Question

[Cao Mạnh Hùng]

Giúp mình với huhu, xin cảm ơn trc ạ :3

Answer 1

[Victorique de Blois]

b8 :

\(ab+bc+ca=5\Rightarrow a^2+5=a^2+ab+bc+ca\)

\(\Rightarrow a^2+5=a\left(a+b\right)+c\left(a+b\right)=\left(a+c\right)\left(a+b\right)\)

tương tự có \(b^2+5=\left(b+c\right)\left(b+a\right)\) và \(c^2+5=\left(c+a\right)\left(c+b\right)\) 

thay vào A ta đc :

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

\(c\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)}{\left(c+a\right)\left(c+b\right)}}\)

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

\(A=a\left|b+c\right|+b\left|c+a\right|+c\left|b+a\right|\)

\(A=ab+ac+bc+ba+cb+ca\) do a;b;c dương

\(A=2\left(ab+bc+ca\right)=2\cdot5=10\)

b9: 

\(x_o=\sqrt[3]{a+\sqrt{a^2+b^3}}-\sqrt[3]{\sqrt{a^2+b^3}-a}\)

\(\Leftrightarrow x_o^3=a+\sqrt{a^2+b^3}-\sqrt{a^2+b^3}+a-3^3\sqrt{\left(a+\sqrt{a^2+b^3}\right)\cdot\left(\sqrt{a^2+b^3}-a\right)}\cdot x_o\)

\(\Leftrightarrow x_o^3=2a-3\sqrt[3]{a^2+b^3-a^2}\cdot x_o\)

\(\Leftrightarrow x_o^3=2a-3bx_O\)

\(\Leftrightarrow x_o^3+3bx_o-2a=0\left(đpcm\right)\)

Answer 2

[Victorique de Blois]

bài 9 dấu <=> thứ nhất do dài quá bị xuống dòng, như này này b 

\(a+\sqrt{a^2+b^3}-\sqrt{a^2+b^3}+a-3\sqrt[3]{\left(\sqrt{a^2+b^3}+a\right)\left(\sqrt{a^2+b^3-a}\right)}\cdot x_o\)