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6: Để P>1 thì P-1>0
\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}>0\)
\(\Leftrightarrow\sqrt{a}-2< 0\)
hay a<4
Kết hợp ĐKXĐ, ta được: \(0\le a< 4\)
5: Để P>0 thì \(x-4\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}-4>0\)
hay x>16
![](https://rs.olm.vn/images/avt/0.png?1311)
1. ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow \sqrt{x-1}=5-\sqrt{x-4}$
$\Rightarrow x-1=25+x-4-10\sqrt{x-4}$
$\Leftrightarrow 22=10\sqrt{x-4}$
$\Leftrightarrow 2,2=\sqrt{x-4}$
$\Leftrightarrow 4,84=x-4\Leftrightarrow x=8,84$
(thỏa mãn)
2. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow (2x-2\sqrt{x})-(5\sqrt{x}-5)=0$
$\Leftrightarrow 2\sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0$
$\Leftrightarrow (\sqrt{x}-1)(2\sqrt{x}-5)=0$
$\Leftrightarrow \sqrt{x}-1=0$ hoặc $2\sqrt{x}-5=0$
$\Leftrightarrow x=1$ hoặc $x=\frac{25}{4}$ (tm)
3. ĐKXĐ: $x\geq 3$
Bình phương 2 vế thu được:
$3x-2+2\sqrt{(2x+1)(x-3)}=4x$
$\Leftrightarrow 2\sqrt{(2x+1)(x-3)}=x+2$
$\Leftrightarrow 4(2x+1)(x-3)=(x+2)^2$
$\Leftrightarrow 4(2x^2-5x-3)=x^2+4x+4$
$\Leftrightarrow 7x^2-24x-16=0$
$\Leftrightarrow (x-4)(7x+4)=0$
Do $x\geq 3$ nên $x=4$
Thử lại thấy thỏa mãn
Vậy $x=4$
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x-\sqrt{x}-6=0\) (1)
\(\Leftrightarrow x+2\sqrt{x}-3\sqrt{x}-6=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\)
\(\Leftrightarrow x=9\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{9\right\}\)
a) \(x-\sqrt{x}-6=0\Leftrightarrow x+2\sqrt{x}-3\sqrt{x}-6=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)=0\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=0\)
\(\left\{{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\) vậy \(x=9\)
mấy câu sau hình như đề sai
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)
\(A=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)
c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)
\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)
\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)
\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)
d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)
\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)
\(D=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) đkxđ: \(\begin{cases}\sqrt{x^2-4}\ge0\\\sqrt{x^2}+4x+4\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}\begin{cases}x-2\ge0\\x+2\ge0\end{cases}\\x+2\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}x\ge2\\x\le-2\end{cases}\) \(\Leftrightarrow-2\ge x\ge2\)
\(\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}+\sqrt{\left(x+2\right)^2}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}=x+2\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x+2\right)^2\)
\(\Leftrightarrow\left(x+2\right)\left(x-2-x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
S={-2}
b) đkxđ: \(\begin{cases}\sqrt{1-x^2}\ge0\\\sqrt{x+1}\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}1-x^2\ge0\\x+1\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}x^2\le1\\x\ge-1\end{cases}\) \(\Leftrightarrow\begin{cases}\begin{cases}x\le1\\x\ge-1\end{cases}\\x\ge-1\end{cases}\) \(\Leftrightarrow-1\le x\le1\)
\(\sqrt{1-x^2}+\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{1-x^2}=-\sqrt{x+1}\)
\(\Leftrightarrow1-x^2=x+1\)
\(\Leftrightarrow-x-x^2=0\)
\(\Leftrightarrow-x\left(1+x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\1+x=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\left(N\right)\\x=-1\left(N\right)\end{array}\right.\)
S={-1;0}
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: 1 ≥ x ≥ -1
Ta có: VT ≥ 0 = VP
Dấu "=" xảy ra khi và chỉ khi
\(\left\{{}\begin{matrix}\sqrt{1-x^2}=0\\\sqrt{1+x}=0\end{matrix}\right.\)
<=> x = -1 (TM)
b) ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
Ta có: VT ≥ 0 = VP
Dấu "=" xảy ra khi và chỉ khi
\(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x^2+4x+4}=0\end{matrix}\right.\)
<=> x = -2 (TM)
c) \(\sqrt{1-x^2}+\sqrt{x+1}=0\)
ĐKXĐ: \(\left\{{}\begin{matrix}1-x^2\ge0\\x+1\ge0\end{matrix}\right.\) \(\Rightarrow\)\(\left\{{}\begin{matrix}1\ge x^2\\x\ge-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge-1\end{matrix}\right.\)
=> -1 \(\le\) x \(\le\) 1
\(\sqrt{1-x^2}+\sqrt{x+1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(1-x\right)\left(1+x\right)}+\sqrt{x+1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(1+x\right)}.\left(\sqrt{1-x}+1\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{1+x}=0\\\sqrt{1-x}=-1\left(voli\right)\end{matrix}\right.\Rightarrow x+1=0\)
=> x = -1 ( thỏa mãn)
d) ĐKXĐ: \(x^2-4\ge0\Rightarrow x^2\ge4\)
\(\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
\(\sqrt{x^2-4}+\sqrt{\left(x+2^2\right)}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2\right)\left(x+2\right)}+\sqrt{\left(x+2^2\right)}=0\)
\(\Leftrightarrow\)\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+2=0\\\sqrt{x-2}=-\sqrt{x+2}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-2\\x-2=x+2\left(voli\right)\end{matrix}\right.\)
Vậy x= -2