c: Ta có: \(\sqrt{2x}=\sqrt{5}\)

\(\Leftrightarrow2x=5\)

hay \(x=\dfrac{5}{2}\)

d: Ta có: \(\sqrt{3x-1}=4\)

\(\Leftrightarrow3x-1=16\)

\(\Leftrightarrow3x=17\)

hay \(x=\dfrac{17}{3}\)

Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}=6\)

\(\Leftrightarrow2\left|x-1\right|=6\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Ta có: \(\sqrt{4x^2-4x+9}=3\)

\(\Leftrightarrow4x^2-4x=0\)

\(\Leftrightarrow4x\left(x-1\right)=0\)

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

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\\frac{-1-\sqrt{5}}{4}\le x\le-\frac{1}{8}\end{matrix}\right.\)(Có thể chưa chính xác)

\(12x^2+16x+1=2\sqrt{24x^3+12x^2-6x}+4\sqrt{x^2-x}+4\sqrt{8x^3+9x^2+x}\)

Áp dụng AM-GM:

\(2\sqrt{24x^3+12x^2-6x}=2\sqrt{6x\left(4x^2+2x-1\right)}\le6x+\left(4x^2+2x-1\right)=4x^2+8x-1\left(1\right)\)

\(4\sqrt{x^2-x}=2\sqrt{1.\left(4x^2-4x\right)}\le4x^2-4x+1\left(2\right)\)

\(4\sqrt{8x^3+9x^2+x}=2\sqrt{\left(4x^2+4x\right)\left(8x+1\right)}\le\left(4x^2+4x\right)+\left(8x+1\right)=4x^2+12x+1\left(3\right)\)

Cộng \(\left(1\right),\left(2\right),\left(3\right)\), ta có: \(VP\le VT\)

Dấu ''='' xảy ra khi :

\(\left\{{}\begin{matrix}4x^2+2x-1=6x\\4x^2-4x=1\\4x^2+4x=8x+1\end{matrix}\right.\)\(\Rightarrow4x^2-4x-1=0\)

\(\Rightarrow x=\frac{1\pm\sqrt{2}}{2}\) (t/m ĐKXĐ)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\\\sqrt{x^2+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\\x^2=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}∃x̸\\x=\pm\sqrt{8}\end{matrix}\right.\)

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

Nghiệm của pt là `x = +- 2 \sqrt{2}`

\(VT=\sqrt{\left(\sqrt{x+2}-1\right)^2+3}+\sqrt{\left(\sqrt{x+2}-3\right)^2}>\sqrt{3}>1\) pt vô nghiệm 

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.

\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\Leftrightarrow2x+1=2x^3+x^2+2x+1\)\(\Leftrightarrow2x^3+x^2=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\left(1\right)\)

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

\(x^2+1\ge1\forall x\Rightarrow2x+1\ge0!2x+1!=2x+1\)

\(\left(1\right)\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)

\(\left(1\right)\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)

\(\left(1\right)\Leftrightarrow2x+1=\left(2x+1\right)\left(x^2+1\right)\Leftrightarrow\left(2x+1\right).\left(1-\left(x^2+1\right)\right)=0\)

\(\hept{\begin{cases}2x+1=0\\-x^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}}\)

Chúc bạn học tốt !!!