Question

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0\)

Answer 1

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0.\)

\(=\left(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x\right)+1\)

\(=x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1\)

\(=x\left(\frac{1\cdot5}{30}+\frac{1\cdot3}{30}-\frac{4\cdot2}{30}\right)+1\)

\(=x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1\)

\(=x\left(\frac{5+3-8}{30}\right)+1\)

\(=x\cdot0+1=1\)

\(\Rightarrow\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0=1\)

Answer 2

trả lời:

  \frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0.61​x+101​x−154​x+1+0.

=\left(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x\right)+1=(61​x+101​x−154​x)+1

=x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=x(61​+101​−154​)+1

=x\left(\frac{1\cdot5}{30}+\frac{1\cdot3}{30}-\frac{4\cdot2}{30}\right)+1=x(301⋅5​+301⋅3​−304⋅2​)+1

=x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1=x(305​+303​−308​)+1

=x\left(\frac{5+3-8}{30}\right)+1=x(305+3−8​)+1

=x\cdot0+1=1=x⋅0+1=1

\Rightarrow\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0=1⇒61​x+101​x−154​x+1+0=1

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