\(\hept{\begin{cases}\sqrt[3]{2y+24}+\sqrt{12-x}=6\left(1\right)\\x^3+2xy^2+X-2yx^2-4y^3-2y=0\left(2\right)\end{cases}}\)

\(\left(1\right)\)ĐK:\(x\le12\)

Đặt \(u=\sqrt[3]{2y+24}\)\(\Rightarrow u^3=2y+24\)

\(v=\sqrt{12-x}\) \(\Rightarrow v^2=12-x\)

Ta có hệ  phương trình :\(\hept{\begin{cases}u+v=6\\u^3+v^2=2y-x+36\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}v=6-u\\u^3+\left(6-u\right)^2=2y-x+36\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}v=6-u\\u^3+u^2+36-12u=2y+x+36\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}v=6-u\\u^3+u^2-12u=2y+x\end{cases}}\)

lop may vay

\(ĐK:x\le12\\ PT\Leftrightarrow\left(\sqrt[3]{x+24}-3\right)+\left(\sqrt{12-x}-3\right)=0\\ \Leftrightarrow\dfrac{x-3}{\sqrt[3]{\left(x+24\right)^2}+3\sqrt[3]{x+24}+9}-\dfrac{x-3}{\sqrt{12-x}+3}=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\\dfrac{1}{\sqrt[3]{\left(x+24\right)^2}+3\sqrt[3]{x+24}+9}=\dfrac{1}{\sqrt{12-x}+3}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt[3]{\left(x+24\right)^2}+3\sqrt[3]{x+24}+9=\sqrt{12-x}+3\\ \Leftrightarrow\sqrt[3]{x+24}\left(\sqrt[3]{x+24}+3\right)+6-\sqrt{12-x}=0\\ \Leftrightarrow\dfrac{\left(x+24\right)\left(\sqrt[3]{x+24}+3\right)}{\sqrt[3]{\left(x+24\right)^2}}+\dfrac{x+24}{6+\sqrt{12-x}}=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-24\left(tm\right)\\\dfrac{\sqrt[3]{x+24}+3}{\sqrt[3]{\left(x+24\right)^2}}=\dfrac{-1}{6+\sqrt{12-x}}\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\dfrac{\sqrt[3]{x+24}+3}{\sqrt[3]{x+24}}+\dfrac{1}{\sqrt[3]{x+24}}+\dfrac{1}{6+\sqrt{12-x}}-\dfrac{1}{\sqrt[3]{x+24}}=0\\ \Leftrightarrow\dfrac{\sqrt[3]{x+24}+4}{\sqrt[3]{x+24}}+\dfrac{\sqrt[3]{x+24}+4-10-\sqrt{12-x}}{\sqrt[3]{x+24}\left(6+\sqrt{12-x}\right)}=0\\ \Leftrightarrow\dfrac{x+88}{\sqrt[3]{x+24}\left(\sqrt[3]{\left(x+24\right)^2}-4\sqrt[3]{x+24}+16\right)}+\dfrac{\sqrt[3]{x+24}+4-10-\sqrt{12-x}}{\sqrt[3]{x+24}\left(6+\sqrt{12-x}\right)}=0\)

Xét \(\sqrt[3]{x+24}+4-10-\sqrt{12-x}=\dfrac{x+88}{\sqrt[3]{\left(x+24\right)^2}-4\sqrt[3]{x+24}+16}-\dfrac{x+88}{10+\sqrt{12-x}}=0\)

\(=\left(x+88\right)\left(\dfrac{1}{\sqrt[3]{\left(x+24\right)^2}-4\sqrt[3]{x+24}+16}-\dfrac{1}{10+\sqrt{12-x}}\right)\)

Thay vào PT (2) ta đặt đc nhân tử chung là \(x+88\)

Và ngoặc lớn còn lại vô nghiệm

\(\Leftrightarrow x+88=0\Leftrightarrow x=-88\left(tm\right)\)

Vậy PT có nghiệm \(x\in\left\{-88;-24;3\right\}\)

P/s mình thấy giải theo PP đặt ẩn phụ dễ hơn á ;-;

Lag tí -.-'

`ĐK:2<=x<=6`

BP 2 vế ta có:

