Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
~ Học tốt ~
Bài 1:
1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)
\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)
\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)
\(=3^2=9\)
2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)
\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)
\(=2^7:2^3:\dfrac{1}{2^4}\)
\(=2^4.2^4=256\)
3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)
\(=\dfrac{43}{48}\)
4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)
\(=-3-1+\dfrac{1}{8}\)
\(=-4+\dfrac{1}{8}\\ \)
\(=-\dfrac{31}{8}\)
5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)
Chúc bạn học tốt 
\(\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-0,75:1\dfrac{1}{2}-1,25^2\)
\(=\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-\dfrac{3}{4}:\dfrac{3}{2}-\dfrac{25}{16}\) \(=\left(\dfrac{31}{30}-\dfrac{9}{10}\right).\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2}{15}.\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\left(-\dfrac{1}{50}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{8}{25}\)
Toàn câu dễ nên bạn tự làm đi.
Trong lúc bạn đánh xong bài này thì bạn có thể làm xong rồi đó.
Đừng có ỷ lại vào người khác ,động não lên.
a: \(=\dfrac{-3}{4}\left(31+\dfrac{11}{23}+8+\dfrac{12}{23}\right)=\dfrac{-3}{4}\cdot40=-30\)
b: \(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\)
\(=\dfrac{35}{6}:\dfrac{-175+132}{42}+\dfrac{15}{2}\)
\(=\dfrac{35}{6}\cdot\dfrac{42}{-43}+\dfrac{15}{2}\)
\(=\dfrac{35\cdot7}{-43}+\dfrac{15}{2}\)
\(=\dfrac{-70\cdot7+15\cdot43}{86}=\dfrac{155}{86}\)
c: \(=\dfrac{-7}{5}\left(4+\dfrac{5}{9}+5+\dfrac{4}{9}\right)=\dfrac{-7}{5}\cdot10=-14\)
d: \(=4+\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}\cdot\dfrac{64}{125}\cdot\dfrac{-8}{27}\right)\)
\(=\dfrac{89}{16}+25\cdot\dfrac{-32}{375}\)
\(=\dfrac{89}{16}-\dfrac{32}{15}=\dfrac{823}{240}\)
e: \(=\dfrac{2}{3}-4\cdot\left(\dfrac{2}{4}+\dfrac{3}{4}\right)=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
a)
\(\left(\dfrac{1}{12}\right)^{27}\) và \(\left(\dfrac{1}{12}\right)^{18}\)
Ta có:
\(\left(\dfrac{1}{12}\right)^{27}\)=\(\left(\left(\dfrac{1}{12}\right)^3\right)^9=\left(\dfrac{1}{18}\right)^9\)(1)
\(\left(\dfrac{1}{12}\right)^{18}\)\(=\left(\left(\dfrac{1}{12}\right)^3\right)^6=\left(\dfrac{1}{8}\right)^6\)(2)
Từ (1) và (2) ta có:
\(\left(\dfrac{1}{18}\right)^9>\left(\dfrac{1}{8}\right)^6\)
Vậy \(\left(\dfrac{1}{12}\right)^{27}\)>\(\left(\dfrac{1}{12}\right)^{18}\)
b)
\(\left(10\right)^{20}\) và \(\left(-9\right)^{10}\)
Ta có:
\(10^{20}=\left(\left(10\right)^4\right)^5\) (1)
\(\left(-9\right)^{10}=\left(\left(-9\right)^2\right)^5\) (2)
Từ (1) và (2), ta có:
\(\left(10000\right)^5>\left(81\right)^5\)
Vậy \(\left(10\right)^{20}\)> \(\left(-9\right)^{10}\)
c)
\(64^3\) và \(16^2\)
Ta có:
\(64^3=\left(\left(4^{ }\right)^3\right)^3\)(1)
\(16^2=\left(\left(4\right)^2\right)^2\)(2)
Từ (1) và (2), ta có:
\(\left(\left(4\right)^3\right)^3>\left(\left(4\right)^2\right)^2\)
Vậy \(64^3\)> \(16^2\)
d)
\(\left(-16\right)^3\) và \(\left(-64\right)^2\)
Ta có:
\(\left(-16\right)^3=\left(\left(-4\right)^2\right)^3\)(1)
\(\left(-64\right)^2=\left(\left(-4\right)^3\right)^2=\left(\left(-4\right)^2\right)^3\) (2)
Từ (1) và (2), ta có:
\(16^3=16^3\)
Vậy \(\left(-16\right)^3\)= \(\left(-64\right)^2\)
Chúc bạn học tốt!
chỗ câu a) mình nhầm nha bạn,1/12 mình nhầm mất (1/12)^3=1/1728. Bạn sửa lại sẽ dễ hiểu hơn :))
Bạn sửa hộ mình nhé, cảm ơn!
a, \(\left(0,8.7+0,64\right).\left(1,25.7-\dfrac{4}{5}.1,25\right)\)
\(=6,24.1,25\left(7-\dfrac{4}{5}\right)=7,8.6,2=48,36\)
b, \(\dfrac{1}{4}-\dfrac{\dfrac{1}{4}+\dfrac{1}{9}}{\dfrac{1}{9}}=\dfrac{1}{4}-\dfrac{\dfrac{13}{36}}{\dfrac{1}{9}}=\dfrac{1}{4}-\dfrac{13}{4}=-3\)
c, \(2\dfrac{1}{2}-3,4\left(12\right)-\dfrac{4}{3}+\dfrac{1}{3}.\left(\dfrac{1}{2}+0,5-3\dfrac{1}{2}\right)\)
\(=\dfrac{5}{2}-\dfrac{563}{165}-\dfrac{4}{3}+\dfrac{1}{3}.\left(-\dfrac{5}{2}\right)\)
\(=-\dfrac{301}{330}-\dfrac{4}{3}-\dfrac{5}{6}=-\dfrac{508}{165}\)
Chúc bạn học tốt!!!
5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
=\(4+6-3+5\)
=\(12\)
2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)
=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)
=\(\dfrac{11}{25}.\left(-100\right)\)
=\(-44\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
\(\left(\dfrac{1}{2}\right)^{12}=\left(\dfrac{1}{8}\right)^3\\ \left(\dfrac{1}{3}\right)^9=\left(\dfrac{1}{27}\right)^3\\ Ta\text{ }có:\dfrac{1}{8}>\dfrac{1}{27}\\ Vậy:\left(\dfrac{1}{2}\right)^{12}>\left(\dfrac{1}{3}\right)^9\)
\(\left(\dfrac{1}{2}\right)^{12}=\dfrac{1}{4096};\left(\dfrac{1}{3}\right)^9=\dfrac{1}{19683}\\ \Rightarrow\dfrac{1}{4096}>\dfrac{1}{19683}\\ \Rightarrow\left(\dfrac{1}{2}\right)^{12}>\left(\dfrac{1}{3}\right)^9\)