Timf x, bieets
a) \(\frac{-2,6}{12}=\frac{x}{-42}\)
b) \(\frac{1,4}{-5}=\frac{2,1}{x}\)
c) \(\frac{3}{x-1}=\frac{21}{16}\)
d) \(\frac{1,2}{30}=\frac{3x+4}{50}\)
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a) \(-\frac{3}{x}=\frac{15}{7}\)
=> -3.7 = 15x
=> 15x = -21
=> x = -21:15
=> x = -1,4
Vậy x = -1,4
b) \(\frac{x+3}{4}=\frac{5}{20}\)
\(\Rightarrow\frac{x+3}{4}=\frac{1}{4}\)
=> x + 3 = 1
=> x = 1 - 3
=> x = -2
Vậy x = -2
d) \(\frac{x-1}{3}=\frac{x+1}{5}\)
=> 5(x - 1) = 3(x + 1)
=> 5x - 5 = 3x + 3
=> 5x - 3x = 5 + 3
=> 2x = 8
=> x = 8:2
=> x = 4
Vậy x = 4
\(a,\frac{-3}{x}=\frac{15}{7}\)
=> -21 = 15x
=> \(x=-\frac{21}{15}=-\frac{7}{5}\)
b,
\(\frac{x+3}{4}=\frac{5}{20}\)
=> \(\frac{5(x+3)}{20}=\frac{5}{20}\)
=> 5\((x+3)\)= 5
=> x + 3 = 1
=> x = -2
\(c,\frac{1,2}{30}=\frac{3x+4}{50}\)
=> \(\frac{\frac{12}{10}}{30}=\frac{3x+4}{50}\)
=> \(\frac{\frac{6}{5}}{30}=\frac{3x+4}{50}\)
=> \(\frac{2}{50}=\frac{3x+4}{50}\)
=> 3x + 4 = 2
=> 3x = -2
=> x = -2/3
\(d,\frac{x-1}{3}=\frac{x+1}{5}\)
=> 5[x - 1] = 3[x + 1]
=> 5x - 5 = 3x + 3
=> 5x - 5 - 3x = 3
=> 5x - 3x - 5 = 3
=> 2x = 8
=> x = 4
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
c) x=-2 nha
d) =\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+......+\(\frac{1}{11.12}\)
=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{11}\)-\(\frac{1}{12}\)
=\(\frac{1}{5}\)-\(\frac{1}{12}\)= \(\frac{7}{60}\)
\(\Rightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}-\frac{1}{3}\)
\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{-2}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{-2}{21}\div\frac{1}{42}\)
\(\Rightarrow\frac{x}{3}=-4\)
\(\Rightarrow\frac{x}{3}=\frac{-12}{3}\)
\(\Rightarrow x=-12\)
\(\Leftrightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\frac{5}{14}\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\frac{x}{3}=\frac{2}{3}\)
\(\Leftrightarrow x=2\)
\(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{7}{14}-\frac{2}{14}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{5}{14}.\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{21}:\frac{5}{14}\)
\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( 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) =
\(\frac{-2,6}{12}=\frac{x}{-42}\Leftrightarrow x=\frac{\left(-2,6\right)\left(-42\right)}{12}=9,1\)
\(\frac{1,4}{-5}=\frac{2,1}{x}\Leftrightarrow x=\frac{2,1.\left(-5\right)}{1,4}=-7,5\)
\(\frac{3}{x-1}=\frac{21}{16}\Leftrightarrow x-1=\frac{3.16}{21}=\frac{16}{7};x=\frac{23}{7}\)
\(\frac{1,2}{30}=\frac{3x+4}{50}\Leftrightarrow3x+4=\frac{50.1,2}{30}=2;3x=-1;x=-\frac{1}{3}\)
a) \(\frac{-2,6}{12}=\frac{x}{42}\)
\(x=\frac{-2,6.\left(-42\right)}{12}=9,1\)
b) \(\frac{1.4}{-5}=\frac{2,1}{x}\)
\(x=\frac{-5.2,1}{1,4}=-7,5\)
c) \(\frac{3}{x-1}=\frac{21}{16}\)
\(x-1=\frac{3.16}{21}=\frac{16}{7}\)
\(x=\frac{16}{7}+1=\frac{23}{7}\)
d) \(\frac{1,2}{30}=\frac{3x+4}{50}\)
\(3x+4=\frac{12.50}{30}=20\)
\(3x=20-4=16\)
\(x=16:3=\frac{16}{3}\)