![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK:\(-\frac{1}{2}\le x\le4\)
\(\sqrt{4-x}+\sqrt{2x+1}=3\)
\(\Leftrightarrow\sqrt{4-x}-\left(\frac{1}{2}x-2\right)+\sqrt{2x+1}-\left(-\frac{1}{2}x-1\right)=0\)
\(\Leftrightarrow\frac{4-x-\left(\frac{1}{2}x-2\right)^2}{\sqrt{4-x}+\frac{1}{2}x-2}+\frac{2x+1-\left(-\frac{1}{2}x-1\right)^2}{\sqrt{2x+1}+\frac{1}{2}x-1}=0\)
\(\Leftrightarrow\frac{\frac{-\left(x^2-4x\right)}{4}}{\sqrt{4-x}+\frac{1}{2}x-2}+\frac{\frac{-\left(x^2-4x\right)}{4}}{\sqrt{2x+1}+\frac{1}{2}x-1}=0\)
\(\Leftrightarrow\frac{-x\left(x-4\right)}{4}\left(\frac{1}{\sqrt{4-x}+\frac{1}{2}x-2}+\frac{1}{\sqrt{2x+1}+\frac{1}{2}x-1}\right)=0\)
Thấy: \(\frac{1}{\sqrt{4-x}+\frac{1}{2}x-2}+\frac{1}{\sqrt{2x+1}+\frac{1}{2}x-1}>0\)
\(\Rightarrow\frac{-x\left(x-4\right)}{4}=0\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x+5\sqrt{x}-6=0\)
\(\Leftrightarrow\sqrt{x}-1=0\)
hay x=1
b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)
hay \(x=\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x+5\sqrt{x}-6=0\)
\(\Leftrightarrow\sqrt{x}-1=0\)
hay x=1
b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)
hay \(x=\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x+5\sqrt{x}-6=0\)
\(\Leftrightarrow\sqrt{x}-1=0\)
hay x=1
b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)
hay \(x=\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3+\sqrt{2x-3}=x\) (ĐKXĐ: x \(\ge\)1,5)
\(\Leftrightarrow\sqrt{2x-3}=x-3\)
\(\Leftrightarrow2x-3=x^2-6x+9\)
\(\Leftrightarrow-x^2+8x-12=0\)
\(\Leftrightarrow-\left(x^2-8x+12\right)=0\)
\(\Leftrightarrow x^2-6x-2x+12=0\)
\(\Leftrightarrow x.\left(x-6\right)-2.\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=6\\x=2\end{cases}\left(\text{TMĐK}\right)}\)
Vậy ...
\(\sqrt{x+1}+\sqrt{2x+3}=5\)(*)
đkxđ \(\hept{\begin{cases}x+1\ge0\\2x+3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-1\\x\ge-\frac{3}{2}\end{cases}}\Leftrightarrow x\ge-1\)
(*) \(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2=25\)\(\Leftrightarrow x+1+2x+3+2\sqrt{\left(x+1\right)\left(2x+3\right)}=25\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+3x+2x+3}=25\)\(\Leftrightarrow2\sqrt{2x^2+5x+3}=21-3x\)
\(\Leftrightarrow\left(2\sqrt{2x^2+5x+3}\right)^2=\left(21-3x\right)^2\)\(\Leftrightarrow4\left(2x^2+5x+3\right)=441-126x+9x^2\)
\(\Leftrightarrow8x^2+20x+12=441-126x+9x^2\)\(\Leftrightarrow x^2-146x+429=0\)
\(\Leftrightarrow x^2-3x-143x+429=0\)\(\Leftrightarrow x\left(x-3\right)-143\left(x-3\right)=0\)\(\Leftrightarrow\left(x-3\right)\left(x-143\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=3\\x=143\end{cases}}\)(nhận)
Chẳng hiểu làm sai chỗ nào mà x = 143, trong khi x = 143 thì VT = 29 \(\ne\)5. Chỉ có x = 3 thỏa thôi.