Viết các biểu thức sau đây dưới dạng lập phương của 1 tổng hoặc hiệu:
a)x^3+9x^2+27x+27
b)x^3/8+3/4x^2y^2+3/2xy^4+y^6
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Bài làm:
Ta có: \(\frac{x^3}{8}+\frac{3}{4}x^2y^2+\frac{3}{2}xy^4+y^6\)
\(=\left(\frac{x}{2}\right)^3+3.\left(\frac{x}{2}\right)^2.y^2+3.\frac{x}{2}.\left(y^2\right)^2+\left(y^2\right)^3\)
\(=\left(\frac{x}{2}+y^2\right)^3\)
Bổ sung đề :
\(\dfrac{x^3}{8}+\dfrac{3}{4}x^2y^2+\dfrac{3}{2}xy^4+y^6=\left(\dfrac{x}{2}\right)^3+3.\left(\dfrac{x}{2}\right)^2+y^2+3\left(\dfrac{x}{2}\right)y^4+y^6=\left(\dfrac{x}{2}+y^2\right)^3\)
\(27x^3-9x^2+x-\frac{1}{27}=\left(3x\right)^3-3.3^2.\frac{1}{3}x^2+3.3.\left(\frac{1}{3}\right)^2x-\left(\frac{1}{3}\right)^2\)
\(=\left(3x-\frac{1}{3}\right)^3\)
1a/ z2 - 6z + 5 - t2 - 4t = z2 - 2 . 3z + 32 - 4 - t2 - 4t = (z2 - 2 . 3z + 32) - (22 + 2 . 2t + t2) = (z - 3)2 - (2 + t)2
b/ x2 - 2xy + 2y2 + 2y2 + 1 = x2 - 2xy + y2 + y2 + 2y + 1 = (x2 - 2xy + y2) + (y2 + 2y + 1) = (x - y)2 + (y + 1)2
c/ 4x2 - 12x - y2 + 2y + 8 = (2x)2 - 12x - y2 + 2y + 32 - 1 = [ (2x)2 - 2 . 3 . 2x + 32 ] - (y2 - 2y + 1) = (2x - 3)2 - (y - 1)2
2a/ (x + y + 4)(x + y - 4) = x2 + xy - 4x + xy + y2 - 4y + 4x + 4y + 16 = x2 + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y2 + 16
= x2 + 2xy + y2 + 42 = (x + y)2 + 42
b/ (x - y + 6)(x + y - 6) = x2 + xy - 6x - xy - y2 + 6y + 6x + 6y - 36 = x2 + (xy - xy) + (-6x + 6x) + (6y + 6y) - y2 - 36
= x2 - y2 + 12y - 62 = x2 - (y2 - 12y + 62) = x2 - (y2 - 2 . 6y + 62) = x2 - (y - 6)2
c/ (y + 2z - 3)(y - 2z - 3) = y2 -2yz - 3y + 2yz - 4z2 - 6z - 3y + 6z + 9 = y2 + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z2 + 9
= y2 - 6y - 4z2 + 9 = (y2 - 6y + 9) - 4z2 = (y - 3)2 - (2z)2
d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x2 + 4y2 + 6yz - 2xy + 6yz + 9z2 - 3xz = (2xy - 2xy) + (3xz - 3xz) - x2 + (6yz + 6yz) + 9z2 + 4y2
= -x2 + 4y2 + 12yz + 9z2 = (4y2 + 12yz + 9z2) - x2 = [ (2y)2 + 2 . 2 . 3yz + (3z)2 ] - x2 = (2y + 3z)2 - x2
a, x^3 +3.3.x^2+3.3^2.x+3^3
= (x+3)^3.
b , 23-3*x*22+3*x2*2-x3
<=> (2-x)3
c, (x2)3-3*(x2)2*x+3*x2*x2-x3
<=> (x2-x)3
b,\(\dfrac{4}{9}x^2+4x+9=\left(\dfrac{2}{3}x\right)^2+2.\dfrac{2}{3}x.3+3^2=\left(\dfrac{2}{3}x+3\right)^2\)
c, \(x^3+9x^2+27x+27=x^3+3.x^2.3+3.x.3^2+3^3=\left(x+3\right)^3\)
d, \(\dfrac{1}{8}-\dfrac{3}{4}x+\dfrac{3}{2}x^2-x^3=\left(\dfrac{1}{2}\right)^3-3.\left(\dfrac{1}{2}\right)^2.x+3.\dfrac{1}{2}.x^2-x^3=\left(\dfrac{1}{2}-x\right)^3\)
TK MIK
Bài 1:
a) \(x^2+10x+26+y^2+2y=(x^2+10x+25)+(y^2+2y+1)\)
..................................................= \(\left(x+5\right)^2+\left(y+1\right)^2\)
b) \(z^2-6z+5-t^2-4t=(z^2-6t+9)-(t^2+4t+4)\)
............................................= \(\left(z-3\right)^2-\left(t+2\right)^2\)
c) \(x^2-2xy+2y^2+2y+1=(x^2-2xy+y^2)+(y^2+2y+1)\)
..................................................= \(\left(x-y\right)^2+\left(y+1\right)^2\)
d) \(4x^2-12x-y^2+2y+8=\left(4x^2-12x+9\right)-\left(y^2-2y+1\right)\)
.................................................= \(\left(2x-3\right)^2-\left(y-1\right)^2\)
Bài 2:
a) \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-16\)
b) \(\left(x-y+6\right)\left(x+y-6\right)=x^2-\left(y-6\right)^2\)
c) \(\left(y+2z-3\right)\left(y-2z+3\right)=y^2-\left(2z-3\right)^2\)
d) \(\left(x+2y+3z\right)\left(2y+3z-x\right)=\left(2y+3z\right)^2-x^2\)