Question

a, x - 128 = ( 4\(\frac{20}{21}\) - 5 ) : ( \(\frac{4141}{4242}\) - 1 ) : ( \(\frac{636363}{646464}\) - 1)

 b,  A= \(\frac{-7}{4}\) . ( \(\frac{33}{12}\) + \(\frac{3333}{2020}\) + \(\frac{333333}{303030}\) + \(\frac{33333333}{42424242}\) )

Answer 1

Ta có : \(x-128=\left(4\frac{20}{21}-5\right)\left(\frac{4141}{4242}-1\right):\left(\frac{636363}{646464}-1\right)\\ =>x-128=\left(-\frac{1}{21}\right):\left(-\frac{1}{42}\right):\left(-\frac{1}{64}\right)\\ =>x-128=-128\\ =>x=0\)

Answer 2

a, Ta có x - 128 =( \(4\frac{20}{21}-5\)):\(\left(\frac{4141}{4242}-1\right):\left(\frac{636363}{646464}-1\right)\)

\(\Rightarrow\)x-128= \(\left(\frac{104}{21}-5\right):\left(\frac{41.101}{42.101}-1\right):\left(\frac{63.10101}{64.10101}-1\right)\)

\(\Rightarrow\)x-128=\(\left(\frac{-1}{21}\right):\left(\frac{-1}{42}\right):\left(\frac{-1}{64}\right)\)

\(\Rightarrow x-128=-128\)

\(\Rightarrow x=\left(-128\right)+128\)

\(\Rightarrow x=0\)