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Bài 1:
a) Để A,B có nghĩa \(\Leftrightarrow\begin{cases}2x+3\ge0\\x-3>0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge-\frac{3}{2}\\x>3\end{cases}\)\(\Leftrightarrow x>3\)
b) Để A= B
\(\Leftrightarrow\sqrt{\frac{2x+3}{x-3}}=\frac{\sqrt{2x+3}}{\sqrt{x-3}}\)
\(\Leftrightarrow\sqrt{\frac{2x+3}{x-3}}-\sqrt{\frac{2x+3}{x-3}}=0\)
\(\Leftrightarrow0x=0\) (thỏa mãn với mọi x>3)
Vậy x>3 thì A=B
a, ĐKXĐ A: \(\frac{2x+3}{x-3}\)\(\frac{2x+3}{x-3}\ge0\Rightarrow\left[\begin{array}{nghiempt}\hept{\begin{cases}2x+3\ge0\\x-3>0\end{array}\right.\\\hept{\begin{cases}2x-3\le0\\x-3< 0\end{array}\right.\end{cases}\Rightarrow\left[\begin{array}{nghiempt}\hept{\begin{cases}x\ge-\frac{3}{2}\\x>3\end{array}\right.\\\hept{\begin{cases}x\le-\frac{3}{2}\\x< 3\end{array}\right.\end{cases}\Rightarrow}\left[\begin{array}{nghiempt}x>-\frac{3}{2}\\x< 3\end{array}\right.}\)
ĐKXĐ B: \(\begin{cases}2x+3\ge0\\x-3>0\end{cases}\Rightarrow\begin{cases}x\ge-\frac{3}{3}\\x>3\end{cases}}\)
\(\sqrt{\dfrac{a}{b}}\)=\(\dfrac{\sqrt{a}}{\sqrt{b}}\) với a,b<0
Ta có : \(\sqrt{\dfrac{-49}{-81}}\)=\(\sqrt{\dfrac{49}{81}}\)=\(\dfrac{7}{9}\)
\(\sqrt{ab}=\sqrt{-a}.\sqrt{-b}\) (vì a<0 , b<0)
Áp dụng : \(\sqrt{\left(-25\right).\left(-64\right)}=\sqrt{-\left(-25\right)}.\sqrt{-\left(-64\right)}=\sqrt{25}.\sqrt{64}=5.8=40\)
a: \(=\sqrt{3a}:\sqrt{b}\)
b: \(=\sqrt{a}:\sqrt{xy}\)
a: ĐKXĐ: 2x-10>=0
=>2x>=10
=>x>=5
b: \(\sqrt{A^2B}=\sqrt{A^2}\cdot\sqrt{B}=\left|A\right|\cdot\sqrt{B}\)
\(\sqrt{72}=\sqrt{36\cdot2}=6\sqrt{2}\)
c: \(A=\sqrt{16}+\sqrt{81}=4+9=13\)
\(B=\sqrt{\dfrac{\left(15\sqrt{5}+5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}}\)
\(=\sqrt{\dfrac{15}{\sqrt{2}}+5\sqrt{20}-3\sqrt{45}}\)
\(=\sqrt{\dfrac{15\sqrt{2}+2\sqrt{5}}{2}}=\sqrt{\dfrac{30\sqrt{2}+4\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{30\sqrt{2}+4\sqrt{5}}}{2}\)
\(C=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\left(2+\sqrt{3}\right)\)
\(=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}-\left(2+\sqrt{3}\right)+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(=2+\sqrt{3}-2-\sqrt{3}+\sqrt{2}=\sqrt{2}\)
1) Khi x = 49 thì:
\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)
2) Ta có:
\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)
\(B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)
Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)
\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)
Vậy x = 4
Với \(\begin{cases}a< 0\\b< 0\end{cases}\) thì \(\sqrt{\frac{a}{b}}=\sqrt{-a}:\sqrt{-b}\)
Áp dụng \(\sqrt{\frac{-49}{-81}}=\sqrt{-\left(-49\right)}:\sqrt{-\left(-81\right)}=\sqrt{49}:\sqrt{81}=7:9=\frac{7}{9}\)