x2+4xy−4z2+4y2x2+4xy−4z2+4y2

=x2+4xy+4y2−4z2=x2+4xy+4y2−4z2

=(x+2y)2−4z2=(x+2y)2−4z2

=(x+2y−2z)(x+2y+2z)=(x+2y−2z)(x+2y+2z)

x2+2x−15x2+2x−15

=x2+2x+1−16=x2+2x+1−16

=(x+1)2−16=(x+1)2−16

=(x+1−4)(x+1+4)=(x+1−4)(x+1+4)

=(x−3)(x+5)

đề bài?chứ cái này m thấy tối giản r(có lẽ)

\(M=\frac{xy+2x+1}{xy+x+y+1}+\frac{yz+2y+1}{yz+y+z+1}+\frac{zx+2z+1}{zx+x+z+1}\)

\(=\frac{x\left(y+1\right)+x+1}{\left(x+1\right)\left(y+1\right)}+\frac{y\left(z+1\right)+y+1}{\left(y+1\right)\left(z+1\right)}+\frac{z\left(x+1\right)+z+1}{\left(z+1\right)\left(x+1\right)}\)

\(=\frac{x}{x+1}+\frac{1}{y+1}+\frac{y}{y+1}+\frac{1}{z+1}+\frac{z}{z+1}+\frac{1}{x+1}\)

\(=\left(\frac{x}{x+1}+\frac{1}{x+1}\right)+\left(\frac{y}{y+1}+\frac{1}{y+1}\right)+\left(\frac{z}{z+1}+\frac{1}{z+1}\right)\)

\(=3\)

Câu n trước nhé:

n) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(x^2+5x+5=t\)

\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

\(=\left(t-1\right)\left(t+1\right)-24\)\(=t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)

\(=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)=\left(x^2+5x\right)\left(x^2+5x+10\right)=x\left(x+5\right)\left(x^2+5x+10\right)\)

Câu j nè:

\(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)\)\(=\left(x-2\right)\left(x-3\right)\)

a) \(\left(3x-4\right)^2=\left(3x\right)^2-2.3x.4+4^2=9x^2-24x+16\)

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

c)  \(\left(3x-2\right)\left(3x+2\right)=9x^2-4\)

d) \(\left(3x+1\right)^2=9x^2+6x+1\)

e) \(\left(2x-3\right)^2=4x^2-12x+9\)

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