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a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
Bài 1 và Bài 2 dễ, bn có thể tự làm được!
Bài 3:
a) ta có: 1020 = (102)10 = 10010
=> 10010>910
=> 1020>910
b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)
(-3)50 = 350 = (35)10= 24310
=> 12510 < 24310
=> (-5)30 < (-3)50
c) ta có: 648 = (26)8= 248
1612 = ( 24)12 = 248
=> 648 = 1612
d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(\frac{1}{2}-x=\frac{57}{28}\)
\(x=-\frac{43}{28}\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Bài 1
\(\frac{12^4\cdot10^2}{3^4\cdot4^5\cdot5^2}\)
\(=\frac{3^4\cdot2^8\cdot2^2\cdot5^2}{3^4\cdot2^{10}\cdot5^2}=1\)
Bài 2
\(\left(2^3:4\right)\cdot2^{x+1}=64\)
\(2\cdot2^{x+1}=64\)
\(2^{x+2}=2^6\)
\(x+2=6\)
\(x=4\)
\(b,\left|3-2x\right|-3=\left(-3\right)\)
\(\left|3-2x\right|=0\)
\(3-2x=0\)
\(2x=3\)
\(x=\frac{3}{2}\)
\(c,4-3x=-\frac{5}{3}\)
\(-3x=-4-\frac{5}{3}\)
\(3x=\frac{17}{3}\)
\(x=\frac{17}{9}\)
\(d,\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\frac{1}{4}:x=-\frac{7}{20}\)
\(x=-\frac{5}{7}\)
Bài 3
\(a,4\frac{1}{3}:x=-6:0,3\)
\(\frac{13}{3}:x=-20\)
\(x=-\frac{13}{60}\)
\(b,\left|3-\frac{3}{4}\right|-9=\frac{1}{2}\)
không có x đề sai
1.\(\frac{12^4\cdot10^2}{3^4\cdot4^5\cdot5^2}=\frac{2^8\cdot3^4\cdot2^2\cdot5^2}{3^4\cdot2^{10}\cdot5^2}=\frac{2^{10}\cdot3^4\cdot5^2}{3^4\cdot2^{10}\cdot5^2}=1\)
2. a,\(\left(2^3:4\right)\cdot2^{x+1}=64\)
\(\left(8:4\right)\cdot2^{x+1}=64\)
\(2\cdot2^{x+1}=64\)
\(2^{x+2}=2^6\)
\(\Rightarrow x+2=6\)
\(x=4\)
Vậy....
b, \(\left|3-2x\right|-3=\left(-3\right)\)
\(\left|3-2x\right|=\left(-3\right)+3\)
\(\left|3-2x\right|=0\)
\(3-2x=0\)
\(2x=3\)
\(x=\frac{3}{2}\)
c, \(4-3x=-\frac{5}{3}\)
\(3x=4-\left(-\frac{5}{3}\right)\)
\(3x=\frac{17}{3}\)
\(x=\frac{17}{9}\)
Vậy...
d, \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)
\(\frac{1}{4}:x=-\frac{7}{20}\)
\(x=\frac{1}{4}:\left(-\frac{7}{20}\right)\)
\(x=-\frac{5}{7}\)
Vậy...
3. a, \(4\frac{1}{3}:x=-6:0,3\)
\(\frac{13}{3}:x=-20\)
\(x=\frac{13}{3}:\left(-20\right)\)
\(x=-\frac{13}{60}\)
Vậy...
b, x đâu bn 'v' ?