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\(A=\dfrac{3\left(\sqrt{x}+1\right)-2}{2\left(\sqrt{x}+1\right)}=\dfrac{3}{2}-\dfrac{1}{\sqrt{x}+1}\)
Ta có \(\sqrt{x}+1\ge1\Leftrightarrow-\dfrac{1}{\sqrt{x}+1}\ge-1\)
\(\Leftrightarrow A\ge\dfrac{3}{2}-1=\dfrac{1}{2}\)
Dấu \("="\Leftrightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\)
\(x-3\sqrt{x}=0\)
\(\Rightarrow\left(\sqrt{x}\right)^2-3\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}.\left(\sqrt{x}-3\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\sqrt{x}=0\\\sqrt{x}-3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\\sqrt{x}=3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)
Vậy : \(x\in\left\{0;9\right\}\)
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\(a,2\sqrt{x}+3=0\)
\(\Leftrightarrow2\sqrt{x}=-3\)
\(\Leftrightarrow\sqrt{x}=-\frac{3}{2}\)( loại )
\(b,\frac{5}{12}\sqrt{x}-\frac{1}{6}=\frac{1}{3}\Leftrightarrow\frac{5}{12}\sqrt{x}=\frac{1}{2}\Leftrightarrow\sqrt{x}=\frac{6}{5}\Leftrightarrow x=\frac{36}{25}\)
\(c,\sqrt{x+3}+3=0\Leftrightarrow\sqrt{x+3}=-3\)( loại )
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x-3√x = 0
⇒√x2-3√x=0
⇒√x(√x-3)=0
⇒ hoặc √x=0⇒x=0
hoặc √x-3=0 ⇒√x=3 ⇒ x=+-√3
vậy x={0;-√3;√3}
Ta có: \(x-3\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2\sqrt{x}-10=20\left(ĐKXD:x\ge0\right)\)
\(\Leftrightarrow2\sqrt{x}=30\Leftrightarrow\sqrt{x}=15\)
\(\Leftrightarrow x=225\)
b) \(2x-\sqrt{x}=0\left(ĐKXĐ:x\ge0\right)\)
\(\Leftrightarrow2x=\sqrt{x}\Leftrightarrow4x^2=x\Leftrightarrow4x^2-x=0\Leftrightarrow x\left(4x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}}\)
Vậy ....
c) \(x+3\sqrt{x}=0\left(ĐKXĐ:x\ge0\right)\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+3\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x\in\varnothing\end{cases}}}\)
Vậy x = 0
d) \(\left(x-1\right)\left(x^2+1\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x^2=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x\in\varnothing\end{cases}}}\)
Vậy x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{\left(x-3\sqrt{5}\right)^2}+\sqrt{\left(y+3\sqrt{5}\right)^2}+\left|x+y+z\right|=0\)
\(\Leftrightarrow\left|x-3\sqrt{5}\right|+\left|y+3\sqrt{5}\right|+\left|x+y+z\right|=0\)
\(\Leftrightarrow\begin{cases}x-3\sqrt{5}=0\\y+3\sqrt{5}=0\\x+y+z=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x=3\sqrt{5}\\y=-3\sqrt{5}\\z=-x-y=-3\sqrt{5}+3\sqrt{5}=0\end{cases}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề :\(\sqrt{x-\sqrt{5}}+\sqrt{y+\sqrt{3}}+\left(x+y+z\right)^2=0\)
\(\sqrt{x-\sqrt{5}};\sqrt{y+\sqrt{3}};\left(x+y+z\right)^2\ge0\)nên vế trái không âm và bằng 0 (theo gt) chỉ khi :
\(\sqrt{x-\sqrt{5}}=\sqrt{y+\sqrt{3}}=\left(x+y+z\right)^2=0\Rightarrow\hept{\begin{cases}x-\sqrt{5}=0\\y+\sqrt{3}=0\\x+y+z=0\left(1\right)\end{cases}}\)
\(\Rightarrow x=\sqrt{5};y=-\sqrt{3}\)và kết hợp với 1,ta có\(z=\sqrt{3}-\sqrt{5}\)
Mk nghĩ các bt trong căn với (x+ y+z) phải có 2 nữa , xem lại đề
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\(đk:x\ge0\)
\(\left(\sqrt{x}+1\right).\left(\sqrt{x}-3\right)=0\)
=> \(\sqrt{x}-3=0\)( vì \(\sqrt{x}+1>0\)với mọi x>_ 0 )
=> x = 9
\(x-3\sqrt{x}\)\(=0\)
\(\Leftrightarrow\sqrt{x}\)\((\sqrt{x}\)\(-3)\)\(=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x=0}\\\sqrt{x-3=0}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\text{9}\end{cases}}\)
ngu
ngu