Bạn tham khảo lời giải tại đây:

https://hoc24.vn/cau-hoi/giai-pt-sqrtx-2sqrt4-x2x2-5x-1.219493072549

a, ĐK: \(x\ge2\)

\(\sqrt{2x+1}-\sqrt{x-2}=x+3\)

\(\Leftrightarrow\dfrac{x+3}{\sqrt{2x+1}+\sqrt{x-2}}=x+3\)

\(\Leftrightarrow\left(x+3\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x-2}}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\sqrt{2x+1}+\sqrt{x-2}=1\left(vn\right)\end{matrix}\right.\)

Phương trình vô nghiệm.

b, ĐK: \(x\ge-1\)

\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)

\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{\left(x+3\right)\left(x+1\right)}\)

\(\Leftrightarrow-\sqrt{x+3}\left(\sqrt{x+1}-1\right)+2x\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2x\\\sqrt{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

c, ĐK: \(x\ge-3\)

\(2\sqrt{x+3}=9x^2-x-4\)

\(\Leftrightarrow x+3+2\sqrt{x+3}+1=9x^2\)

\(\Leftrightarrow\left(\sqrt{x+3}+1\right)^2=9x^2\)

\(\Leftrightarrow\left(\sqrt{x+3}+1-3x\right)\left(\sqrt{x+3}+1+3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=3x-1\\\sqrt{x+3}=-3x-1\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}3x-1\ge0\\x+3=9x^2-6x+1\end{matrix}\right.\Leftrightarrow...\)

TH2: \(\left\{{}\begin{matrix}-3x-1\ge0\\x+3=9x^2+6x+1\end{matrix}\right.\Leftrightarrow...\)

Tự giải nha, t kh có máy tính ở đây.

Lời giải:

1. ĐKXĐ: $x\geq \frac{-5+\sqrt{21}}{2}$

PT $\Leftrightarrow x^2+5x+1=x+1$

$\Leftrightarrow x^2+4x=0$

$\Leftrightarrow x(x+4)=0$

$\Rightarrow x=0$ hoặc $x=-4$

Kết hợp đkxđ suy ra $x=0$

2. ĐKXĐ: $x\leq 2$

PT $\Leftrightarrow x^2+2x+4=2-x$

$\Leftrightarrow x^2+3x+2=0$

$\Leftrightarrow (x+1)(x+2)=0$

$\Leftrightarrow x+1=0$ hoặc $x+2=0$

$\Leftrightarrow x=-1$ hoặc $x=-2$
3.

ĐKXĐ: $-2\leq x\leq 2$

PT $\Leftrightarrow \sqrt{2x+4}=\sqrt{2-x}$

$\Leftrightarrow 2x+4=2-x$

$\Leftrightarrow 3x=-2$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

a,ĐKXĐ:\(x\ge2\)

\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)

b,ĐKXĐ:\(x\in R\)

\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

c, ĐKXĐ:\(x\ge0\)

\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)

a/ đk: \(\left[{}\begin{matrix}x\le\frac{-5-3\sqrt{5}}{10}\\x\ge\frac{-5+3\sqrt{5}}{10}\end{matrix}\right.\)\(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)

\(\Leftrightarrow\sqrt{x^2+x+1}+\sqrt{3\left(x^2+x+1\right)-1}=\sqrt{5\left(x^2+x+1\right)-6}\)

đặt\(x^2+x+1=t\left(t>0\right)\)

\(\sqrt{t}+\sqrt{3t-1}=\sqrt{5t-6}\)

bình phương 2 vế pt trở thành:

\(t+3t-1+2\sqrt{t\left(3t-1\right)}=5t-6\)

\(\Leftrightarrow2\sqrt{3t^2-t}=t-5\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left(2\sqrt{3t^2-t}\right)^2=\left(t-5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\11t^2+6t-25=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left[{}\begin{matrix}t=\frac{-3+2\sqrt{71}}{11}\\t=\frac{-3-2\sqrt{71}}{11}\end{matrix}\right.\end{matrix}\right.\)=> không có gtri t nào t/m

vậy pt vô nghiệm

a/ ĐKXĐ: ...

Đặt \(x^2+x+1=a>0\)

\(\sqrt{a}+\sqrt{3a-1}=\sqrt{5a-6}\)

\(\Leftrightarrow4a-1+2\sqrt{3a^2-a}=5a-6\)

\(\Leftrightarrow2\sqrt{3a^2-a}=a-5\) (\(a\ge5\))

\(\Leftrightarrow4\left(3a^2-a\right)=a^2-10a+25\)

\(\Leftrightarrow11a^2+6a-25=0\)

Nghiệm xấu quá, chắc bạn nhầm lẫn đâu đó

b/

Đặt \(x^2+x+1=a>0\)

\(\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)

\(\Leftrightarrow2a+3+2\sqrt{a^2+3a}=2a+7\)

\(\Leftrightarrow\sqrt{a^2+3a}=2\)

\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x^2+x+1=1\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{3x^2-5x+7}=3-\sqrt{3x^2-7x+2}\)

\(\Rightarrow3x^2-5x+7=3x^2-7x+11-6\sqrt{3x^2-7x+2}\)

\(\Leftrightarrow3\sqrt{3x^2-7x+2}=2-x\) (\(x\le2\))

\(\Leftrightarrow9\left(3x^2-7x+2\right)=x^2-4x+4\)

\(\Leftrightarrow26x^2-59x+14=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{26}\end{matrix}\right.\)

Do biến đổi ko tương đương nên cần thay lại nghiệm vào pt ban đầu kiểm tra

d/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{x^2+3x+2}+\sqrt{x^2+6x+5}=\sqrt{2x^2+9x+7}\)

\(\Leftrightarrow2x^2+9x+7+2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}=2x^2+9x+7\)

\(\Leftrightarrow\sqrt{\left(x+1\right)^2\left(x+2\right)\left(x+5\right)}=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\left(l\right)\\x=-5\end{matrix}\right.\)

cái j zị

đề bị sao r đó

theo kinh nghiệm lâu năm của tui thì đề là;

\(\sqrt{x-2}-\sqrt{x+1}+\sqrt{2x-5}=2x^2-5x\) nhưng sao là hệ nhỉ