Ta có: \(a+b+c+\sqrt{abc}=4\)

\(\Rightarrow4a+4b+4c+4\sqrt{abc}=16\)

\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\)

\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)

\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=\left|2a+\sqrt{abc}\right|=2a+\sqrt{abc}\)

Tương tự: 

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\\\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\end{matrix}\right.\)

\(\Rightarrow A=\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-c\right)\left(4-a\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)

Ta có \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(a+c+\sqrt{abc}\right)\left(4-c\right)}\)

\(=\sqrt{\left(a^2+ac+a\sqrt{abc}\right)\left(4-c\right)}\\ =\sqrt{4a^2+ac\left(4-\sqrt{abc}-a-c\right)+4a\sqrt{abc}}\\ =\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}\\ =2a+\sqrt{abc}\left(a,b,c>0\right)\)

Cmtt \(\sqrt{b\left(4-c\right)\left(4-a\right)}=2b+\sqrt{abc};\sqrt{c\left(4-b\right)\left(4-a\right)}=2c+\sqrt{abc}\)

\(\Rightarrow A=2\left(a+b+c\right)+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c\right)+2\sqrt{abc}\\ A=2\left(a+b+c+\sqrt{abc}\right)=2\cdot4=8\)

ta có \(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4a+4\sqrt{abc}\)

=> \(4a+4\sqrt{abc}=16-4b-4c\Leftrightarrow4a+4\sqrt{abc}+bc=16-4b-4c+bc\)

=> \(\left(2\sqrt{a}+\sqrt{bc}\right)^2=\left(4-b\right)\left(4-c\right)\Rightarrow a\left(4-b\right)\left(4-c\right)=a\left(2\sqrt{a}+\sqrt{bc}\right)^2\)

=> \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a}\left(2\sqrt{a}+\sqrt{bc}\right)=2a+\sqrt{abc}\)

tương tự như thế thay vào , thì A=8

Ta có:

\(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4c+4\sqrt{abc}\)

\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\Leftrightarrow4a+4\sqrt{abc}+bc=16-4b-4c+bc\)

\(\Rightarrow\left(2\sqrt{a}+\sqrt{bc}\right)^2=\left(4-b\right)\left(4-c\right)\Rightarrow a\left(4-b\right)\left(4-c\right)=a\left(2\sqrt{a}+\sqrt{bc}\right)^2\)

\(\Rightarrow\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a}\left(2\sqrt{a}+\sqrt{bc}\right)=2a+\sqrt{abc}\)

Tương tự như thế thay vào, thì A = 8

\(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4c+4\sqrt{abc}=16\Rightarrow16-4b-4c=4a+4\sqrt{abc}\)

\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)

\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=2a+\sqrt{abc}\)

Tương tự : \(\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\); \(\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\)

\(\Rightarrow A=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)

chưa có cách lm sao pn

Nguyễn Bùi Đại Hiệp phục bạn này lần nào hỏi cũng chép sai đề.

\(a+b+c+\sqrt{abc}=4\)

\(\Leftrightarrow4\left(a+b+c\right)+4\sqrt{abc}=16\)(*)

\(A=\Sigma\left(\sqrt{a\left(4-b\right)\left(4-c\right)}\right)-\sqrt{abc}\)

\(A=\Sigma\left(\sqrt{a\left(16-4b-4c+bc\right)}\right)-\sqrt{abc}\)

Thay (*) vào A ta được :

\(A=\Sigma\left(\sqrt{a\left(4a+4b+4c+4\sqrt{abc}-4b-4c+bc\right)}\right)-\sqrt{abc}\)

\(A=\Sigma\left(\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\right)-\sqrt{abc}\)

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

\(A=\Sigma\left[\sqrt{a}\cdot\left(2\sqrt{a}+\sqrt{bc}\right)\right]-\sqrt{abc}\)

\(A=\Sigma\left(2a+\sqrt{abc}\right)-\sqrt{abc}\)

\(A=2\left(a+b+c\right)+3\sqrt{abc}-\sqrt{abc}\)

\(A=2\left(a+b+c\right)+2\sqrt{abc}\)

\(A=2\left(a+b+c+\sqrt{abc}\right)\)

\(A=2\cdot4=8\)

Vậy....

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Câu hỏi của Hoa Hồng Nhung - Toán lớp 9 | Học trực tuyến

yêu cầu tính hả bạn

cái này bạn nhân giả thiết với 4 rồi chuyển làm sao để pt thành nhân tử có chứa như cái trong căn ấy

có \(a+b+c+\sqrt{abc}=4\Rightarrow4-b=a+c+\sqrt{abc};\)\(4-a=b+c+\sqrt{abc};\)\(4-c=a+b+\sqrt{abc}\)

\(\Rightarrow\left(4-b\right)\left(4-c\right)=\left(a+c+\sqrt{abc}\right)\left(a+b+\sqrt{abc}\right)\)\(=a^2+ab+ac+bc+2a\sqrt{abc}+b\sqrt{abc}+c\sqrt{abc}\)

\(=a\left(a+b+c+\sqrt{abc}\right)+\sqrt{abc}\left(a+b+c+\sqrt{abc}\right)+bc\)
\(=4a+4\sqrt{abc}+bc\)\(\Rightarrow a\left(4-b\right)\left(4-c\right)=4a^2+4a\sqrt{abc}+abc=\left(2a+\sqrt{abc}\right)^2\)\(\Rightarrow\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=2a+\sqrt{abc}\)

tương tự \(\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\)\(;\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\)

\(\Rightarrow\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-c\right)\left(4-a\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}\)\(-\sqrt{abc}\)\(=2\left(a+b+c+\sqrt{abc}\right)=2.4=8\)

Đề phải là số thực không âm mới đúng