Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
-\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\left|x\right|-6x+9\)
\(x< 0\)
\(--->x+3-x-6x+9\)
\(=\left(x-x\right)-6x+3+9\)
\(=-6x+\left(3+9\right)=-6x+12\)
\(x>0\)
\(--->3+x+x-6x+9\)
\(=\left(x+x-6x\right)+\left(3+9\right)\)
\(=\left(2x-6x\right)+12\)
\(=4x+12\)
a. \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{x-1}{-\left(x+4\right)}=\dfrac{4\sqrt{x}}{x+4}\)
b. \(\:B=\dfrac{4\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}+4}=\dfrac{4+4\sqrt{2}}{7+2\sqrt{2}}=\dfrac{\left(4+4\sqrt{2}\right).\left(7-2\sqrt{2}\right)}{\left(7+2\sqrt{2}\right).\left(7-2\sqrt{2}\right)}=\dfrac{12+20\sqrt{2}}{41}\)
a: \(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
b: \(A-2=\dfrac{2-2x-2\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{-2x-2\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{-2\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}< =0\)
=>A<=2
Vì \(x+\sqrt{x}+1>0\) nên A>0
\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{x}{x-1}\right):\left(\sqrt{x}-\dfrac{\sqrt{x}}{\sqrt{x+1}}\right)\)
\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x^2+x}-\sqrt{x}}{\sqrt{x+1}}\)
\(M=\dfrac{\sqrt{x^2+x}-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x+1}}{\sqrt{x^2+x}-\sqrt{x}}\)
\(M=\dfrac{\sqrt{x+1}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
Để \(M\le0\)
\(\Rightarrow\dfrac{\sqrt{x+1}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\le0\)
mà \(\sqrt{x+1}\ge0\)
\(\Rightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\le0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1\ge0\\\sqrt{x}+1\le0\end{matrix}\right.\)\(\Rightarrow x\ge1\)
\(\left\{{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}+1\ge0\end{matrix}\right.\) \(\Rightarrow x\ge-1\)
Vậy nghiệm của pt : \(x\ge1\) ; \(x\ge-1\)
---------
P/s: Hm..Câu b mình không chắc chắn lắm /_/
a, \(x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+|x-3|=x+3-x+3=6\)
b, \(\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)
\(=|x+2|-|x|=x+2+x=2x+2\)
Lời giải:
Đặt \(\sqrt{(1-x)+(1-x)\sqrt{1-x^2}}=a; \sqrt{(1-x)-(1-x)\sqrt{1-x^2}}=b\)
Khi đó: \(P=a+b\geq 0\)
Ta có:
\(a^2+b^2=(1-x)+(1-x)\sqrt{1-x^2}+(1-x)-(1-x)\sqrt{1-x^2}=2(1-x)\)
Và:
\(ab=(1-x)\sqrt{(1+\sqrt{1-x^2})(1-\sqrt{1-x^2})}\)
\(=(1-x)\sqrt{1-(1-x^2)}=(1-x)\sqrt{x^2}=|x|(1-x)\)
\(\Rightarrow P^2=(a+b)^2=a^2+b^2+2ab=2(1-x)+2|x|(1-x)\)
\(=2(1-x)(1+|x|)=\frac{4036}{2017}.\frac{2018}{2017}\)
\(\Rightarrow P=\sqrt{\frac{4036.2018}{2017^2}}=\frac{\sqrt{4036.2018}}{2017}\)
Lời giải:
Với mọi $1\geq x\geq 0$ thì $x+\sqrt{x}+1\geq 1$
$\Rightarrow E=\frac{5}{x+\sqrt{x}+1}\leq \frac{5}{1}=5$
Vậy $E_{\max}=5$ khi $x=0$