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

chưa có cách lm sao pn

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

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

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

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

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

Ta có:

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

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

Ta lại có:

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

= (4a² + \(4a\sqrt{abc}\)+ abc)

= (\(2a+\sqrt{abc}\))²

Tương tự ta có

b(4 - c)(4 - a) = (\(2b+\sqrt{abc}\))²

c(4 - a)(4 - b) = (\(2c+\sqrt{abc}\))²

Từ đây ta có

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

\(=8\)     

Nhầm 

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

Thành

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

Mà thôi cũng làm tương tự thôi nên bạn tự làm lại nhé

Cho đa thức P(x)= \(\frac{x^5}{5}\)+\(\frac{x^3}{3}\)+\(\frac{7x}{15}\), CMR P(x)  luôn nguyên khi x nguyên.

Lời giải:

Ta có:

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

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

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

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

Hoàn toàn tương tự với các biểu thức còn lại, suy ra:

\(M=2a+\sqrt{abc}+2b+\sqrt{abc}+2c+\sqrt{abc}-\sqrt{abc}\)

\(=2(a+b+c+\sqrt{abc})=2.4=8\)

Ta co:

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

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

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

Tương tự ta cũng co:

\(\hept{\begin{cases}\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\\\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\end{cases}}\)

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