\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)

\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xz^2+xyz\right)+\left(y^2z+yz^2+xyz\right)\)

\(=xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(xy+xz+yz\right)\)

nick ko hay rồi tcn còn ko hay nữa

\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz.\)

\(=x^2.\left(y+z\right)+yz.\left(y+z\right)+x\left(y^2+z^3\right)+2xyz\)

\(=\left(y+z\right).\left(x^2+yz\right)+x\left(y^{^2}+z^2+2yz\right)\)

\(=\left(y+z\right).\left[x.\left(x+2\right)+y.\left(x+2\right)\right]\)

\(=\left(y+z\right).\left(x+z\right).\left(x+y\right)\)

       \(xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)

\(=\left(xyz-xy-xz+x\right)-yz+y+z-1\)

\(=x\left(yz-y-z+1\right)-\left(yz-y-z+1\right)\)

\(=\left(x-1\right)\left(yz-y-z+1\right)\)

\(=\left(x-1\right)\left[y\left(z-1\right)-\left(z-1\right)\right]\)

\(=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)

      \(49\left(y-4\right)^2-9y^2-36y-36\)

\(=\left[7\left(y-4\right)\right]^2-\left[\left(3y\right)^2+2.3y.6+6^2\right]\)

\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)

\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)=\left(4y-34\right)\left(10y-22\right)=4\left(2y-17\right)\left(5y-11\right)\)

\(49\left(y-4\right)^2-9y^2-36y-36=49\left(y-4\right)^2-9\left(y^2+4y+4\right)\)\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+4\right)\right]^2=\left(7y-28-3y-12\right)\left(7y-28+3y+12\right)\)\(=\left(4y-40\right)\left(10y-16\right)=4\left(y-20\right)\left(5y-8\right)\)

    9x² + 90x + 225 - ( x - 7 )²

= ( 3x + 15 )² - ( x - 7 )²

= ( 3x + 15 + x - 7 )( 3x + 15 - x + 7 )

= ( 4x + 8 )( 2x + 22 )

= 4( x + 2 ) 2 ( x + 11 )

= 8( x + 2 )( x + 11 )

Hk tốt

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

                                                                            \(=\left(-x-4y+5\right)\left(3x+2y+3\right)\)

Cách 1: Thực hiện phép chia: \(f\left(x\right):g\left(x\right)=x-2\)

Cách 2:

 \(f\left(x\right)=x^3-x^2+x^2-2x-12x+24\)

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

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

Khi đó: \(f\left(x\right):g\left(x\right)=x-2\)

a, \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)

\(\Leftrightarrow\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=a+b+c\)

\(\Leftrightarrow\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(a+c\right)}{c+a}+\frac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)

\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)

\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\) (đpcm)

b, Từ \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\) hay ayz+bxz+cxy=0

Từ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}\right)=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\cdot\frac{cxy+ayz+bzx}{abc}=1\)

Mà ayz+bxz+cxy=1

=>\(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\) (đpcm)

sửa lại Mà ayz+bzx+cxy=0 nhé