Em xin phép làm bài EZ nhất :)

4,ĐK :\(\forall x\in R\)

Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))

\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)

\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)

\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)

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

Vậy ....

b sai \(\sqrt{x^2+6x+9}\) ms đúng

Giai đi nha

a) \(x^3+1=2\sqrt[3]{2x-1}\) (1)

Đặt \(\sqrt[3]{2x-1}=a\Rightarrow a^3=2x-1\)

\(\Rightarrow1=2x-a^3\)

Phương trình (1) khi đó trở thành :

\(x^3+2x-a^3=2a\)

\(\Leftrightarrow\left(x^3-a^3\right)+2\left(x-a\right)=0\)

\(\Leftrightarrow\left(x-a\right)\left(x^2+ax+a^2+2\right)=0\)

\(\Leftrightarrow x=a\)

Do đó : \(x=\sqrt[3]{2x-1}\Leftrightarrow x^3-2x+1=0\)

\(\Leftrightarrow\left(x-1\right).\left(x^2-x-1\right)=0\)

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

a) \(\sqrt{1-4x+4x^2}=5\) 

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

\(\Leftrightarrow\left|1-2x\right|=5\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b) \(\sqrt{x^2+6x+9}=3x-1\)

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

\(\Leftrightarrow\left|x+3\right|=3x-1\)

\(\Leftrightarrow x+3=3x-1\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

\(a,\sqrt{1-4x+4x^2}=5\\ \Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\\ \Leftrightarrow\left|1-2x\right|=5\)

\(TH_1:x\le\dfrac{1}{2}\)

\(1-2x=5\\ \Leftrightarrow x=-2\left(tm\right)\)

\(TH_2:x\ge\dfrac{1}{2}\)

\(-1+2x=5\\ \Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{-2;3\right\}\)

\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left|x+3\right|=3x-1\)

\(TH_1:x\ge-3\\ x+3=3x-1\\ \Leftrightarrow-2x=-4\Leftrightarrow x=2\left(tm\right)\)

\(TH_2:x< 3\\ -x-3=3x-1\\ \Leftrightarrow-4x=2\\ \Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

Vậy \(S=\left\{2;-\dfrac{1}{2}\right\}\)

f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)

\(\Leftrightarrow\left|x+1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

đk: \(x\ge\frac{-5}{4}\)

pt \(\Rightarrow\left(2x-2\right)^2=\left(\sqrt{4x+5}+1\right)^2\)

+) \(\sqrt{4x+5}=2x-3\). giải ra được \(x=2+\sqrt{3}\)

+) \(\sqrt{4x+5}=-2x+1\). giải ra được: \(x=1-\sqrt{2}\)

Vậy pt có 2 nghiệm trên.

ĐK : \(x\ge-\frac{5}{4}\)

\(2x^2-6x-1=\sqrt{4x+5}\)

\(\Leftrightarrow4x^2-12x-2-2\sqrt{4x+5}=0\)

\(\Leftrightarrow\left(4x^2-8x+4\right)-\left(4x+6+2\sqrt{4x+5}\right)=0\)

\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(4x+5+2\sqrt{4x+5}+1\right)=0\)

\(\Leftrightarrow4\left(x-1\right)^2-\left(\sqrt{4x+5}+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-1\right)+\left(\sqrt{4x+5}+1\right)\right]\left[2\left(x-1\right)-\left(\sqrt{4x+5}+1\right)\right]=0\)

\(\Leftrightarrow\left(2x-2+\sqrt{4x+5}+1\right)\left(2x-2-\sqrt{4x+5}-1\right)=0\)

\(\Leftrightarrow\left(2x+\sqrt{4x+5}-1\right)\left(2x-\sqrt{4x+5}-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+\sqrt{4x+5}-1=0\\2x-\sqrt{4x+5}-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1-\sqrt{2}\\x=2+\sqrt{3}\end{cases}}\)( tmđk )

Vậy pt có nghiệm là x = 1 -\(\sqrt{2}\) ; x = 2 +\(\sqrt{3}\)