a/\(x^2+8x+17=\left(x^2+8x+16\right)+1=\left(x+4\right)^2+1\)

\(\left(x+4\right)^2\ge0\Rightarrow\left(x+4\right)^2+1\ge1>0\)

b/\(x^2-10x+29=\left(x^2-10x+25\right)+4=\left(x-5\right)^2+4\)

\(\left(x-5\right)^2\ge0\Rightarrow\left(x-5\right)^2+4\ge4>0\)

c/

Ta có 

\(\left(2m-a\right)^2+\left(3m-b\right)^2+\left(3m-c\right)^2=\)

\(=4m^2-4ma+a^2+9m^2-6mb+b^2+9m^2-6mc+c^2=\)

\(=22m^2-2m\left(2a+3b+3c\right)+a^2+b^2+c^2=\)

\(=22m^2-2m.11m+a^2+b^2+c^2=a^2+b^2+c^2\)

đúng nha bạn

\(\left(x+y\right)^2=x^2+y^2\)

\(\Leftrightarrow x^2+2xy+y^2=x^2+y^2\)

\(\Leftrightarrow2xy=0\)

Vậy điều trên chỉ đúng khi một trong hai số bằng 0

\(35^2-15^2\)

\(=\left(35+15\right)\left(35-15\right)\)

\(=50.20\)

\(=1000\)

\(x+5x^2=0\)

\(\Leftrightarrow x\left(5x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)

\(25x^2-9=0\)

\(25x^2=9\)

\(x^2=\frac{25}{9}\)

\(x^2=\left(\frac{5}{3}\right)^2=\left(-\frac{5}{3}\right)^2\)

\(\Rightarrow x=\frac{5}{3};x=-\frac{5}{3}\)

\(25x^2-9=0\)

\(\Leftrightarrow\)\(25x^2=0+9\)

\(\Leftrightarrow\)\(25x^2=9\)

\(\Leftrightarrow\)\(x^2=\frac{9}{25}\)

\(\Leftrightarrow\)\(x=-\frac{3}{5};\frac{3}{5}\)

Đặt \(n=2k\)

\(\Rightarrow A=\frac{8k^3}{24}+\frac{4k^2}{8}+\frac{2k}{12}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}=\frac{2k^3+3k^2+k}{6}=\frac{k\left(2k^2+3k+1\right)}{6}=\)

\(=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\)

A nguyên khi \(k\left(k+1\right)\left(2k+1\right)⋮6\) Tức là \(k\left(k+1\right)\left(2k+1\right)\) đồng thời chia hết cho 2 và 3

+ Với k chẵn \(\Rightarrow k⋮2\Rightarrow k\left(k+1\right)\left(2k+1\right)⋮2\)

+ Với k lẻ \(\Rightarrow k+1⋮2\Rightarrow k\left(k+1\right)\left(2k+1\right)⋮2\)

\(\Rightarrow k\left(k+1\right)\left(2k+1\right)⋮2\forall k\in N\)

+ Nếu \(k⋮3\Rightarrow k\left(k+1\right)\left(2k+1\right)⋮3\)

+ Nếu k chia 3 dư 1 \(\Rightarrow k-1⋮3\Rightarrow2k-2⋮3\Rightarrow2k-2+3=2k+1⋮3\)

+ Nếu k chia 3 dư 2 \(\Rightarrow k+1⋮3\Rightarrow k\left(k+1\right)\left(2k+1\right)⋮3\)

\(\Rightarrow k\left(k+1\right)\left(2k+1\right)⋮3\forall k\in N\)

\(\Rightarrow k\left(k+1\right)\left(2k+1\right)⋮6\forall k\in N\)

Điều này chứng tỏ rằng A là số nguyên với mọi n chẵn