Cậu là boy hay girl vậy không lẽ là giới tính thứ 3

Cậu gửi lại thui đấy chứ. Gửi thêm mấy cái khác đê

Bạn ơi mình tìm GTLN nhé

\(\dfrac{2^{15}\cdot9^4}{6^8\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^8\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^8\cdot3^8\cdot2^9}=\dfrac{2^{15}\cdot3^8}{2^{17}\cdot3^8}=\dfrac{1}{2^2}=\dfrac{1}{4}\)

Một acquy có suất điện động ξ = 2 V , có dung lượng q=240A.h. Điện năng của acquy là

A. 480 J

B. 1728 kJ

C. 480 kJ

D. 120 J

Ta có : \(\dfrac{22}{33}=\dfrac{2\cdot11}{3\cdot11}=\dfrac{2}{3}=\dfrac{2\cdot30}{3\cdot30}=\dfrac{60}{90}\)

Mà \(\dfrac{60}{90}< \dfrac{60}{71}\Rightarrow\dfrac{22}{33}< \dfrac{60}{71}\)

Vậy \(\dfrac{22}{33}< \dfrac{60}{71}\)

\(B=\left(x^4+5\right)^2\\ \text{Ta có : }x^4\ge0\\ \Rightarrow x^4+5\ge5\\ \Rightarrow\left(x^4+5\right)^2\ge25\\ \text{Dấu }"="\text{ xảy ra khi : }x^4=0\Leftrightarrow x=0\\ \text{Vậy }B_{\left(Min\right)}=25\text{ khi }x=0\\ \)

\(C=\left(x-1\right)^2+\left(y+2\right)^2\\ \text{Ta có: }\left(x-1\right)^2\ge0\\ \left(y+2\right)^2\ge0\\ \Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\\ \text{Dấu }"="\text{ xảy ra khi : }\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\ \text{ Vậy }C_{\left(Min\right)}=0\text{ khi }x=1\text{ và }y=-2\)

Câu 2 :

\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)

Sorry . Mình chỉ biết đến đây thôi

Câu 1 :

\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)

\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)

\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)