\(M=\left(\frac{1}{3}x-y\right)\left(x^2+3xy+9y^2\right)+9y^3-\frac{1}{3}x^3\)

\(=\frac{1}{3}x^3+x^2y+3xy^2-x^2y-3xy^2-9y^3+9y^3-\frac{1}{3}x^3\)

\(=\left(\frac{1}{3}x^3-\frac{1}{3}x^3\right)+\left(x^2y-x^2y\right)+\left(3xy^2-3xy^2\right)-\left(9y^3-9y^3\right)\)

\(=0\)

Vậy : Giá trị của M ko phụ thuộc vào biến x,y

=.= hk tốt!!

\(2x^2+98+28x-8y^2\)

\(=2\left(x^2+14x+49-4y^2\right)\)

\(=2\left[\left(x^2+2\cdot x\cdot7+7^2\right)-\left(2y\right)^2\right]\)

\(=2\left[\left(x+7\right)^2-\left(2y\right)^2\right]\)

\(=2\left(x-2y+7\right)\left(x+2y+7\right)\)

\(ay^2-4ay+4a-by^2+4by-4b\)

\(=\left(ay^2-4ay+4a\right)-\left(by^2-4by+4b\right)\)

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

\(=a\left(y-2\right)^2-b\left(y-2\right)^2\)

\(=\left(y-2\right)^2\left(a-b\right)\)

=(ay2-by2)+(-4ay+4by)+(4a-4b)

=y2(a-b)+ 4(a-b)

=(a-b)×(y2+4)

\(\frac{4x^4-20x^3+13x^2+30x+9}{\left(4x^2-1\right)^2}\)

\(=\frac{4x^3\left(x-3\right)-8x^2\left(x-3\right)-11x\left(x-3\right)-3\left(x-3\right)}{\left(4x^2-1\right)^2}\)

\(=\frac{\left(x-3\right)\left(4x^3-8x^2-11x-3\right)}{\left(4x^2-1\right)^2}\)

\(=\frac{\left(x-3\right)\left[4x^2\left(x-3\right)+4x\left(x-3\right)+\left(x-3\right)\right]}{\left[\left(2x-1\right)\left(2x+1\right)\right]^2}\)

\(=\frac{\left(x-3\right)^2\left(4x^2+4x+1\right)}{\left(2x-1\right)^2\left(2x+1\right)^2}=\frac{\left(x-3\right)^2\left(2x+1\right)^2}{\left(2x-1\right)^2\left(2x+1\right)^2}=\frac{\left(x-3\right)^2}{\left(2x-1\right)^2}\)

\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)

  \(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{n^2-1}{n^2}\right)\)

\(=\text{[}\frac{\left(2-1\right)\left(2+1\right)}{2^2}\text{]}.\text{[}\frac{\left(3-1\right)\left(3+1\right)}{3^2}\text{]}.\text{[}\frac{\left(4-1\right)\left(4+1\right)}{4^2}\text{]}...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)

\(=\left(\frac{1.3}{2^2}\right).\left(\frac{2.4}{3^2}\right).\left(\frac{3.5}{4^2}\right)...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)

\(=\frac{\text{[}1.2.3...\left(n-1\right)\text{]}.\text{[}3.4.5...\left(n+1\right)\text{]}}{\text{[}2.3.4...n\text{]}.\text{[}2.3.4...n\text{]}}\)

\(=\frac{1}{n}.\frac{n+1}{2}\)

\(=\frac{n+1}{2n}\)

\(\frac{x}{x^2-2x}=\frac{B}{4x^2-16}\Leftrightarrow\frac{x}{x\left(x-2\right)}=\frac{B}{\left(2x+4\right)\left(2x-4\right)}\)

\(\Leftrightarrow x\left(2x+4\right)\left(2x-4\right)=x\left(x-2\right).B\)

\(\Rightarrow B=\frac{x.\left[2\left(x+2\right)\right].\left[2\left(x-2\right)\right]}{x\left(x-2\right)}=\frac{x.2\left(x+2\right).2\left(x-2\right)}{x\left(x-2\right)}\)

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

\(\frac{x}{x^2-2x}=\frac{B}{4x^2-16}\)

\(\frac{x}{x\left(x-2\right)}=\frac{B}{4.\left(x^2-4\right)}\)

\(\frac{1}{x-2}=\frac{B}{4.\left(x^2-4\right)}\)

\(\Rightarrow B.\left(x-2\right)=4.\left(x-2\right)\left(x+2\right)\)

\(B=4.\left(x+2\right)\)

\(B=4x+8\)

Do E là điểm bất kì trên AB, mà E đối xứng với F qua O => F nằm trên DC⇒ D,F,C thẳng hàng