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ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
a) M\(=\frac{x-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}:\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}-1}:\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)\(=\frac{x}{\sqrt{x}+1}\)
b) Khi \(x=7+4\sqrt{3}\Rightarrow\frac{7+4\sqrt{3}}{\sqrt{\left(2+\sqrt{3}\right)^2}+1}=\frac{7+4\sqrt{3}}{3+\sqrt{3}}\)
c)\(M=\frac{1}{2}\Leftrightarrow\frac{x}{\sqrt{x}+1}=\frac{1}{2}\Leftrightarrow\sqrt{x}=2x-1\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\x^2=4x^2-4x+1\Leftrightarrow3x^2-4x+1=0\Leftrightarrow\left(3x-1\right)\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\\left[{}\begin{matrix}x=\frac{1}{3}\left(l\right)\\x=1\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
a, Thay x = 25 => \(\sqrt{x}=5\)vào biểu thức A ta được :
\(A=\frac{25+6}{21}=\frac{31}{21}\)
b, Với \(x>0;x\ne4;x\ne16\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}-2}+\frac{5\sqrt{x}-8}{2\sqrt{x}-x}=\frac{\sqrt{x}-1}{\sqrt{x}-2}+\frac{5\sqrt{x}-8}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)\
\(=\frac{-x+6\sqrt{x}-8}{-\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{-\sqrt{x}+4}{-\sqrt{x}}=\frac{\sqrt{x}-4}{\sqrt{x}}\)
c, số xấu quá check lại phần trên hộ mình
Bài 2:
a)
\(\sqrt{x-3}+\sqrt{x+2}\)
Biểu thức trên được xác định khi và chỉ khi:
\(\left\{{}\begin{matrix}x-3\ge0\\x+2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x\ge-2\end{matrix}\right.\)
b)
\(\sqrt{x+4}-\frac{1}{\sqrt{x-3}}\)
Biểu thức trên được xác định khi và chỉ khi:
\(\left\{{}\begin{matrix}x+4\ge0\\x+3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x>-3\end{matrix}\right.\)
Bài 3:
\(\sqrt{x^2-2x+1}\le3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}\le3\)
\(\Leftrightarrow x-1\le3\)
\(\Leftrightarrow x\le4\)
b)\(\frac{2}{3}.\sqrt{4x^2-20}+2\sqrt{\frac{x^2-5}{9}}-3\sqrt{x^2-5}=2\)
\(< =>\frac{2}{3}.\sqrt{4\left(x^2-5\right)}+2\cdot\frac{\sqrt{x^2-5}}{3}-3\sqrt{x^2-5}=2\)
\(< =>\frac{2}{3}.2\sqrt{\left(x^2-5\right)}+2\cdot\frac{\sqrt{x^2-5}}{3}-3\sqrt{x^2-5}=2\)
\(< =>\frac{4}{3}\sqrt{\left(x^2-5\right)}+\frac{2}{3}.\sqrt{x^2-5}-3\sqrt{x^2-5}=2\)
\(< =>-\sqrt{\left(x^2-5\right)}=2\)
\(< =>\sqrt{\left(x^2-5\right)}=-2\)(vô nghiệm)
a)\(\sqrt{25x-25}-\frac{15}{2}\sqrt{\frac{x-1}{9}}=6+\frac{3}{2}\sqrt{x-1}\)
\(< =>\sqrt{25\left(x-1\right)}-\frac{15}{2}.\frac{\sqrt{x-1}}{3}-\frac{3}{2}\sqrt{x-1}=6\)
\(< =>5\sqrt{x-1}-\frac{5}{2}.\sqrt{x-1}-\frac{3}{2}\sqrt{x-1}=6\)
\(< =>\sqrt{x-1}=6\)
\(< =>x-1=36\)
\(< =>x=37\)
vậy ...
ĐKXĐ tất cả các câu bạn tự tìm
\(C=\frac{4\left(\sqrt{x}+3\right)+3}{\sqrt{x}+3}=4+\frac{3}{\sqrt{x}+3}\le4+\frac{3}{3}=5\)
\(C_{max}=5\) khi \(x=0\)
\(A=\frac{2\left(\sqrt{x}+2\right)-17}{\sqrt{x}+2}=2-\frac{17}{\sqrt{x}+2}\ge2-\frac{17}{2}=-\frac{13}{2}\)
\(A_{min}=-\frac{13}{2}\) khi \(x=0\)
\(B=\frac{x+2\sqrt{x}+1+9}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)^2+9}{\sqrt{x}+1}=\sqrt{x}+1+\frac{9}{\sqrt{x}+1}\)
\(B\ge2\sqrt{\frac{9\left(\sqrt{x}+1\right)}{\sqrt{x}+1}}=6\Rightarrow B_{min}=6\) khi \(\sqrt{x}+1=3\Leftrightarrow x=4\)
\(A=\frac{2\left(\sqrt{x}+2\right)+1}{\sqrt{x}+2}=2+\frac{1}{\sqrt{x}+2}\)
Để A nguyên \(\Rightarrow\sqrt{x}+2=Ư\left(1\right)=\left\{-1;1\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+2=-1\\\sqrt{x}+2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-3\left(l\right)\\\sqrt{x}=-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x nguyên để A nguyên
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}< 1\)
Mặt khác \(A+2=\frac{\sqrt{x}-2}{\sqrt{x}+1}+2=\frac{\sqrt{x}-2+2\sqrt{x}+2}{\sqrt{x}+1}=\frac{3\sqrt{x}}{\sqrt{x}+1}\ge0\)
\(\Rightarrow A\ge-2\Rightarrow-2\le A< 1\)
Mà A nguyên \(\Rightarrow A=\left\{-2;-1;0\right\}\)
- Với \(A=-2\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}=-2\Rightarrow\sqrt{x}-2=-2\sqrt{x}-2\)
\(\Rightarrow3\sqrt{x}=0\Rightarrow x=0\)
- Với \(A=-1\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}=-1\Rightarrow\sqrt{x}-2=-\sqrt{x}-1\)
\(\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)
- Với \(A=0\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}=0\Rightarrow\sqrt{x}-2=0\Rightarrow x=4\)
Vậy \(x=\left\{0;\frac{1}{4};4\right\}\)
Bài 1:
a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
=\(\sqrt{xy}\)
b.ĐK: x ≠ 1
Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)
*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)
⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)
⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)
c.Ta có:
Bài 2:
a: \(A=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{x-9}=\dfrac{x-3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{x-25}=\dfrac{\sqrt{x}}{\sqrt{x}-5}\)
b: \(P=A:B=\dfrac{\sqrt{x}}{\sqrt{x}+3}:\dfrac{\sqrt{x}}{\sqrt{x}-5}=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(P-1=\dfrac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{-8}{\sqrt{x}+3}< 0\)
=>P<1
\(B=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
TH1: \(x\ge2\Rightarrow B=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
TH2: \(1\le x< 2\Rightarrow B=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(A^2=2x+2\sqrt{x^2-\left(x^2-4\right)}=2x+4\)
\(\Rightarrow A=\sqrt{2x+4}\)