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Tính :
a) ( x - 2y )( 3xy + 6y^2 + x )
b) [ 4 ( x-y )^5 + 2( x - y )^3 - 3( x - y )^2 ] : ( y - x )^2
a)\(\left(x-2y\right)\left(3xy+6y^2+x\right)=x\left(3xy+6y^2+x\right)-2y\left(3xy+6y^2+x\right)\)\(=3x^2y+6xy^2+x^2-6xy^2-12y^3-2xy\)
\(=3x^2y+x^2-12y^3-2xy\)
b)\(\text{[}4\left(x-y\right)^5+2\left(x-y\right)^3-3\left(x-y\right)^2\text{]}:\left(y-x\right)^2\)
=\(\text{[}4\left(x-y\right)^5+2\left(x-y\right)^3-3\left(x-y\right)^2\text{]}:\left(x-y\right)^2\)
=\(^{=4\left(x-y\right)^3+2\left(x-y\right)-3}\)
Bài 2:
a) \(=x^2-36y^2\)
b) \(=x^3-8\)
Bài 3:
a) \(=x^2+2x+1-x^2+2x-1-3x^2+3=-3x^2+4x+3\)
b) \(=6\left(x-1\right)\left(x+1\right)=6x^2-6\)
Ta có a + b + 8 = 0
=> x3 + 3x2 + 6x + y3 + 3y2 + 6y + 8 = 0
=> (x3 + 3x2 + 3x + 1) + (y3 + 3y2 + 3y + 1) + (3x + 3y + 6) = 0
=> (x + 1)3 + (y + 1)3 + 3(x + y + 2) = 0
=> (x + y + 2)[(x + 1)2 + (x + 1)(y + 1) + (y + 1)2 + 3] = 0
Vì (x + 1)2 + (x + 1)(y + 1) + (y + 1)2 + 3 \(>0\forall x;y\)
=> x + y + 2 = 0
=> x + y = -2
Vậy A = -2
xyz bạn ơi! tại sao từ dòng 3 lại thành dòng 4 vậy
thank you bạn!!! <3
(x2 -y2 +6y -9) : (x-y+3)
= [x2-(y2-6y+9)] : (x-y+3)
=[x2-(y-3)2 ] : (x-y+3)
=[(x-y+3)(x+y-3)] :(x-y+3)
=x+y-3
\(A=x^2+2xy+y^2-6-6y-5=\left(x+y\right)^2-6\left(x+y\right)-5=2^2-6\cdot2-5=-13\)
\(B=3\left(x^2+y^2\right)-\left(x^3+y^3\right)+1\)
\(=3x^2+3y^2-\left(x+y\right)\left(x^2+xy+y^2\right)+1\)
\(=3x^2+3y^2-2\left(x^2+xy+y^2\right)+1\)
\(=3x^2+3y^2-2x^2+2xy-2y^2+1=x^2+2xy+y^2+1\\ =\left(x+y\right)^2+1=2^2+1=5\)
TL:
1)\(\left(y^2-6y+9\right)-\left(3-y\right)^2=\left(y-3\right)^2-\left(3-y\right)^2\)
\(=\left(y-3+3-y\right)\left(y-3-3+y\right)=0.\left(2y-6\right)=0\)
2)\(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=\left(x-3\right)^2-x^2+16\)
\(=\left(x-3+x\right)\left(x-3-x\right)+16=\left(2x-3\right).\left(-3\right)+16=-6x+9+16\)
\(=-6x+25\)
hc tốt
\(1,\left(y^2-6x+9\right)-\left(3-y\right)^2\)
\(=\left(y-3\right)^2-\left(y-3\right)^2=0\)
\(2,\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16=-6x+21\)
\(3...\)\(< ->1\)
1, a,= (x+2)^2/3.(x+2) = x+2/3
b, = 3x.(x+4)/2x.(x+4) = 3/2
k mk nha
Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Rightarrow\left(x+\frac{1}{6}y+3\right)^2=x^2+\left(\frac{y}{6}\right)^2+3^2+2.x.\frac{y}{6}+2.\frac{y}{6}.3+2.x.3\)
\(=x^2+\frac{y^2}{36}+9+\frac{xy}{3}+y+6x\)
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