Question

cho a,b,c > 0 thỏa mãn a+b+c+\(2\sqrt{abc}\)=1 . Tính giá trị biểu thức 

P=\(\sqrt{a\left(1-b\right)\left(1-c\right)}\)+\(\sqrt{b\left(1-c\right)\left(1-a\right)}\)+\(\sqrt{c\left(1-a\right)\left(1-b\right)}\)- \(\sqrt{abc}\)+2016 

Answer 1

ĐỀ thi hsg toán 9 hải phòng năm 2016-2017

Answer 2

thiệt ah

Answer 3

Ta có:\(\sqrt{a\left(1-b\right)\left(1-c\right)}=\sqrt{a\left(1-b-c+ab\right)}\)

\(=\sqrt{a\left(a+2\sqrt{abc}+bc\right)}=\sqrt{a\left(\sqrt{a}+\sqrt{bc}\right)^2}\)

\(=\sqrt{\left(a+\sqrt{abc}\right)^2}=a+\sqrt{abc}\)

Tương tự ta CM dc:

\(\sqrt{b\left(1-c\right)\left(1-a\right)}=b+\sqrt{abc};\sqrt{c\left(1-a\right)\left(1-b\right)}=c+\sqrt{abc}\)

\(\Rightarrow P=a+\sqrt{abc}+b+\sqrt{abc}+c+\sqrt{abc}-\sqrt{abc}+2016\)

\(P=a+b+c+2\sqrt{abc}+2016\)

\(P=1+2016=2017\)

Answer 4

From August, 2020:

Ta có: \(\sqrt{a\left(1-b\right)\left(1-c\right)}=\sqrt{a\left(1-b-c+bc\right)}\)

\(=\sqrt{a\left(a+2\sqrt{abc}+bc\right)}=\sqrt{a\left(\sqrt{a}+\sqrt{bc}\right)^2}\)

\(=\sqrt{\left(a+\sqrt{abc}\right)^2}=a+\sqrt{abc}\)

Tương tự ta CM được:

\(\sqrt{b\left(1-c\right)\left(1-a\right)}=b+\sqrt{abc}\) ; \(\sqrt{c\left(1-a\right)\left(1-b\right)}=c+\sqrt{abc}\)

=> \(P=a+\sqrt{abc}+b+\sqrt{abc}+c+\sqrt{abc}-\sqrt{abc}+2016\)

\(P=a+b+c+2\sqrt{abc}+2016\)

\(P=1+2016=2017\)