Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
13)\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}\right)+\dfrac{45}{8}\left(\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}+\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left[\left(\dfrac{4}{15}+\dfrac{11}{15}\right)-\left(\dfrac{7}{8}-\dfrac{9}{8}\right)\right]\)
=\(\dfrac{45}{8}.\dfrac{5}{4}\)=\(\dfrac{225}{32}\)
14)\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right):\dfrac{7}{15}\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right).\dfrac{15}{7}\)
=\(\dfrac{15}{7}\left[\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right)\right]\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}-\dfrac{2}{5}+\dfrac{27}{4}\right)\)
=\(\dfrac{15}{7}.\dfrac{35}{8}\)=\(\dfrac{75}{8}\)
a) \(9^8=\left(3^2\right)^8=3^{16}\)
\(27^3=\left(3^3\right)^3=3^9\)
\(81^2=\left(3^4\right)^2=3^8\)
\(\Leftrightarrow3^{16}:\left(3^9.3^8\right)\)
\(\Leftrightarrow3^{16}:3^{17}\)
\(\Leftrightarrow3^{-1}=\frac{1}{3}\)
a . ( -1/3 ) . 9/11 + ( -8/9 ) . 27
= 9/3 . ( -1/11 ) + 27/9 . ( -8 )
= 3 . ( -1/11 ) + 3 . ( -8 )
= 3 . ( -1/11 + ( -8 ) )
= 3 . ( -89/11 )
= -267/11
b . ( 1/2 - 13/14 ) : 5/7 - ( - - 2/21 + 1/7 ) : 5/7
= -3/7 : 5/7 - 5/21 : 5/7
= ( -3/7 - 5/21 ) : 5/7
= -2/3 : 5/7
= -14/15
Bài 1:
a.\(12+\left\{45-\left[36:\left(12-9\right)^2\right]\right\}\)
\(=12+\left\{45-\left[36:3^2\right]\right\}\)
\(=12+\left\{45-\left[36:9\right]\right\}\)
\(=12+\left\{45-4\right\}\)
\(=12+41\)
\(=53\)
b.\(24:\left[48-\left(42:7\right)^2\right]\)
\(=24:\left[48-6^2\right]\)
\(=24:\left[48-36\right]\)
\(=24:12=2\)
Bài 2 :
C1 : { 0;1;2;3;4;5;6;7;8;9;10}
C2 : \(\left\{x\in N|x\le10\right\}\)
pha ngoac ra la dc
con bai 2 C1 {1:2:3:4:5:6:7:8:9:0}
C2 {nt huoc N /n<10}
a) \(\left(\frac{1}{81}\right)^x\cdot27^{2x}=\left(-9\right)^4\)
\(\Leftrightarrow\frac{1}{3^{4x}}\cdot3^{6x}=9^4\)
\(\Leftrightarrow\frac{3^{6x}}{3^{4x}}=3^8\)
\(\Leftrightarrow3^{2x}=3^8\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)
b) \(5^x\cdot\left(5^3\right)^2=625\)
\(\Leftrightarrow5^{x+6}=5^4\)
\(\Leftrightarrow x+6=4\)
\(\Leftrightarrow x=-2\)
c) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
\(\Leftrightarrow\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\\left(4x-1\right)^{10}=1=\left(\pm1\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=\frac{1}{2}\\x=0\end{matrix}\right.\)
Vậy....
a) \(1+\left(-3\right)+5+\left(-7\right)+9+\left(-11\right)\\ =\left(1+5+7\right)-\left(3+7+11\right)\\ =14-21\\ =-7\)
b) \(\left(-2\right)+4+\left(-6\right)+8+\left(-10\right)+12\\ =\left(4+8+12\right)-\left(2+6+10\right)\\ =24-18\\ =6\)
\(A=\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)
\(A=\frac{38}{45}-\frac{8}{45}+\frac{17}{51}+\frac{3}{11}\)
\(A=\left(\frac{38}{45}-\frac{8}{45}\right)+\frac{1}{3}+\frac{3}{11}\)
\(A=\left(\frac{2}{3}+\frac{1}{3}\right)+\frac{3}{11}\)
\(A=1+\frac{3}{11}\)
\(A=\frac{11}{11}+\frac{3}{11}=\frac{14}{11}\)
\(B=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
\(B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}...\frac{100}{99}\)
\(B=\frac{3.4.5.6...100}{2.3.4.5...99}\)
\(B=\frac{100}{2}\)
\(B=50\)
a) Cách 1: \(73 - \left( {2 - 9} \right) = 73 - 2 + 9 = 80\);
Cách 2: \(73 - \left( {2 - 9} \right) = 73 -(-7)=73+7 = 80\)
b) Cách 1: \(\left( { - 45} \right) - \left( {27 - 8} \right) = \left( { - 45} \right) - 27+8 =-72+8=- 64\)
Cách 2: \(\left( { - 45} \right) - \left( {27 - 8} \right) = \left( { - 45} \right) - 19 = - 64\)