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a)= \(\left(3+\sqrt{5}\right)\left(\sqrt{\left(3-\sqrt{5}\right)^2}\right)\)=\(\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=9-5=4\)
b)= \(\frac{2\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}+\frac{\sqrt{2^2.7}}{2}-2\)=\(\frac{2\left(3-\sqrt{7}\right)}{9-7}+\sqrt{7}-2\)=1
c) =\(\frac{3}{3\left(\sqrt{7}-2\right)}-\frac{3}{3\left(\sqrt{7}+2\right)}\)=\(\frac{1}{\sqrt{7}-2}-\frac{1}{\sqrt{7}+2}=\frac{\sqrt{7}+2-\left(\sqrt{7}-2\right)}{\left(\sqrt{7}+2\right)\left(\sqrt{7}-2\right)}\)=\(\frac{4}{7-4}=\frac{4}{3}\)
d) =\(\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)^{ }\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\frac{\left(88-44\sqrt{3}\right)}{25-3}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\frac{22\left(4-2\sqrt{3}\right)}{22}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(1+\sqrt{3}\right)\left(\sqrt{3}-1\right)\)=3-1 = 2
e) = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{7\sqrt{x}-3}{x-9}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\)= \(\frac{x-4\sqrt{x}+3}{x-9}+\frac{7\sqrt{x}-3}{x-9}+\sqrt{x}\)= \(\frac{x+3\sqrt{x}}{x-9}+\sqrt{x}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\sqrt{x}\)= \(\frac{\sqrt{x}}{\sqrt{x}-3}+\sqrt{x}=\frac{x-2\sqrt{x}}{\sqrt{x}-3}\)
Lời giải:
a)
\(\left\{\begin{matrix} x\geq 0\\ 3-\sqrt{x}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x\leq 9\end{matrix}\right.\Leftrightarrow 0\leq x\leq 9\)
b)
\(\left\{\begin{matrix} x-1\geq 0\\ 2-\sqrt{x-1}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x-1\leq 4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\leq 5\end{matrix}\right.\)
\(\Leftrightarrow 1\leq x\leq 5\)
c)
\(-7+3x>0\Leftrightarrow x>\frac{7}{3}\)
d)
\(\left\{\begin{matrix} x-1\geq 0\\ 5-x>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x< 5\end{matrix}\right.\Leftrightarrow 1\leq x< 5\)
e) \(x\in\mathbb{R}\)
f) \(\left\{\begin{matrix} 2-x>0\\ x-5\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x< 2\\ x\geq 5\end{matrix}\right.\) (vô lý)
Do đó không tồn tại $x$ để hàm số tồn tại
g)
\(\left[\begin{matrix} \left\{\begin{matrix} 3x-6-2x\geq 0\\ 1-x>0\end{matrix}\right.\\ \left\{\begin{matrix} 3x-6-2x\leq 0\\ 1-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x\geq 6\\ x< 1\end{matrix}\right.(\text{vô lý})\\ \left\{\begin{matrix} x\leq 6\\ x>1 \end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow 1< x\leq 6\)
mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)
1.
\(DK:x\in\left[-4;5\right]\)
\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)
\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)
Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)
\(\Rightarrow\sqrt{x-5}=0\)
\(x=5\left(n\right)\)
Vay nghiem cua PT la \(x=5\)
2.
\(DK:x\ge0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)
Ta co:
\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)
Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)
TH2:(loai)
Vay nghiem cua PT la \(x\in\left[4;9\right]\)
a/
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{7-x}=a\\\sqrt[3]{x-5}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3+b^3=3\\a^3-b^3=2\left(6-x\right)\end{matrix}\right.\) với \(a+b\ne0\)
Ta có hệ:
\(\left\{{}\begin{matrix}a^3+b^3=2\\\frac{a-b}{a+b}=\frac{a^3-b^3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3=2\\\frac{a-b}{a+b}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{2}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}a^3+b^3=2\\a-b=0\end{matrix}\right.\) \(\Rightarrow a=b=1\Rightarrow\left\{{}\begin{matrix}\sqrt[3]{7-x}=1\\\sqrt[3]{x-5}=1\end{matrix}\right.\) \(\Rightarrow x=6\)
TH2: \(\left\{{}\begin{matrix}a^3+b^3=2\\\frac{1}{a+b}=\frac{a^2+ab+b^2}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3=2\\\frac{1}{a+b}=\frac{a^2+ab+b^2}{a^3+b^3}=\frac{a^2+ab+b^2}{\left(a+b\right)\left(a^2-ab+b^2\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3=2\\\frac{a^2+ab+b^2}{a^2-ab+b^2}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3=2\\ab=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b^3=2\end{matrix}\right.\\\left\{{}\begin{matrix}b=0\\a^3=2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)
b/
Lập phương 2 vế:
\(\left(\sqrt[3]{x+1}+\sqrt[3]{x-1}\right)^3=5x\)
\(\Leftrightarrow x+1+x-1+3\sqrt[3]{\left(x^2-1\right)}\left(\sqrt[3]{x+1}+\sqrt[3]{x-1}\right)=5x\)
\(\Leftrightarrow2x+3\sqrt[3]{x^2-1}\left(\sqrt[3]{5x}\right)=5x\)
\(\Leftrightarrow x=\sqrt[3]{5x\left(x^2-1\right)}\)
\(\Leftrightarrow x^3=5x\left(x^2-1\right)\)
\(\Leftrightarrow x\left(5\left(x^2-1\right)-x^2\right)=0\)
\(\Leftrightarrow x\left(4x^2-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{\sqrt{5}}{2}\\x=-\frac{\sqrt{5}}{2}\end{matrix}\right.\)