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a) \(12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)
\(=12\cdot\dfrac{4}{9}+\dfrac{4}{3}\)
\(=\dfrac{12\cdot4}{9}+\dfrac{4}{3}\)
\(=\dfrac{16}{3}+\dfrac{4}{3}\)
\(=\dfrac{16+4}{3}\)
\(=\dfrac{20}{3}\)
b) \(\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}\cdot\left(-\dfrac{1}{2}\right)^2\right]\)
\(=\dfrac{9}{4}-\left(\dfrac{1}{2}:2-9\cdot\dfrac{1}{4}\right)\)
\(=\dfrac{9}{4}-\left(\dfrac{1}{4}-9\cdot\dfrac{1}{4}\right)\)
\(=\dfrac{9}{4}-\dfrac{1}{4}\cdot\left(1-9\right)\)
\(=\dfrac{9}{4}+\dfrac{8}{4}\)
\(=\dfrac{17}{4}\)
c) \(\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\)
\(=-\dfrac{1}{12}:\dfrac{5}{11}+\dfrac{1}{12}\)
\(=\dfrac{1}{12}\cdot-\dfrac{11}{5}+\dfrac{1}{12}\)
\(=\dfrac{1}{12}\cdot\left(-\dfrac{11}{5}+1\right)\)
\(=\dfrac{1}{12}\cdot-\dfrac{6}{5}\)
\(=-\dfrac{1}{10}\)
d) \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)
\(=-\dfrac{1}{15}+\dfrac{4}{9}:\left(2+\dfrac{2}{3}\right)-\dfrac{5}{6}\)
\(=-\dfrac{1}{15}+\dfrac{4}{9}:\dfrac{8}{3}-\dfrac{5}{6}\)
\(=-\dfrac{9}{10}+\dfrac{1}{6}\)
\(=-\dfrac{11}{15}\)
e) \(\dfrac{3^7\cdot8^6}{6^6\cdot\left(-2\right)^{12}}\)
\(=\dfrac{3^7\cdot\left(2^3\right)^6}{2^6\cdot3^6\cdot2^{12}}\)
\(=\dfrac{3^7\cdot2^{18}}{2^{6+12}\cdot3^6}\)
\(=\dfrac{2^{18}\cdot3^7}{2^{18}\cdot3^6}\)
\(=3^{7-6}\)
\(=3\)
\(a,12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\\ =12\cdot\dfrac{4}{9}+\dfrac{4}{3}\\ =\dfrac{16}{3}+\dfrac{4}{3}\\ =\dfrac{20}{3}\\ b,\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}.\left(-\dfrac{1}{2}\right)^2\right]\\ =\dfrac{9}{4}-\left(\dfrac{1}{2}\cdot\dfrac{1}{2}-9\cdot\dfrac{1}{4}\right)\\ =\dfrac{9}{4}-\left(\dfrac{1}{4}-\dfrac{9}{4}\right)\\ =\dfrac{9}{4}-\left(-\dfrac{8}{4}\right)\\ =\dfrac{17}{4}\)
\(c,\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\\ =\left(-\dfrac{9}{12}+\dfrac{8}{12}\right)\cdot\dfrac{11}{5}+\left(-\dfrac{3}{12}+\dfrac{4}{12}\right)\\ =-\dfrac{1}{12}\cdot\dfrac{11}{5}+\dfrac{1}{12}\\ =-\dfrac{11}{60}+\dfrac{1}{12}\\ =-\dfrac{1}{10}\)
\(d,\dfrac{-1^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left(-\dfrac{5}{6}\right)\\ =-\dfrac{1}{15}+\dfrac{4}{9}\cdot\dfrac{3}{8}+\dfrac{5}{6}\\ =-\dfrac{1}{15}+\dfrac{1}{6}+\dfrac{5}{6}\\ =\dfrac{1}{10}+\dfrac{5}{6}\\ =\dfrac{14}{15}\)
`e,` Không hiểu đề á c: )
Bài 1:
ta có: 333<3333; 444<4444
=> 333444<33334444
Bài 2:
