Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
\(\sqrt {\dfrac{2}{{9 - x}}}\) có nghĩa khi \(\left\{ \begin{array}{l} \dfrac{2}{{9 - x}} \ge 0\\ 9 - x \ne 0 \end{array} \right. \Leftrightarrow 9 - x > 0 \Leftrightarrow - x > - 9 \Leftrightarrow x < 9\)
\(\sqrt {{x^2} + 2x + 1} \) có nghĩa khi: \({x^2} + 2x + 1 = {\left( {x + 1} \right)^2} > 0\forall x \in R\)
\(\sqrt{9-x^2}\) có nghĩa khi: \(9 - {x^2} \ge 0 \Leftrightarrow - {x^2} \ge - 9 \Leftrightarrow {x^2} \le 9 \Leftrightarrow \left| x \right| \le 9\)
\(\Leftrightarrow x\ge3\) hoặc \(x\ge-3\)
\(\sqrt {\dfrac{1}{{{x^2} - 4}}} \) có nghĩa khi: \(\left\{ \begin{array}{l} \dfrac{1}{{{x^2} - 4}} \ge 0\\ {x^2} - 4 \ne 0 \end{array} \right. \Leftrightarrow {x^2} - 4 > 0 \Leftrightarrow \left| x \right| > 4\)
\(\Leftrightarrow x>2\) hoặc \(x>-2\)
\(a,\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\frac{20+6\sqrt{10}-5\sqrt{10}-9}{16-10}.\)
\(=\frac{11-\sqrt{10}}{6}\)
\(b,=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{54-8}\)
\(=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{46}\)
Câu 1:
\(\left\{{}\begin{matrix}\frac{x-1}{x+3}\ge0\\x+3\ne0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge1\\x< -3\end{matrix}\right.\)
b/
\(\left\{{}\begin{matrix}\frac{x-1}{4-x}\ge0\\4-x\ne0\end{matrix}\right.\) \(\Rightarrow1\le x< 4\)
c/
\(\left\{{}\begin{matrix}\frac{a^3}{b^2}\ge0\\b^2\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3\ge0\\b\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a\ge0\\b\ne0\end{matrix}\right.\)
Câu 2:
\(\sqrt{64+6\sqrt{7}}=\sqrt{63+2\sqrt{63}+1}=\sqrt{\left(\sqrt{63}+1\right)^2}=1+\sqrt{63}=1+3\sqrt{7}\)
\(\sqrt{16+8\sqrt{3}}=\sqrt{12+2\sqrt{12.4}+4}=\sqrt{\left(\sqrt{12}+\sqrt{4}\right)^2}=\sqrt{12}+\sqrt{4}=2+2\sqrt{3}\)
\(\sqrt{9-2\sqrt{14}}=\sqrt{7-2\sqrt{7.2}+2}=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}=\sqrt{7}-\sqrt{2}\)
\(\sqrt{9}=3\)
\(\sqrt{\frac{4}{9}}=\frac{2}{3}\)
\(\sqrt{0,25}=0,5\)
\(\sqrt{2}=1,4141....\)
Lê Nguyên Hạo hình như cả -3 nữa á bn