Question

[Huê An]

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Answer 1

[Toru]

Bài 1:

a) $x^2+6x+9$

$=x^2+2.x.3+3^2$

$=(x+3)^2$

b) $9x^2-6x+1$

$=(3x)^2-2.3x.1+1^2$

$=(3x-1)^2$

c) $x^2y^2+xy+\frac14$

$=(xy)^2+2.xy.\frac12+\left(\frac12\right)^2$

$=\left(xy+\frac12\right)^2$

d) $(x-y)^2+6(x-y)+9$

$=(x-y)^2+2.(x-y).3+3^2$

$=(x-y+3)^2$

Bài 2: 

a) $-x^3+3x^2-3x+1$

$=1^3-3.1^2.x+3.1.x^2-x^3$

$=(1-x)^3$

b) $x^3+x^2+\frac13 x+\frac{1}{27}$

$=x^3+3.x^2.\frac13+3.x.\left(\frac13\right)^2+\left(\frac13\right)^3$

$=\left(x+\frac13\right)^3$

c) $x^6-3x^4y+3x^2y^2-y^3$

$=(x^2)^3-3.(x^2)^2.y+3.x^2.y^2-y^3$

$=(x^2-y)^3$

d) $(x-y)^3+(x-y)^2+\frac13 (x-y)+\frac{1}{27}$

$=(x-y)^3+3.(x-y)^2.\frac13+3.(x-y).\left(\frac13\right)^2+\left(\frac13\right)^3$

$=\left(x-y+\frac13\right)^3$

Bài 3:

a) $x^3+27$

$=x^3+3^3$

$=(x+3)(x^2-x.3+3^2)$

$=(x+3)(x^2-3x+9)$

b) $x^3-\frac18$

$=x^3-\left(\frac12\right)^3$

$=\left(x-\frac12\right)\left[x^2-x.\frac12+\left(\frac12\right)^2\right]$

$=\left(x-\frac12\right)\left(x^2-\frac12 x+\frac14\right)$

c) $8x^3+y^3$

$=(2x)^3+y^3$

$=(2x+y)[(2x)^2-2x.y+y^2]$

$=(2x+y)(4x^2-2xy+y^2)$

d) $8x^3-27y^3$

$=(2x)^3-(3y)^3$

$=(2x-3y)[(2x)^2+2x.3y+(3y)^2]$

$=(2x-3y)(4x^2+6xy+9y^2)$

Answer 2

[Toru]

Bài 4:

a) \(101^2=\left(100+1\right)^2\)

\(=100^2+2.100.1+1^2\)

\(=10000+200+1=10201\)

b) \(75^2-50.75+25^2\)

\(=75^2-2.75.25+25^2\)

\(=\left(75-25\right)^2\)

\(=50^2=2500\)

c) \(103.97\)

\(=\left(100+3\right).\left(100-3\right)\)

\(=100^2-3^2\\ =10000-9=9991\)

Bài 5:

a) \(\left(x+3y\right)^2-\left(x-3y\right)^2\)

\(=\left(x+3y-x+3y\right)\left(x+3y+x-3y\right)\\ =6y.2x=12xy\)

b) \(Q=\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y\right)^2\)

\(=\left(x-y\right)^2-2.\left(x-y\right).2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\\ =\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\\ =\left(x-y-2x-4y\right)^2\\ =\left(-x-5y\right)^2\)

c) \(A=\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(=\left(x+2+x-2\right)\left[\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\right]-2x\left(x^2+12\right)\\ =2x\left(x^2+4x+4-x^2+4+x^2-4x+4\right)-2x\left(x^2+12\right)\\ =2x\left(x^2+12\right)-2x\left(x^2+12\right)=0\)

d) \(B=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)

\(=\left(xy+2\right)^3-3.\left(xy+2\right)^2.2+3.\left(xy+2\right).2^2-2^3\\ =\left(xy+2-2\right)^3\\ =\left(xy\right)^3=x^3y^3\)

e) \(A=\left(x-3\right)\left(x^2+3x+9\right)-\left(x^3+3\right)\)

\(=\left(x-3\right)\left(x^2+x.3+3^2\right)-x^3-3\\ =x^3-3^3-x^3-3\\ =-27-3=-30\)

Bài 6:

\(a,VT=\left(a-b\right)^2=a^2-2ab+b^2\\ =\left(a^2+2ab+b^2\right)-4ab\\ =\left(a+b\right)^2-4ab=VP\\ b,VT=\left(x+y\right)^2+\left(x-y\right)^2\\ =x^2+2xy+y^2+x^2-2xy+y^2\\ =2x^2+2y^2\\ =2\left(x^2+y^2\right)=VP\\ c,VT=\left(x+y\right)^2-\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]\\ =\left(x+y-x+y\right)\left(x+y+x-y\right)\\ =2y.2x=4xy=VP\\ d,VT=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)\\ =\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\\ =\left[\left(x-y\right)+\left(x+y\right)\right]^2\\ =\left(x-y+x+y\right)^2\\ =\left(2x\right)^2=4x^2=VP\)