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\(\frac{a}{27}=\frac{-5}{9}=\frac{-45}{b}\)
\(=>\frac{a.3}{81}=\frac{-45}{81}=\frac{-45}{b}\)
=>b=81
a.3=-45
a=-15
Vậy a=-15;b=81
1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
a, \(x\) \(\times\) \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) + \(\dfrac{3}{4}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{19}{12}\) : \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{19}{6}\)
b, \(x\) : \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\)
\(x\): \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) + \(\dfrac{3}{4}\)
\(x\) : \(\dfrac{1}{2}\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{19}{12}\) \(\times\) \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{19}{24}\)
c, \(x\) \(\times\) \(\dfrac{3}{4}\) + \(x\) \(\times\) \(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
\(x\) \(\times\) ( \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\)) = \(\dfrac{7}{8}\)
\(x\) \(\times\) 1 = \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{7}{8}\)
d, \(x\times\) \(\dfrac{3}{4}\) - \(x\) \(\times\) \(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
\(x\) \(\times\) ( \(\dfrac{3}{4}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{8}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{7}{8}\) : \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{7}{4}\)
Bài 1:
a)12,5 x (-5/7) + 1,5 x (-5/7)
=-5/7*(12,5+1,5)
=-5/7*14
=-10
b)(-1/4) x (6|2/11) + 3|9/11 x (-1/4)
=-1/4*(68/11+42/11)
=-1/4*10
=-5/2
c tương tự
d)\(\frac{9^8\cdot4^3}{27^4\cdot6^5}=\frac{\left(3^2\right)^8\cdot\left(2^2\right)^3}{\left(3^3\right)^4\cdot\left(2\cdot3\right)^5}=\frac{3^{16}\cdot2^6}{3^{12}\cdot2^5\cdot3^5}=\frac{3^{16}\cdot2^5\cdot2}{3^{16}\cdot3^1\cdot2^5}=\frac{2}{3}\)
Bài 2:
a)Ta có:
2800=(28)100=256100
8200=(82)100=64100
Vì 256100>64100 =>2800>8200
b)Ta có:
1245=(123)15=172815
Vì 62515<172815 =>62515<1245
a)
\(x+\dfrac{5}{6}=7,5\\ \Rightarrow x+\dfrac{5}{6}=\dfrac{15}{2}\\ \Rightarrow x=\dfrac{15}{2}-\dfrac{5}{6}\\ \Rightarrow x=\dfrac{20}{3}\)
Vậy...
b)
\(x.\dfrac{3}{5}-1\dfrac{1}{2}=\dfrac{5}{6}\\ \Rightarrow x.\dfrac{3}{5}-\dfrac{3}{2}=\dfrac{5}{6}\\ \Rightarrow x.\dfrac{3}{5}=\dfrac{5}{6}+\dfrac{3}{2}\\ \Rightarrow x.\dfrac{3}{5}=\dfrac{7}{3}\\ \Rightarrow x=\dfrac{7}{3}:\dfrac{3}{5}\\ \Rightarrow x=\dfrac{35}{9}\)
Vậy...
c) Đề bài thiếu (\(x:\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{3}=?\))
d)
\(12,5-x=6,75\\ \Rightarrow\dfrac{25}{2}-x=\dfrac{27}{4}\\ \Rightarrow x=\dfrac{25}{2}-\dfrac{27}{4}\\ \Rightarrow x=\dfrac{23}{4}\)
Vậy...