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Lời giải:
a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$
$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.
$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$
$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$
d.
$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$
$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$
$=-x^2y+4x^2-2xy^2-10x$
$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$
1) \(\left(x+1\right)^2=x^2+2x+1\)
2) \(\left(2x+1\right)^2=4x^2+4x+1\)
3) \(\left(2x+y\right)^2=4x^2+4xy+y^2\)
4) \(\left(2x+3\right)^2=4x^2+12x+9\)
5) \(\left(3x+2y\right)^2=9x^2+12xy+4y^2\)
6) \(\left(2x^2+1\right)^2=4x^4+4x^2+1\)
7) \(\left(x^3+1\right)^2=x^6+2x^3+1\)
8) \(\left(x^2+y^3\right)^2=x^4+2x^2y^3+y^6\)
9) \(\left(x^2+2y^2\right)^2=x^4+4x^2y^2+4y^4\)
10) \(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)
a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
2a) pt <=> (x + 6)^2 = 0
<=> x = -6
b) pt <=> (4x - 1)^2 = 0
<=> x = 1/4
c) pt<=> (x + 1)^3 = 0
<=> x = -1
Bài 1:
a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)
\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)
\(=32x^2+18y^2\)
b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)
\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)
\(=-12x^2-24\)
Bài 2:
a: Ta có: \(x^2+12x+36=0\)
\(\Leftrightarrow x+6=0\)
hay x=-6
b: Ta có: \(16x^2-8x+1=0\)
\(\Leftrightarrow4x-1=0\)
hay \(x=\dfrac{1}{4}\)
Bài 1:
a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)
\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)
\(=32x^2+18y^2\)
b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)
\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)
\(=-12x^2-24\)
c: Ta có: \(C=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+\left(x-2y\right)^2\)
\(=\left(x+2y+x-2y\right)^2\)
\(=4x^2\)
\(b,\left(x+2\right)^2-25\)
\(=\left(x+2\right)^2-5^2\)
\(=\left(x-3\right)\left(x+7\right)\)
\(c,36\left(x-y\right)^2\)
\(=36\left(x^2-2xy+y^2\right)\)
\(=36x^2-72xy+36y^2\)
\(d,x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)
\(=x^2+2.x.\dfrac{1}{4}+\dfrac{1}{4}^2\)
\(=\left(x+\dfrac{1}{4}\right)^2\)
\(e,2x^4y^3-3x^2y^4+5x^3y^4\)
\(=x^2y^3\left(2x^2-3y+5xy\right)\)
Các câu còn lại làm tương tự, chú ý sd HĐT
`#3107.101107`
`D = x^3 - y^3 - 3xy` biết `x - y - 1 = 0`
Ta có:
`x - y - 1 = 0`
`=> x - y = 1`
`D = x^3 - y^3 - 3xy`
`= (x - y)(x^2 + xy + y^2) - 3xy`
`= 1 * (x^2 + xy + y^2) - 3xy`
`= x^2+ xy + y^2 - 3xy`
`= x^2 - 2xy + y^2`
`= x^2 - 2*x*y + y^2`
`= (x - y)^2`
`= 1^2 = 1`
Vậy, với `x - y = 1` thì `D = 1`
________
`E = x^3 + y^3` với `x + y = 5; x^2 + y^2 = 17`
`x + y = 5`
`=> (x + y)^2 = 25`
`=> x^2 + 2xy + y^2 = 25`
`=> 2xy = 25 - (x^2 + y^2)`
`=> 2xy = 25 - 17`
`=> 2xy = 8`
`=> xy = 4`
Ta có:
`E = x^3 + y^3`
`= (x + y)(x^2 - xy + y^2)`
`= 5 * [ (x^2 + y^2) - xy]`
`= 5 * (17 - 4)`
`= 5 * 13`
`= 65`
Vậy, với `x + y = 5; x^2 + y^2 = 17` thì `E = 65`
________
`F = x^3 - y^3` với `x - y = 4; x^2 + y^2 = 26`
Ta có:
`x - y = 4`
`=> (x - y)^2 = 16`
`=> x^2 - 2xy + y^2 = 16`
`=> (x^2 + y^2) - 2xy = 16`
`=> 2xy = (x^2 + y^2) - 16`
`=> 2xy = 26 - 16`
`=> 2xy = 10`
`=> xy = 5`
Ta có:
`F = x^3 - y^3`
`= (x - y)(x^2 + xy + y^2)`
`= 4 * [ (x^2 + y^2) + xy]`
`= 4 * (26 + 5)`
`= 4*31`
`= 124`
Vậy, với `x - y = 4; x^2 + y^2 = 26` thì `F = 124.`