\(ĐK:x\in R\)

Đặt \(\sqrt{x^2+3}=t\left(t\ge0\right)\)

\(PT\Leftrightarrow2t^2-\left(7x+1\right)t+3x^2+3x=0\\ \Delta=\left(7x+1\right)^2-4\cdot2\left(3x^2+3x\right)=25x^2-10x+1=\left(5x-1\right)^2\ge0\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{7x+1-5x+1}{4}\\t=\dfrac{7x+1+5x-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{2x+2}{4}=\dfrac{x+1}{2}\\t=\dfrac{12x}{4}=3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=\dfrac{x+1}{2}\\\sqrt{x^2+3}=3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+3=\dfrac{x^2+2x+1}{4}\\x^2+3=9x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x^2-2x+11=0\\x^2=\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\Delta=4-132< 0\\\left[{}\begin{matrix}x=\dfrac{\sqrt{6}}{4}\\x=-\dfrac{\sqrt{6}}{4}\end{matrix}\right.\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{\sqrt{6}}{4};\dfrac{\sqrt{6}}{4}\right\}\)

ĐKXĐ: \(x\ge\dfrac{2}{7}\)

\(\sqrt{5x^2-5x+3}-\left(x+1\right)+2x-\sqrt{7x-2}+4x^2-7x+2=0\)

\(\Leftrightarrow\dfrac{4x^2-7x+2}{\sqrt{5x^2-5x+3}+\left(x+1\right)^2}+\dfrac{4x^2-7x+2}{2x+\sqrt{7x-2}}+4x^2-7x+2=0\)

\(\Leftrightarrow\left(4x^2-7x+2\right)\left(\dfrac{1}{\sqrt{5x^2-5x+3}+\left(x+1\right)^2}+\dfrac{1}{2x+\sqrt{7x-2}}+1\right)=0\)

Ta có \(\dfrac{1}{\sqrt{5x^2-5x+3}+\left(x+1\right)^2}+\dfrac{1}{2x+\sqrt{7x-2}}+1>0\)

\(\Rightarrow4x^2-7x+2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7-\sqrt{17}}{8}\\x=\dfrac{7+\sqrt{17}}{8}\end{matrix}\right.\)

\(\)

cai nay la hag dag thuc phan tih ra la dk

pt<=>căn((x-1/2)^2+75/4)+căn(2(x-1/2)^2+3(x+2)^2)+căn((x-1/2)^2+3(2x+3/2)^2)>=3*căn3(x+2)

dấu = xãy ra khi x=1/2

\(2x^4+7x^3+x^2-7x-3=0\)

\(\Leftrightarrow2x^4+7x^3+3x^2-2x^2-7x-3=0\)

\(\Leftrightarrow\left(2x^4+7x^3+3x^2\right)-\left(2x^2+7x+3\right)=0\)

\(\Leftrightarrow x^2\left(2x^2+7x+3\right)-\left(2x^2+7x+3\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(2x^2+7x+3\right)=0\)

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

\(\Leftrightarrow x\in\left\{\pm1;\frac{-1}{2};-3\right\}\)

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

đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\)

\(\Rightarrow\frac{ab+1}{a+b}=1\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)

thay vào là được

bạn có thể giải rõ hơn ko