2y^2}\)
\(=16x^2y-4y^3+2\)
Ta có : x4 - y4
= (x2)2 - (y2)2
= (x2 - y2)(x2 + y2)
= (x - y)(x + y)(x2 + y2)
b) 9(x - y)2 - 4(x + y)2
= [3(x - y) - 4(x + y)][3(x - y) + 4(x + y)]
= [3x - 3y - 4x - 4y][3x - 3y + 4x + 4y]
= (-x - 7y)(x + y)
e.\(x^4+2x^2+1=\left(x^2+1\right)^2\)
c.\(x^2-9y^2=\left(x-3y\right)\left(x+3y\right)\)
f.\(-x^2-2xy-y^2+1=-\left[\left(x+y\right)^2-1\right]=-\left(x+y-1\right)\left(x+y+1\right)=\left(x-y+1\right)\left(x+y+1\right)\)
g.\(x^3-x^2-x+1==x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)=\left(x-1\right)^2\left(x+1\right)\)
h.\(\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
i.\(\left(x+y\right)^3-x^3-y^3=\left(x+y\right)^3-\left(x^3+y^3\right)=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-\left(x^2-xy+y^2\right)\right]=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)
\(=3xy\left(x+y\right)\)
tíck mình nha bn thanks !!!!!
a ) 36x2 - ( 3x - 2 )2
= ( 6x - 3x + 2 ) ( 6x + 3x - 2 )
= ( 3x + 2 ) ( 9x - 2 )
b ) 16.( 4x + 5 )2 - 25. ( 2x + 2 )2
= [ 4.( 4x + 5 ) + 5. ( 2x + 2 ) ] [ 4 .( 4x + 5 ) - 5. ( 2x + 2 ) ]
= ( 16x + 5 + 10x + 10 ) ( 16x + 5 - 10x - 10 )
= ( 26x + 15 ) ( 6x - 5 )
a ) 36x2 - ( 3x - 2 )2
= ( 6x - 3x + 2 ) ( 6x + 3x - 2 )
= ( 3x + 2 ) ( 9x - 2 )
b ) 16.( 4x + 5 )2 - 25. ( 2x + 2 )2
= [ 4.( 4x + 5 ) + 5. ( 2x + 2 ) ] [ 4 .( 4x + 5 ) - 5. ( 2x + 2 ) ]
= ( 16x + 5 + 10x + 10 ) ( 16x + 5 - 10x - 10 )
= ( 26x + 15 ) ( 6x - 5 )
bài 2
Giải:x6+y6)-3(x4+y4)
2(x6+y6)−3(x4+y4)2(x6+y6)−3(x4+y4)
⇔2(x2+y2)(x4−x2y2+y4)−3x4−3y4⇔2(x2+y2)(x4−x2y2+y4)−3x4−3y4
⇔2(x4−x2y2+y4)−3x4−3y4⇔2(x4−x2y2+y4)−3x4−3y4
⇔2x4−2x2y2+2y4−3x4−3y4⇔2x4−2x2y2+2y4−3x4−3y4
⇔−2x2y2−x4−y4⇔−2x2y2−x4−y4
⇔−(x4+2x2y2+y4)⇔−(x4+2x2y2+y4)
⇔−(x2+y2)2⇔−(x2+y2)2
⇔−1
bài 1
bạn thay vào hết và tính ra là được
\(\Leftrightarrow\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)
\(\Leftrightarrow3x^3+3y^3+3xy\left(x+y\right)-3x^3-3y^3-3xy\left(x+y\right)=0\)(điều phải c/m)
1,Ta có: \(A=2\left[\left(x^2\right)^3+\left(y^2\right)^3\right]-3x^4-3y^4\)
\(=2\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4\)
Thay \(x^2+y^2=1,\) ta có:
\(A=2.1\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4\)
\(=2x^4-2x^2y^2+2y^4-3x^4-3y^4\)
\(=-\left(x^4+2x^2y^2+y^4\right)=-\left(x^2+y^2\right)^2=-1\)
2,Ta có: \(B=\left(x^2-y^2\right)\left(x^2+y^2\right)+\left(x^4+x^2y^2\right)+3y^2\)
\(=\left(x^2-y^2\right).1+x^2\left(x^2+y^2\right)+3y^2\)
\(=x^2-y^2+x^2+3y^2=2\left(x^2+y^2\right)=2\)