a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)

a) \(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-7x^2-9x+4+x^3+3x^2+4x+2=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-\left(7x^2+9x-4\right)+\left(x+1\right)^3+x+1=\sqrt[3]{7x^2+9x-4}\) (*)

Đặt \(\sqrt[3]{7x^2+9x-4}=a;x+1=b\)

Khi đó (*) \(\Leftrightarrow-a^3+b^3+b=a\)

\(\Leftrightarrow\left(b-a\right).\left(b^2+ab+a^2+1\right)=0\)

\(\Leftrightarrow b=a\)

Hay \(x+1=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow\left(x+1\right)^3=7x^2+9x-4\)

\(\Leftrightarrow x^3-4x^2-6x+5=0\)

\(\Leftrightarrow x^3-4x^2-5x-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1\pm\sqrt{5}}{2}\end{matrix}\right.\)

ta có \(\left(x+3\right)\sqrt{15-x^2}=\left(x-3\right)\left(x+4\right)\)

<=> \(\left(x-3\right)\left(\sqrt{15-x^2}-x-4\right)=0\)

đến đây dễ rồi

x+3 chứ 

Đkxđ: \(x\ge3\)

pt đã cho \(\Leftrightarrow x^2-x-12+3\left(\sqrt{x-3}-1\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x+3+\dfrac{3}{\sqrt{x-3}+1}=0\left(vôlí\right)\end{matrix}\right.\)

Vậy pt đã cho có nghiệm duy nhất \(x=4\)

\(\sqrt{x-3}+\sqrt{4x-12}=6\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{x-3}+2\sqrt{x-3}=6\)

\(\Leftrightarrow3\sqrt{x-3}=6\Leftrightarrow\sqrt{x-3}=2\)

\(\Leftrightarrow x-3=4\Leftrightarrow x=7\left(tm\right)\)

\(\sqrt{x-3}+\sqrt{4x-12}=6\)đk : x >= 3 

\(\Leftrightarrow\sqrt{x-3}+2\sqrt{x-3}=6\Leftrightarrow3\sqrt{x-3}=6\Leftrightarrow\sqrt{x-3}=2\)

\(\Leftrightarrow x-3=4\Leftrightarrow x=7\)

Lời giải:

Đặt $\sqrt[3]{x}=a; \sqrt[3]{2x-3}=b$. Ta có:

\(\left\{\begin{matrix} a+b=\sqrt[3]{4(a^3+b^3)}\\ 2a^3-b^3=3\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} a^3+b^3+3ab(a+b)=4(a^3+b^3)\\ 2a^3-b^3=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a^3+b^3=ab(a+b)\\ 2a^3-b^3=3\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} (a-b)^2(a+b)=0(1)\\ 2a^3-b^3=3(2)\end{matrix}\right.\)

Từ $(1)$ suy ra $a=b$ hoặc $a=-b$.

Nếu $a=b$. Thay vào $(2)$ suy ra $a^3=b^3=3$

$\Leftrightarrow x=2x-3=3$ (thỏa mãn)

Nếu $a=-b$. Thay vào $(2)$ suy ra $a^3=1; b^3=-1$

$\Leftrightarrow x=1; 2x-3=-1$ (thỏa mãn)

Vậy $x=3$ hoặc $x=1$

\(\sqrt[3]{x}+\sqrt[3]{2x-3}=\sqrt[3]{12\left(x-1\right)}\left(1\right)\)

\(\left(1\right)\Leftrightarrow x+2x-3+3.\sqrt[3]{x\left(2x-3\right)}.\left(\sqrt[3]{x}+\sqrt[3]{2x-3}\right)=12x-12\)

\(\Rightarrow\sqrt[3]{12x\left(x-1\right)\left(2x-3\right)}=3x-3\)

\(\Leftrightarrow12x\left(x-1\right)\left(2x-3\right)=[3\left(x-1\right)]^3\)

\(\Leftrightarrow12x\left(2x^2-5x+3\right)=27\left(x^3-3x^2+3x-1\right)\)

\(\Leftrightarrow24x^3-60x^2+36x=27x^3-81x^2+81x-27\)

\(\Leftrightarrow3x^3-21x^2+45x-27=0\)

\(\Leftrightarrow3\left(x-1\right)\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\). Thử lại ta thấy cả x=1;x=3 đều t/m bài toán

Vậy, pt có tập nghiệm là S={1;3}