\(bdt< =>x\left(x+y\right)\le\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{y}< =>x^2-xy+y^2\ge xy\)

\(< =>\left(x-y\right)^2\ge0\)(dpcm)

mình đánh nhầm nha mn

=\(\frac{19}{49}\)chứ ko phải là 4949 đâu nhá

thông cảm nhé mn

a) (x  + y + z)³ - x³ - y³ - z³

= (x + y + z)³ - z³ - (x³ + y³) 

= (x + y + z - z)[(x + y + z)² + (x + y + z).z + z²) - (x + y)(x² - xy + y²)

= (x + y)(x² + y² + z² + 2xy + 2yz + 2zx + 2xz + 2yz + z² + z²) - (x + y)(x² - xy + y²)

= (x + y)(x² + y² + 3z² + 2xy + 4yz + 4zx) - (x + y)(x² - xy + y²)

= (x + y)(3z² + 3xy + 5yz + 4zx) 

b) Sửa đề x⁴ + 2010x² + 2009x + 2010

= (x⁴ + x² + 1) + (2009x² + 2009x + 2009)

= (x⁴ + 2x² + 1 - x²) + 2009(x² + x + 1)

= [(x² + 1)² - x²] + 2009(x² + x + 1)

= (x² + x + 1)(x² - x + 1) + 2009(x² + x + 1)

= (x² + x + 1)(x² - x + 2010)

\(\left(\frac{x+2}{2x-4}-\frac{x-2}{2x+4}+\frac{8}{x^2-4}\right):\frac{4}{x-2}\)

\(=\left(\frac{x+2}{2\left(x-2\right)}-\frac{x-2}{2\left(x+2\right)}+\frac{8}{\left(x-2\right)\left(x+2\right)}\right):\frac{4}{x-2}\)

\(=\left(\frac{\left(x+2\right)^2}{2\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)^2}{2\left(x-2\right)\left(x+2\right)}+\frac{8x}{x\left(x-2\right)\left(x+2\right)}\right):\frac{4}{x-2}\)

\(=\left(\frac{x^2+4x+4-x^2+4x-4+8x}{2\left(x-2\right)\left(x+2\right)}\right):\frac{4}{x-2}\)

\(=\frac{16x}{2\left(x-2\right)\left(x+2\right)}.\frac{x-2}{4}=\frac{16x}{8\left(x+2\right)}=\frac{2x}{x+2}\)

\(\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x+1}\right)\left(1+\frac{1}{x+2}\right)...\left(1+\frac{1}{x+99}\right)\)

\(=\frac{x+1}{x}.\frac{x+2}{x+1}.\frac{x+3}{x+2}...\frac{x+100}{x+99}=\frac{\left(x+1\right)\left(x+2\right)\left(x+3\right)...\left(x+100\right)}{x\left(x+1\right)\left(x+2\right)...\left(x+99\right)}=\frac{x+100}{x}\)