Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\( \left( {\dfrac{x}{{\sqrt x - 2}} - \dfrac{{4x - 4\sqrt x }}{{x - 2\sqrt x }}} \right).\dfrac{6}{{{{\left( {\sqrt x - 2} \right)}^2}}}\\ = \left[ {\dfrac{x}{{\sqrt x - 2}} - \dfrac{{4\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right].\dfrac{6}{{{{\left( {\sqrt x - 2} \right)}^2}}}\\ = \left( {\dfrac{{x - 4\sqrt x + 4}}{{\sqrt x - 2}}} \right).\dfrac{6}{{{{\left( {\sqrt x - 2} \right)}^2}}}\\ = \dfrac{{{{\left( {\sqrt x - 2} \right)}^2}}}{{\sqrt x - 2}}.\dfrac{6}{{{{\left( {\sqrt x - 2} \right)}^2}}} = \dfrac{6}{{\sqrt x - 2}} \)
\(A=\sqrt{\left(x-2\right)\left(x-1\right)x\left(x+1\right)+5}\)
\(=\sqrt{\left(x^2-x-2\right)\left(x^2-x\right)+5}\)
Đặt \(t=x^2-x\) ta đc:
\(A=\sqrt{\left(t-2\right)t+5}=\sqrt{t^2-2t+5}\)
\(=\sqrt{\left(t-1\right)^2+4}\ge\sqrt{4}=2\)
Dấu = khi \(t=1\Leftrightarrow x^2-x=1\Leftrightarrow x=\pm\frac{1}{2}+\frac{\sqrt{5}}{2}\)
Vậy....
b)\(B=\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}\)
\(=\sqrt{\left(x-2\right)^2}+\sqrt{\left(x+3\right)^2}\)
\(=\left|x-2\right|+\left|x+3\right|\)
Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-2\right|+\left|x+3\right|=\left|x-2\right|+\left|-x-3\right|\ge\left|x-2+\left(-x\right)-3\right|=5\)
Dấu = khi \(\left(x-2\right)\left(x+3\right)\ge0\)\(\Rightarrow-3\le x\le2\)
\(\Rightarrow\hept{\begin{cases}-3\le x\le2\\\left(x+3\right)\left(x-2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy....
a, \(\sqrt{\left(\sqrt{2}\right)^2+2\times2\times\sqrt{2}+2^2}\)+ \(\sqrt{2^2+2\times2\times\sqrt{2}+\left(\sqrt{2}\right)^2}\)
= \(\sqrt{\left(\sqrt{2}+2\right)^2}\)+ \(\sqrt{\left(2-\sqrt{2}\right)^2}\)
= \(\sqrt{2}+2+2-\sqrt{2}\)
= 4
Đặt \(x^2+4x+8=a\)(a > 0 vì \(x^2+4x+8=\left(x+2\right)^2+4>0\)) , \(x^2+4x+4=b\left(b\ge0\right)\)
\(\Rightarrow a+b=x^2+4x+8+x^2+4x+4=2\left(x^2+4x+6\right)\)
Khi đó phương trình đã cho trở thành:
\(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\\ \Leftrightarrow a+b+2\sqrt{ab}=a+b\\ \Leftrightarrow2\sqrt{ab}=0\\\Leftrightarrow\sqrt{ab}=0\\ \Leftrightarrow ab=0\\ \Leftrightarrow\left[{}\begin{matrix}a=0\left(kh\text{ô}ngtm\right)\\b=0\end{matrix}\right.\)
\(\Rightarrow x^2+4x+4=0\\ \Leftrightarrow\left(x+2\right)^2=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2.\)
Kl: