\(\dfrac{3^{11}.11+3^{11}.21}{3^9.2^5}\)

=  \(\dfrac{3^{11}.\left(11+21\right)}{3^932}\)

= \(\dfrac{3^{11}32}{3^9.32}\)

= 3²

= 9 

\(=5x^2-4x^2+3x^2-6x=4x^2-6x\)

\(=4\cdot\left(-\dfrac{3}{2}\right)^2-6\cdot\dfrac{-3}{2}\)

\(=4\cdot\dfrac{9}{4}+6\cdot\dfrac{3}{2}\)

=9+9

=18

\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(=\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)

\(=\frac{3^{10}.16}{3^9.2^4}\)

\(=\frac{3^{10}.2^4}{3^9.2^4}=3\)

3^10.(11+5)      =3.16

3^9 . 2^4            1.16                                                                                                                                                                                                     bỏ số 16 thì được kết quả bằng 3

Ta có : \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3^{10}.16}{3^9.16}=3\)3

\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\) \(=\frac{3^{10}.16}{3^9.2^4}=\frac{3^{10}.2^4}{3^9.2^4}=3\)

Tớ đánh sai nha, đáp án là 3.

GTNN là -3 khi x =-2

\(P=\left(x-2\right)^2+\left|y-x\right|+3\)

\(\left(x-2\right)^2>=0\forall x\)

\(\left|y-x\right|>=0\forall x,y\)

Do đó: \(\left(x-2\right)^2+\left|y-x\right|>=0\forall x,y\)

=>\(\left(x-2\right)^2+\left|y-x\right|+3>=3\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\y-x=0\end{matrix}\right.\)

=>x=y=2

\(P=\dfrac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\dfrac{1}{x^2+2x+3}=3+\dfrac{1}{\left(x+1\right)^2+2}\le3+\dfrac{1}{2}=\dfrac{7}{2}\)

\(P_{max}=\dfrac{7}{2}\) khi \(x=-1\)

\(M=\dfrac{2\left(x^2+3x+3\right)+1}{x^2+3x+3}=2+\dfrac{1}{x^2+3x+3}=2+\dfrac{1}{\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}}\le2+\dfrac{1}{\dfrac{3}{4}}=\dfrac{10}{3}\)

\(M_{max}=\dfrac{10}{3}\) khi \(x=-\dfrac{3}{2}\)