\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)

\(20+21+22+23+...+29+30=\frac{\left(20+30\right)\cdot11}{2}=275\)

a)\(ĐK:x\ne-1;x\ne1;x\ne-\frac{1}{2}\)

b) \(A=\left(\frac{1}{x+1}-\frac{2}{x-1}-\frac{x+5}{1-x^2}\right):\frac{2x+1}{x^2-1}\)

\(=\left[\frac{x-1-\left(x+1\right)+x+5}{x^2-1}\right]\cdot\frac{x^2-1}{2x+1}\)

\(=\frac{x-1-x-1+x+5}{x^2-1}\cdot\frac{x^2-1}{2x+1}\)

\(=\frac{x+3}{2x+1}\)

Xét ΔABC vuông tại A(gt)

=>\(BC^2=AB^2+AC^2\)(theo định lý ptago)

=>\(BC^2=10^2+8^2=164\)

=>\(BC\approx12,8\)

Áp dụng hệ thức giữa cạnh góc vuông và hình chiếu của nó trên cạnh huyền ta có:

\(AB^2=BH\cdot BC\Rightarrow BH=\frac{AB^2}{BC}=\frac{8^2}{12,8}=5\)

\(AC^2=HC\cdot BC\Rightarrow HC=\frac{AC^2}{BC}=\frac{10^2}{12,8}\approx7,8\)

Áp dụng hệ thức liên quan tới đường cao ta có:

\(AH^2=BH\cdot CH=5\cdot7,8=39\)

\(\Rightarrow AH\approx6,2\)

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

Vì: \(\left(\sqrt{a}-\frac{1}{2}\right)^2\ge0\) với mọi \(a\ge0\)

=> \(\left(\sqrt{a}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy GTNN của biểu thức trên là \(\frac{3}{4}\) khi \(x=\frac{1}{4}\)

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

Vì: \(\left(\sqrt{x}-1\right)^2\ge0\) với mọi \(x\ge0\)

=> \(-\left(\sqrt{x}-1\right)^2\le0\)

=> \(-\left(\sqrt{x}-1\right)^2-2\le-2\)

Vậy GTLN của biểu thức trên là -2 khi x=1(tm)

a) \(45+x^3-5x^2-9x\)

\(=\left(x^3-5x^2\right)-\left(9x-45\right)\)

\(=x^2\left(x-5\right)-9\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2-9\right)=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)

a) \(x^4-10x^3+25x^2=0\)

\(\Leftrightarrow x^2\left(x^2-10x+25\right)=0\)

\(\Leftrightarrow x^2\left(x-5\right)^2=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2=0\\\left(x-5\right)^2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)

b) \(x^3+3x^2+3x+1=0\)

\(\Leftrightarrow\left(x+1\right)^3=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

a) \(\left(4x-8\right)\left(\frac{1}{2}-x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}4x-8=0\\\frac{1}{2}-x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{1}{2}\end{array}\right.\)

b) \(2x^2-32=0\)

\(\Leftrightarrow2\left(x^2-16\right)=0\)

\(\Leftrightarrow x^2-16=0\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-4\end{array}\right.\)

3) \(\left(x-1\right)\left(x+1\right)^2-\left(2x-1\right)\left(x^2+2x+1\right)=0\)

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\\left(x+1\right)^2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)