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\(x^2y+xy^2+x^2z+y^2z+2xyz\)
\(=x^2z+2xyz+y^2z+x^2y+xy^2\)
\(=\left(x^2z+2xyz+y^2z\right)+\left(x^2y+xy^2\right)\)
\(=z.\left(x^2+2xy+y^2\right)+xy.\left(x+y\right)\)
\(=z.\left(x+y\right)^2+xy.\left(x+y\right)\)
\(=\left(x+y\right).\left[z.\left(x+y\right)+xy\right]\)
\(=\left(x+y\right).\left(xz+yz+xy\right)\)
Chúc bạn học tốt!
Áp dụng BĐT AM-GM ta có:
\(x+y+z+xy+yz+zx\le\frac{x^2+1}{2}+\frac{y^2+1}{2}+\frac{z^2+1}{2}+xy+yz+xz=\frac{x^2+y^2+z^2+2xy+2yz+2zx+3}{2}=\frac{\left(x+y+z\right)^2+3}{2}\)\(\Leftrightarrow6\le\frac{\left(x+y+z\right)^2+3}{2}\Leftrightarrow\left(x+y+z\right)^2+3\ge12\Leftrightarrow\left(x+y+z\right)^2\ge9\Leftrightarrow x+y+z\ge3\)
Áp dụng BĐT Bunhiacopxki ta có:
\(3A=\left(1+1+1\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\ge3^2=9\)
\(\Leftrightarrow A\ge3\)
Dấu " = " xảy ra <=> \(x=y=z=1\)
Vậy \(A_{min}=3\Leftrightarrow x=y=z=1\)
\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)
\(49\left(y-4\right)^2-9y^2-36y-36\)
\(=49\left(y-4\right)^2-\left(9y^2+36y+36\right)\)
\(=49\left(y-4\right)^2-\left(3y+6\right)^2\)
\(=[7\left(y-4\right)]^2-\left(3y+6\right)^2\)
\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)
\(=\left(7y-28+3y+6\right)\left(7y-28-3y-6\right)\)
\(=\left(10y-22\right)\left(4y-34\right)\)
\(x^2y+xy^2+x^2z+y^2z+2xyz=z\left(x^2+2xy+y^2\right)+xy\left(x+y\right)=z\left(x+y\right)^2+xy\left(x+y\right)=\left(x+y\right)\left[z\left(x+y\right)+xy\right]=\left(x+y\right)\left(zx+zy+xy\right)\)
x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz
=x^2y+xy^2+xyz+x^2z+xz^2+xyz+y^2z+yz^2
=xy(x+y+z)+zx(x+y+z)+yz(y+z)
=x(y+z)(x+y+z)+yz(y+z)
=(y+z)(x^2+xy+zx+yz)
=(x+y)(y+z)(z+x)
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