Câu 2
\(\Leftrightarrow2x\sqrt{x}=\left(1-\sqrt{x\left(1-x\right)}\right)\left(\sqrt{x}+\sqrt{1-x}\right)\)
Đặt \(u=\sqrt{x},v=\sqrt{1-x}\) được
\(\left\{{}\begin{matrix}u^2+v^2=1\\2u^3=\left(u+v\right)\left(1-uv\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u^2+v^2=1\\2u^3=\left(u+v\right)\left(1-uv\right)=\left(u+v\right)\left(u^2+v^2-uv\right)=u^3+v^3\end{matrix}\right.\)
\(\Rightarrow u=v=1\) (u = 1 hoặc v = 1)
\(\Rightarrow x=1\)
Câu 1
(b)
Có \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\dfrac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=\dfrac{6^2-14}{2}=11\)
\(\Rightarrow a+11=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{b}\right)\)
Tương tự \(b+11=\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{b}\right)\)
\(c+11=\left(\sqrt{c}+\sqrt{b}\right)\left(\sqrt{c}+\sqrt{a}\right)\)
Có: \(\dfrac{\sqrt{a}}{a+11}+\dfrac{\sqrt{b}}{b+11}+\dfrac{\sqrt{c}}{c+11}\)
\(=\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{c}+\sqrt{a}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{c}+\sqrt{a}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{c}+\sqrt{a}\right)}\)
\(=\dfrac{11\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{c}+\sqrt{a}\right)^2}}\)
\(=\dfrac{11.11}{\sqrt{\left[\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\right]\left[\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{b}\right)\right]\left[\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)\right]}}\)
\(=\dfrac{22}{\sqrt{\left(a+11\right)\left(b+11\right)\left(c+11\right)}}\)
Câu 1
(a)
\(M=\dfrac{x\left(x+1\right)+5\sqrt{x+1}\sqrt{x-1}}{5\left(x-1\right)+x\sqrt{x+1}\sqrt{x-1}}\)
\(M=\dfrac{x\sqrt{x+1}\sqrt{x+1}+5\sqrt{x+1}\sqrt{x-1}}{5\sqrt{x-1}\sqrt{x-1}+x\sqrt{x+1}\sqrt{x-1}}\)
\(M=\dfrac{\sqrt{x+1}\left(x\sqrt{x+1}+5\sqrt{x-1}\right)}{\sqrt{x-1}\left(5\sqrt{x-1}+x\sqrt{x+1}\right)}=\dfrac{\sqrt{x+1}}{\sqrt{x-1}}\)
Vì x có giá trị là \(-1-\sqrt{2}\) nên không thể xác định được giá trị M
