\(x^3y^2-x^2y+2xy-2x^2y^3+2xy^2-4y^2\)

(x²y²-xy+2y)(x-2y)=x³y²-x²y+2xy-2x²y³+2xy²-4y²

(x³ - 2x² + x - 1)(5 - x) = (x³ - 2x² + x - 1).5 - [ (x³ - 2x² + x - 1).x ] = 5x³ - 10x² + 5x - 5 - (x⁴ - 2x³ + x² - x)

= 5x³ - 10x² + 5x - 5 - x⁴ + 2x³ - x² + x = x⁴ + (5x³ + 2x³) + (-10x² - x²) + (5x + x) - 5 = x⁴ + 7x³ - 11x²+ 6x - 5

Cho tứ giác ABCD, 2 đường chéo vuông góc tại O. Biết BC=15cm, CD=24cm và AD=20cm. Tính độ dài AB

Đáp Số hình như là 7 cm còn cách giải thì ???

BC=15cm;CD=24CM;AD=20CM.

(15+24+20):3=19CM

19CM. 

K CHO MINH NHA

ko phai

\(-6x+7=2x+1\)

\(\Leftrightarrow-6x+7-2x-1=0\)

\(\Leftrightarrow\left(-6x-2x\right)+\left(7-1\right)=0\)

\(\Leftrightarrow-8x+6=0\)

Phương trình này có dạng ax + b = 0 nên đây là phương trình bậc nhất một ẩn

ko có GT thỏa mãn

a=0

Do: A=1^4+2.1^2-1^3+1-0=0

B=1-1=0. mà 0 thì chia hết cho 0.

Đặt \(P=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)
\(P=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)

\(=x^4\left(y-z\right)+y^4z-y^4x+z^4x-z^4y\)

\(=x^4\left(y-z\right)+y^4z-z^4y-y^4x+z^4x\)

\(=x^4\left(y-z\right)+yz\left(y^3-z^3\right)-x\left(y^4-z^4\right)\)

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

\(=\left(y-z\right)\left[x^4+yz\left(y^2+yz+z^2\right)-x\left(y^3+y^2z+yz^2+z^3\right)\right]\)

\(=\left(y-z\right)\left(x^4+y^3z+y^2z^2+yz^3-xy^3-xy^2z-xyz^2-xz^3\right)\)

\(=\left(y-z\right)\left(x^4-xz^3-xy^3+y^3z-xy^2z+y^2z^2-xyz^2+yz^3\right)\)

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

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

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

\(=\left(y-z\right)\left(x-z\right)\left(x^3+x^2z+xz^2-y^3-y^2z-yz^2\right)\)

\(=\left(y-z\right)\left(x-z\right)\left(x^3-y^3+x^2z-y^2z+xz^2-yz^2\right)\)

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

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

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

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

Đặt \(A=x^2+xy+y^2+xz+yz+z^2\)

\(A=\frac{2\left(x^2+xy+y^2+xz+yz+z^2\right)}{2}=\frac{2x^2+2xy+2y^2+2xz+2yz+2z^2}{2}\)

\(=\frac{\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(x^2+2xz+z^2\right)}{2}\)

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

=>\(P=\left(y-z\right)\left(x-z\right)\left(x-y\right).\frac{\left(x+y\right)^2+\left(y+z\right)^2+\left(x+z\right)^2}{2}\)

Ta có: \(x>y>z< =>\hept{\begin{cases}x>y\\y>z\\x>z\end{cases}}< =>\hept{\begin{cases}x-y>0\\y-z>0\\x-z>0\end{cases}}\)

Dễ thấy \(\left(x+y\right)^2\ge0;\left(y+z\right)^2\ge0;\left(x+z\right)^2\ge0\) với mọi x;y;z

\(=>P>0\) (đpcm)