Các bn giúp mk với

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

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

→ \(2x+1=x^2-6x+9\)

→ \(2x+1-x^2+6x-9=0\)

→ \(-x^2+8x-8=0\rightarrow x^2-8x+8=0\)

→ \(x_1=4+2\sqrt{2}\)

\(x_2=4-2\sqrt{2}\)

ĐK: \(2x+1\ge0\Leftrightarrow x\ge-\frac{1}{2}\)

\(pt\Leftrightarrow2x+1=\left(x-3\right)^2\\ \Leftrightarrow2x+1=x^2-6x+9\\ \Leftrightarrow x^2-8x+8=0\\ \Leftrightarrow x^2-2.x.4+4^2-4^2+8=0\\ \Leftrightarrow\left(x-4\right)^2-8=0\\ \Leftrightarrow\left(x-4-2\sqrt{2}\right)\left(x-4+2\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4-2\sqrt{2}=0\\x-4+2\sqrt{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4+2\sqrt{2}\\x=4-2\sqrt{2}\end{matrix}\right.\)

Vậy...............................

\(\sqrt{2x+1}=x-3\) (ĐKXĐ:\(x\ge\frac{1}{2}\))

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

\(\Leftrightarrow2x+1=x^2-6x+9\)

\(\Leftrightarrow-x^2+8x-8=0\)

\(\Leftrightarrow-\left(x^2-8x+16-8\right)=0\)

\(\Leftrightarrow-\left(\left(x-4\right)^2-\left(2\sqrt{2}\right)^2\right)=0\)

\(\Leftrightarrow-\left(x-4-2\sqrt{2}\right)\left(x-4+2\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4-2\sqrt{2}=0\\x-4+2\sqrt{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4+2\sqrt{2}\\x=4-2\sqrt{2}\end{matrix}\right.\)

Ta có \(x=4+2\sqrt{2}\) thỏa mãn ĐKXĐ

\(x=4-2\sqrt{2}\) không thỏa mãn ĐKXĐ

nên nghiệm của pt trên là \(x=4+2\sqrt{2}\)

\(\left|2x-\frac{1}{2}\right|+1=3x\)

\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}=3x-1\\2x-\frac{1}{2}=1-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1+\frac{1}{2}\\2x+3x=1+\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)

Chị cũng đang ko biét ở bài này nè. Hu hu

ĐK: \(x\ge\frac{1}{2}\)

Đặt \(t=\sqrt{2x-1}\Leftrightarrow x=\frac{t^2+1}{2}\)(ĐK: \(t\ge0\)) thay vao phương trình ta được:

\(\sqrt{\frac{t^2+1}{2}+4+3t}\)+\(\sqrt{\frac{t^2+1}{2}+12-5t}=7\sqrt{2}\)

\(\Leftrightarrow\sqrt{\frac{t^2+6t+9}{2}}+\sqrt{\frac{t^2-10t+25}{2}}=7\sqrt{2}\)

\(\Leftrightarrow\frac{\sqrt{\left(t+3\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(t-5\right)^2}}{\sqrt{2}}=7\sqrt{2}\)

\(\Leftrightarrow\frac{\left|t+3\right|+\left|t-5\right|}{\sqrt{2}}=7\sqrt{2}\)

\(\Leftrightarrow t+3+\left|t-5\right|=14\)(vì \(t\ge0\Rightarrow t+3>0\))

\(\Leftrightarrow t+\left|t-5\right|=11\)

Xét TH: \(t-5\ge0\Leftrightarrow t\ge5\) thì ta có:

\(t+t-5=11\)

\(\Leftrightarrow2t=16\)

\(\Leftrightarrow t=8\)(chọn)

Xét TH: \(t-5< 0\Leftrightarrow t< 5\) thì ta có:

\(t-t+5=11\)

\(\Leftrightarrow5=11\)(vô lí nên loại)

Lại có: \(t=8\)

\(\Leftrightarrow\sqrt{2x-1}=8\)

\(\Leftrightarrow2x-1=64\)

\(\Leftrightarrow2x=63\)

\(\Leftrightarrow x=\frac{63}{2}=31\frac{1}{2}\)

Vậy nghiệm của phương trình là x=31\(\frac{1}{2}\)

32,5

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

\(\Rightarrow\sqrt{x^2+2x+9}=-\left(2x+1\right)\)

\(\Rightarrow\left\{{}\begin{matrix}-\left(2x+1\right)\ge0\\x^2+2x+9=\left[-\left(2x+1\right)\right]^2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{2}\\x^2+2x+9=4x^2+4x+1\end{matrix}\right.\)

\(\Rightarrow4x^2+4x+1-x^2-2x-9=0\)

\(\Rightarrow3x^2+2x-8=0\)

\(\Rightarrow3x^2+6x-4x-8=0\)

\(\Rightarrow3x\left(x+2\right)-4\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(3x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\left(KTMĐK\right)\\x=\frac{4}{3}\left(TMĐK\right)\end{matrix}\right.\)

Vậy nghiệm của phương trình là 4/3