2/ a/ 

\(\hept{\begin{cases}x-\sqrt{y+\sqrt{y-\frac{1}{4}}}=\frac{1}{2}\\y-\sqrt{x+\sqrt{x-\frac{1}{4}}}=\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-\sqrt{\left(\sqrt{y-\frac{1}{4}}+\frac{1}{2}\right)^2}=\frac{1}{2}\\y-\sqrt{\left(\sqrt{x-\frac{1}{4}}+\frac{1}{2}\right)^2}=\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-\sqrt{y-\frac{1}{4}}-\frac{1}{2}=\frac{1}{2}\\y-\sqrt{x-\frac{1}{4}}-\frac{1}{2}=\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-\sqrt{y-\frac{1}{4}}=1\\y-\sqrt{x-\frac{1}{4}}=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2-2x+1=y-\frac{1}{4}\left(1\right)\\y^2-2y+1=x-\frac{1}{4}\left(2\right)\end{cases}}\)

Lấy (1) - (2) ta được

\(\Rightarrow\left(x-y\right)\left(x+y-1\right)=0\)

Làm nốt

Câu 2/b Hệ chỉ có 2 cái thôi hả

đk tự giải nhé 

với x tjỏa mãn đk ta có 

\(\sqrt{\frac{x^2+3}{x}}=\frac{x^2+7}{2\left(x+1\right)}\Leftrightarrow\sqrt{x^3+3}=\frac{x^3+7x}{2\left(x+1\right)}\)

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

đặt \(\sqrt{x^3+3x}=a\)

ta có pt<=> \(a=\frac{a^2+4x}{2\left(x+1\right)}\Leftrightarrow2a\left(x+1\right)=a^2+4x\)

\(\Leftrightarrow2ax+2a=a^2+4x\Leftrightarrow a^2+4ax-2a-2ax=0\)

\(\Leftrightarrow\left(a^2-2ax\right)-\left(2a-4x\right)=0\Leftrightarrow a\left(a-2x\right)-2\left(a-2x\right)=0\)

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

đến đây tự làm nhé

b, Đặt \(\sqrt[3]{x}=t\)

Ta có: \(\sqrt[3]{x^2}-8\sqrt[3]{x}=20\)

\(\Leftrightarrow t^2-8t=20\Leftrightarrow t^2-8t-20=0\)

\(\Leftrightarrow\left(t+2\right)\left(t-10\right)=0\)

\(\orbr{\begin{cases}t=-2\\t=10\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt[3]{x}=-2\\\sqrt[3]{x}=10\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=-8\\x=1000\end{cases}}\)

Câu 1: ĐKXĐ: ...

\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)

\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)

\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)

\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)

\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow...\)

Câu 2:

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2-4\right)=y^3+2y\\x^2-4=-3y^2\end{matrix}\right.\)

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

\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)

\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)

\(\Leftrightarrow\left(\frac{y^2-3y+2}{3y}\right)\left(\frac{y^2+3y+2}{3y}\right)=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)

\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)

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

\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)

\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)

\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)

\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)

\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)

Tìm được mỗi nghiệm thôi à :v