a. \(\frac{3}{4}x-\frac{4}{5}.x=\frac{-2}{3}\)
\(\left(\frac{3}{4}-\frac{4}{5}\right)\) \(.x\) = \(\frac{-2}{3}\)
\(\frac{-1}{20}.x=\frac{-2}{3}\)
\(x=\frac{-2}{3}:\frac{-1}{20}\)
a) \(\frac{3}{4}.x\) - \(\frac{4}{5}.x\) = \(\frac{-2}{3}\)
=> \(x.\left(\frac{3}{4}-\frac{4}{5}\right)\) = \(\frac{-2}{3}\)
=> \(x.\frac{-1}{20}\) = \(\frac{-2}{3}\)
=> \(x=\frac{-2}{3}:\frac{-1}{20}\)
=> \(x=\frac{40}{3}\)
sorry bn mk nhầm : x = \(\frac{40}{3}\)
b/ theo bài ra ta có:
\(\frac{2x-1}{6}=\frac{y+3}{4}=\frac{2x-y-4}{4x}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{2x-1}{6}=\frac{y+3}{4}=\frac{2x-y-4}{4x}=\frac{2x-1-y-3}{6-4}=\frac{2x-y-4}{2}\)
=> \(\frac{2x-y-4}{4x}=\frac{2x-y-4}{2}\)
=> 4x = 2 => x = 2:4
=> x = 0,5
thay x =0,5 vào \(\frac{2x-1}{6}=\frac{y+3}{4}\) ta có:
=> \(\frac{2.0,5-1}{6}=\frac{y+3}{4}\\ \Rightarrow0=\frac{y+3}{4}\\ \Rightarrow y+3=0\\ \Rightarrow y=-3\)
vậy x = 0,5; y = -3
mk chưa chắc là mk có có làm đúng ko
a) \(\frac{3}{4}.x\) -\(\frac{4}{5}.x\)= \(-\frac{2}{3}\)
=> x.\(\left(\frac{3}{4}-\frac{4}{5}\right)\)= \(\frac{-2}{3}\)
=> x.\(\frac{-1}{20}\)=\(\frac{-2}{3}\)
=>x= \(\frac{\frac{-2}{3}}{\frac{-1}{20}}\)
=>x= \(\frac{40}{3}\)
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