\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}\)

\(=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^{10}.2^{19}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{29}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{29}.3^{18}\left(5.3-7\right)}\)

\(=\frac{10-9}{15-7}\)

\(=\frac{1}{8}\)

tui 

\(A=x^2-12x+18\)

\(A=x^2-2.x.6+36-36+18\)

\(A=\left(x-6\right)^2-18\)

Vì \(\left(x-6\right)^2\ge0\)

Nên \(\left(x-6\right)^2-18\ge-18\)

Vậy \(A_{MIN}=-18\Leftrightarrow x-6=0\Leftrightarrow x=6\)

\(\Leftrightarrow x^2+5x-6=0\)

\(\Leftrightarrow x^2-x+6x-6=0\)

\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)

\(\nghiempt{\Leftrightarrow\begin{cases}x-1=0\\x+6=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-6\end{array}\right.\)

Vậy tập nghiệm của PT: \(S=\left\{1;-6\right\}\)

c.\(\sqrt{2x-1}=x-2\)

ĐKXĐ: \(\begin{cases}x\ge2\\2x-1\ge0\end{cases}\Leftrightarrow\begin{cases}x\ge2\\x\ge\frac{1}{2}\end{cases}}\)

\(\Rightarrow x\ge2\)

\(\Leftrightarrow2x-1=\left(x-2\right)^2\)

\(\Leftrightarrow2x-1-x^2+4x-4=0\)

\(\Leftrightarrow-x^2+6x-5=0\)

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\5-x=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=1\left(loại\right)\\x=5\left(nhận\right)\end{array}\right.\)

Vậy tập nghiệm của PT: \(S=\left\{5\right\}\)

a.Không làm được

b.\(\sqrt{x^2+5x+3}=3\)

\(\Leftrightarrow x^2+5x+3=9\)

\(x^2-2x-4y^2-4y\)

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

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

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

Ta có: \(2x=3y\Leftrightarrow x=\frac{3y}{2}\)

Ta lại có: \(x+y=12\)

Hay \(\frac{3y}{2}+y=12\)

\(\Leftrightarrow\frac{5y}{2}=12\)

\(\Leftrightarrow y=\frac{12\times2}{5}\)

\(\Leftrightarrow y=4.8\)

Vậy \(x=\frac{3y}{2}=7.2\)

Ta có :\(3^{n+2}+3^n\)

\(=3^n.3^2+3^n\)

\(=3^n\left(3^2+1\right)\)

\(=3^n.10\)

Vì \(10⋮10\)      Nên \(3^n.10⋮10\)

Hay \(3^{n+2}+3^n⋮10\)

\(x^2+5x+8\)

\(=x^2+2x\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+8\)

\(=\left(x+\frac{5}{2}\right)^2+\frac{7}{4}\) 

Vì \(\left(x+\frac{5}{2}\right)^2\ge0\)

Nên \(\left(x+\frac{5}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)

Vậy GTNN của đa thức là: \(\frac{7}{4}\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)