Question

BT1: Tìm x, biết:

7) \(\dfrac{1}{6}.x+\dfrac{1}{10}.x-\dfrac{4}{15}.x+1=0\)

Answer 1

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

=> \(\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right).x+1\)= 0

=> \(0.x+1=0\)

=> 0 . x = 0 - 1

=> 0 . x = -1

=> x = 0 : -1

=> x = 0

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Answer 2

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

\(\Leftrightarrow\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right)x+1=0\)

\(\Leftrightarrow0x+1=0\)\(\)

Vì 0x luôn = 0 => 0x + 1 > 0

<=> x \(\in\varnothing\)