Without loss of generality, assume a=17k, then: 7\cdot 17k\cdot b=17(17k+b) \Rightarrow k=\frac{b}{7b-17} \Rightarrow b\ge 7b-17 > 0 \Rightarrow b\in \emptyset.

Since A^2=I, the minimal polynomial of A has simple roots, and therefore A is diagonalisable. A=P^{-1}DP with P invertible and D diagonal. The equation A^2 = I implies D^2=I so D is ...

You are right: a nondegenerate parabola doesn't admit an equation of the form Ax^2+2Bxy+Cy^2=1 . Indeed, if B^2=AC\ne0 , we can multiply all coefficients by A , finding A^2x^2+2ABxy+B^2y^2=B ...

I found your mistake \scriptsize2\int\frac{dx}{x}+3\int\frac{xdx}{x^2+2x+1}-\int\frac{dx}{x^2+2x+1}\color{red}{\ne}2ln(x) -ln(x^2+2x+1)+\frac{3}{x+1}+3ln(x+1) But: 2\int\frac{dx}{x}=2\ln|x|+c\\ 3\int\frac{xdx}{x^2+2x+1}\stackrel{t=x+1}=3\int\frac{(t-1)dt}{t^2}=3\int\left(\frac1t-\frac1{t^2}\right)dt=3\ln|x+1|+\frac3{x+1}+c\\ -\int\frac{dx}{x^2+2x+1}=\frac1{x+1}+c ...

S=\lim_{n \to \infty} (I_{n-1}+I_{n}+I_{n+1}+\ldots+I_{2n-2}) = \lim_{n\to\infty}\sum_{k=n-1}^{2n-2}I_k=\lim_{n\to\infty}\sum_{k=n-1}^{2n-2}\frac{2}{k+2}\\=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac{2}{k} ...

Turning a comment into an answer: For the first statement, you can remove the for all variables a, b, c, and d part, as the counterexample of [a=1,b=2,c=3,d=4] already states this fact. For the ...