`x-2+6-x+2\sqrt{(x-2)(6-x)}=x^2-8x+24`

`<=>4+2\sqrt{(x-2)(6-x)}=x^2-8x+24`

`<=>2\sqrt{(x-2)(6-x)}=x^2-8x+20`

`<=>2sqrt{-x^2+8x-12}=x^2-8x+20`

`<=>-x^2+8x-20+2sqrt{-x^2+8x-12}=0`

`<=>-x^2+8x-12+2sqrt{-x^2+8x-12}-8=0`

Đặt `sqrt{-x^2+8x-12}=a(a>=0)`

`pt<=>a^2+2a-8=0`

`<=>a=2(tm),a=-4(l)`

`<=>-x^2+8x-12=4`

`<=>x^2-8x+16=0`

`<=>(x-4)^2=0<=>x=4(tmđk)`

Vậy `S={4}`

\(\sqrt{x-2}+\sqrt{6-x}\text{=}\sqrt{x^2-8x+24}\)

\(ĐKXĐ:2\le x\le6\)

Xét VP của pt ta thấy : \(\sqrt{x^2-8x+24}\text{=}\sqrt{x^2-8x+16+8}\)

\(\text{=}\sqrt{\left(x-4\right)^2+8}\)

\(\Rightarrow VP\ge\sqrt{8}\)

Xét VT của pt ta có :

\(VT^2\text{=}x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}\)

\(VT^2\text{=}4+2\sqrt{\left(x-2\right)\left(6-x\right)}\)

Áp dụng BĐT cô si cho 2 số không âm ta có :

\(2\sqrt{\left(x-2\right)\left(6-x\right)}\le\left(\sqrt{x-2}\right)^2+\left(\sqrt{6-x}\right)^2\)

\(\text{=}x-2+6-x\text{=}4\)

\(\Rightarrow VT^2\le8\)

\(\Rightarrow VT\le\sqrt{8}\)

Để \(VT\text{=}VP\) \(\Leftrightarrow\left\{{}\begin{matrix}x-4\text{=}0\\\sqrt{x-2}\text{=}\sqrt{6-x}\end{matrix}\right.\)

\(\Leftrightarrow x=4\left(TM\right)\)

Vậy...........

\(\sqrt{12-x}-3+\sqrt[3]{x+24}-3=0\)

Liên hợp sẽ có:

\(\frac{3-x}{\sqrt{12-x}+3}+\frac{x-3}{\sqrt[3]{x+24}+3}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt[3]{x+24}+3}-\frac{1}{\sqrt{12-x}+3}\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-3=0\\\left(\frac{1}{\sqrt[3]{x+24}+3}-\frac{1}{\sqrt{12-x}+3}\right)=0\end{cases}}\)

\(\Rightarrow x=3\)

Xin dấu k nha

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

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

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

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

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

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

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

ĐK: \(-24\le x\le12\)

Đặt : \(\left\{{}\begin{matrix}a=\sqrt[3]{x+24}\\b=\sqrt{12-x}\end{matrix}\right.\Rightarrow a^3+b^2=36}\)Từ cách đặt => pt trở thành a+b=6

=> Hệ :\(\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=a-6\\a^3+a^2-12a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=a-6\\a\left(a^2+a-12\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=a-6\\\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\end{matrix}\right.\)Xong tìm ra b => thay vào cách đặt tìm ra x,y

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)

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

\(\Rightarrow x=2\)

3.

ĐKXĐ: \(x\ge-1\)

\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\) \(\left(Đk:-3\le x\le3\right)\)

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

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

\(2\sqrt{x+3}+2\sqrt{3-x}=x+2\sqrt{3-x}+4\)

\(2\sqrt{x+3}=x+4\)

\(4\left(x+3\right)=x^2+8x+14\)

\(x^2+4x+2=0\)

\(\Delta=16-8=8\)

\(\Delta>0\)=> phương trình có 2 nghiệm phân biệt

\(\left[{}\begin{matrix}x=\dfrac{-4+2\sqrt{2}}{2}=-2+\sqrt{2}\\x=\dfrac{-4-2\sqrt{2}}{2}=-2-\sqrt{2}\end{matrix}\right.\)