(9x²-6xy+y²)-4²

=(3x-y)²-4²

=(3x-y-4)(3x-y+4)

mình cảm ơn

\(a,4x^2+28x+49=\left(2x\right)^2+2.2x.7+7^2=\left(2x+7\right)^2\\ b,16y^2-8y+1=\left(4y\right)^2-2.4y.1+1^2=\left(4y-1\right)^2=\left(1-4y\right)^2\\ 4a^2+20ab+25b^2=\left(2a\right)^2+2.2a.5b+\left(5b\right)^2=\left(2a+5b\right)^2\\ d,9x^2-6xy+y^2=\left(3x\right)^2-2.3x.y+y^2=\left(3x-y\right)^2=\left(y-3x\right)^2\)

a) \(16x^2+8xy+y^2=\left(4x+y\right)^2\)

b) \(4x^2-2xy+\dfrac{1}{4}y^2=\left(2x-\dfrac{1}{2}y\right)^2\)

c) \(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)

d) \(9x^2-6xy+y^2=\left(3x-y\right)^2\)

a) \(=\left(6x\right)^2-2.6x.1+1=\left(6x-1\right)^2\)

b) \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

c) \(=\left(3x-y\right)^2-25=\left(3x-y-5\right)\left(3x-y+5\right)\)

d) \(=x\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)

\(1.\)

\(a.\)

\(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)

\(=\left(x^3-3^3\right)-\left(54+x^3\right)\)

\(=x^3-27-54-x^3\)

\(=-81\)

\(b.\)

\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=\left(27x^3+y^3\right)-\left(27x^3-y^3\right)\)

\(=27x^3+y^3-27x^3+y^3\)

\(=2y^3\)

\(2.\)

\(a.\)

\(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)

\(b.\)

\(\left(2x-3y\right)\left(4x^2+6xy+9y^3\right)=8x^3-27y^3\)

1) a) \(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)

\(=\left(x^3-3^3\right)-\left(54+x^3\right)\\ =\left(x^3-27\right)-54-x^3\\ =-27-54\\ =-81\)

b) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=\left[\left(3x\right)^3+y^3\right]-\left[\left(3x\right)^3-y^3\right]\\ =2y^3\)

2) a) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)

b) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=8x^3-27y^3\)

Bài 2:

a: \(7x^2-14xy+7y^2\)

\(=7\left(x^2-2xy+y^2\right)=7\left(x-y\right)^2\)

b: xy-3x+2y-6

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

=(x+2)(y-3)

c: \(9x^2+6xy-25+y^2\)

\(=\left(9x^2+6xy+y^2\right)-25\)

\(=\left(3x+y\right)^2-25=\left(3x+y-5\right)\left(3x+y+5\right)\)

Bài 1:

a: 15x+10+4x(3x+2)=0

=>5(3x+2)+4x(3x+2)=0

=>(3x+2)(4x+5)=0

=>\(\left[\begin{array}{l}3x+2=0\\ 4x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac23\\ x=-\frac54\end{array}\right.\)

b: \(2x\left(x-6\right)+x^2-36=0\)

=>2x(x-6)+(x-6)(x+6)=0

=>(x-6)(2x+x+6)=0

=>(x-6)(3x+6)=0

=>(x+2)(x-6)=0

=>\(\left[\begin{array}{l}x+2=0\\ x-6=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=6\end{array}\right.\)

\(a,=\left(x+1\right)^2\\ b,=\left(3x-y\right)^2\\ c,=\left(x-3\right)\left(x+3\right)\\ d,=\left(x+4\right)^3\\ e,=\left(x-2\right)^3\\ f,=\left(x+2\right)\left(x^2-2x+4\right)\\ g,=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)