\(A=\frac{21^5}{81}=\frac{\left(3.7\right)^5}{3^4}=\frac{3^5.7^5}{3^4}=3.7^5=50421\)
\(B=\frac{3^3.\left(0,5\right)^5}{\left(1,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{\left(3.0,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{3^5.\left(0,5\right)^5}=\frac{1}{3^2}=\frac{1}{9}\)
\(C=2^2.\frac{1}{128}.45.2^{-6}=\frac{2^2.45}{128.64}=\frac{2^2.45}{2^7.2^6}=\frac{45}{2^{11}}=\frac{45}{2048}\)
\(D=\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+2^2.3^3+3^3}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}\)\(=3^3.\left(-1\right)=-27\)
a) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2+0,5\)
\(=2,5.\)
b) \(\sqrt{\frac{25}{81}}:2\frac{2}{5}-4\frac{5}{9}:2\frac{2}{5}\)
\(=\frac{5}{9}:\frac{12}{5}-\frac{41}{9}:\frac{12}{5}\)
\(=\left(\frac{5}{9}-\frac{41}{9}\right):\frac{12}{5}\)
\(=\left(-4\right):\frac{12}{5}\)
\(=-\frac{5}{3}.\)
c) \(6.\left(-\frac{-1}{2}\right)^2+\frac{3}{5}\)
\(=6.\left(\frac{1}{2}\right)^2+\frac{3}{5}\)
\(=6.\frac{1}{4}+\frac{3}{5}\)
\(=\frac{3}{2}+\frac{3}{5}\)
\(=\frac{21}{10}.\)
Chúc bạn học tốt!
Tìm x:
a)-23+0,5x=1,5
-8+0,5x = 1,5
0,5x = 1,5-(-8) = 9,5
x = 9,5:0,5 = 19
b)(−3)x81=−27
(-3)x:81 =-27
(-3)x = -27.81 = -2187
(-3)x = (-3)7
=> x=7
c)112.x−4=0,5
1,5.x = 0,5+4 = 4,5
x = 4,5:1,5 = 3
d)123:x4=6:0,3
\(\frac{5}{3}\):\(\frac{x}{4}\) = 20
\(\frac{x}{4}\) = \(\frac{5}{3}\):20 = \(\frac{1}{12}\)
=> x:4 = \(\frac{1}{12}\)
x = \(\frac{1}{12}\).4 = \(\frac{1}{3}\)
a) \(\frac{\left(-3\right)^x}{81}=-27\)
\(\Rightarrow\left(-3\right)^x=\left(-27\right).81\)
\(\Rightarrow\left(-3\right)^x=-2187\)
\(\Rightarrow\left(-3\right)^x=\left(-3\right)^7\)
\(\Rightarrow x=7\)
Vậy \(x=7.\)
b) \(1\frac{1}{2}.x-4=0,5\)
\(\Rightarrow1\frac{1}{2}.x=0,5+4\)
\(\Rightarrow1\frac{1}{2}.x=\frac{9}{2}\)
\(\Rightarrow\frac{3}{2}.x=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{2}:\frac{3}{2}\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
\(\frac{9999^2}{3333^2}+\left(0,5\right)^4.\left(-2\right)^4-\sqrt[]{81}\\ =9+0,0625.16-9\\ =9+1-9\\ =1\)
Tính:
\(\frac{9999^2}{3333^2}+\left(0,5\right)^4.\left(-2\right)^4-\sqrt{81}\)
\(=9+\frac{1}{16}.16-9\)
\(=9+1-9\)
\(=10-9\)
\(=1.\)
Chúc bạn học tốt!
\(\frac{\left(-3\right)^x}{81}=-27\)
(-3)x=(-27).81
(-3)x=(-3)3.(-3)4=(-3)7
=>x=7
\(1\frac{1}{2}x-4=0,5\)
\(\frac{3}{2}x=0,5+4=4,5\)
x=4,5:3/2=3
2x-1=16
2x-1=24
=>x-1=4
=>x=5
(3/2)^2-[0,5:2-căn bậc hai của 81 .(1/2)^2]
=9/4-[0,25-9.1/4]
=9/4-(0,25-9/4)
=9/4-1/4+9/4
=17/4