Question

a/4-a+b/4-b+c/4-c=1, then 1/4-a+1/4-b+1/4-c=?
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Answer 1

Answer:

1/4-a+1/4-b+1/4-c=1

Step-by-step explanation:

given

a/4-a+b/4-b+c/4-c=1a/4−a+b/4−b+c/4−c=1

add 1 on each term both side

\left ( a/4-a \right )+1+\left ( b/4-b \right )+1+\left ( c/4-c \right )+1=1+1+1+1(a/4−a)+1+(b/4−b)+1+(c/4−c)+1=1+1+1+1

then taking LCM

\left ( a+4-a \right )/4-a+\left ( b+4-b \right )/4-b+\left ( c+4-c \right )/4-c=4(a+4−a)/4−a+(b+4−b)/4−b+(c+4−c)/4−c=4

here  a,b,c cancel and 4 is left each term

4\left ( 1/4-a+1/4-b+1/4-c \right )=44(1/4−a+1/4−b+1/4−c)=4

taking 4 as a common and cancel both side  then we get required solution

1/4-a+1/4-b+1/4-c=11/4−a+1/4−b+1/4−c=1

hence proved