Question

Tìm x, biết :

\(\frac{x-18}{74}+\frac{x-20}{72}+\frac{x-22}{70}=3\)

 

Answer 1

<=> (x-18/74 - 1)+(x-20/72 - 1)+(x-22/70 - 1) = 0

<=> x-92/74 + x-92/72 + x-92/70 = 0

<=> (x-92).(1/74+1/72+1/70) = 0

<=> x-92 = 0 ( vì 1/74 + 1/72 + 1/70 > 0 )

<=> x=92

Vậy S = {92}

Tk mk nha

Answer 2

Ta có :

\(\frac{x-18}{74}+\frac{x-20}{72}+\frac{x-22}{70}=3\)

\(\Leftrightarrow\)\(\left(\frac{x-18}{74}-1\right)+\left(\frac{x-20}{72}-1\right)+\left(\frac{x-22}{70}-1\right)=3-3\)

\(\Leftrightarrow\)\(\frac{x-92}{74}+\frac{x-92}{72}+\frac{x-92}{70}=0\)

\(\Leftrightarrow\)\(\left(x-92\right)\left(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\right)=0\)

Vì \(\left(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\right)\ne0\)

\(\Rightarrow\)\(x-92=0\)

\(\Rightarrow\)\(x=92\)

Vậy \(x=92\)

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