Tính giá trị của biểu thức:
P=3^11.11+3^11.21/3^9.2^5
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\(\dfrac{3^{11}.11+3^{11}.21}{3^9.2^5}\)
= \(\dfrac{3^{11}.\left(11+21\right)}{3^932}\)
= \(\dfrac{3^{11}32}{3^9.32}\)
= 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
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\) \(=\frac{3^{10}.16}{3^9.2^4}=\frac{3^{10}.2^4}{3^9.2^4}=3\)
\(P=\left(x-2\right)^2+\left|y-x\right|+3\)
\(\left(x-2\right)^2>=0\forall x\)
\(\left|y-x\right|>=0\forall x,y\)
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 \(x=-\dfrac{3}{2}\)
\(P=\frac{3^{11}\cdot11+3^{11}\cdot21}{3^9\cdot2^5}=\frac{3^{11}\cdot\left(11+21\right)}{3^9\cdot32}=\frac{3^{11}\cdot32}{3^9\cdot32}=3^2=9\)
\(P=\frac{3^{11}.11+3^{11}.21}{3^9.2^5}\)
\(=\frac{3^{11}.\left(11+21\right)}{3^9.32}\)
\(=\frac{3^{11}.32}{3^9.32}\)
\(=3^2=9\)