Đặt a = \(\sqrt{12-x}\), b = \(\sqrt[3]{24+x}\), ta có:

a + b = 6 => a = 6 - b , (a+b)² = 36 (1)

Có a² + b³ = 12 - x + 24 + x = 36 (2)

(1), (2) suy ra (a+b)² = a² + b³

<=> a² + 2ab + b² = a² + b³

<=> 2ab + b² = b³

<=> b³ - b² - 2ab = 0

<=> b(b² - b - 2a)=0

Thay a = 6 - b , pt trở thành:

b(b² - b - 2*6 + 2b) = 0

<=> b(b² + b - 12) = 0

<=> b(b² + 4b - 3b -12) = 0

<=> b(b - 3)(b + 4) = 0

<=> b = 0 => x = -24

       b = 3 => x = 3

       b = -4 => x = -88

Vậy S = {-88;-24;3}

ĐK: \(12-x\ge0\Rightarrow x\le12\)

đặt

\(\hept{\begin{cases}u=\sqrt{12-x}\\v=\sqrt[3]{24+x}\end{cases}}=>\hept{\begin{cases}u^2=12-x\\v^3=24+x\end{cases}}=>\hept{\begin{cases}u^2+v^3=36\left(1\right)\\u+v=6\left(2\right)\end{cases}}\)

từ (2) ta có: \(u=6-v\) thay vào (1) được: \(\left(6-v\right)^2+v^3=36\Leftrightarrow v^3+v^2-12v=0\)

\(\Leftrightarrow v\left(v^2+v-12\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}v=0\\v^2+v-12=0\end{cases}}\Leftrightarrow v=0;v=3;v=-4\)

với \(v=0\Rightarrow u=6\Rightarrow12-x=36\Rightarrow x=-24\)(TM)

với \(v=3\Rightarrow u=3\Rightarrow x=3\left(TM\right)\)

với \(v=-4\Rightarrow u=10\Rightarrow x=-88\left(TM\right)\)

vậy tập nghiệm của PT là S={-24,3,-88}

a) Điều kiện $x \ge -5$. Đặt $\sqrt{x+5}=a$ thì $x=a^2-5$. Thay vào ta có $$\begin{array}{l} (a^2-5)^2-7(a^2-5)=6a-30 \\ \Leftrightarrow a^4-17a^2-6a+90=0 \Leftrightarrow (a^2+6a+10)(a-3)^2=0 \end{array}$$

Vậy $a=3 \Leftrightarrow \boxed{ x= 4}$.

Bài 2:
ĐKXĐ: $6\geq x\geq \frac{-1}{3}$
PT $\Leftrightarrow (\sqrt{3x+1}-4)+(1-\sqrt{6-x})+(3x^2-14x-5)=0$

$\Leftrightarrow \frac{3(x-5)}{\sqrt{3x+1}+4}+\frac{x-5}{\sqrt{6-x}+1}+(3x+1)(x-5)=0$
$\Leftrightarrow (x-5)\left[\frac{3}{\sqrt{3x+1}+4}+\frac{1}{\sqrt{6-x}+1}+(3x+1)\right]=0$

Với $x$ thuộc đkxđ, dễ thấy biểu thức trong ngoặc vuông $>0$

$\Rightarrow x-5=0$

$\Leftrightarrow x=5$

Bài 3:

PT $3x=\sqrt{x^2+12}-\sqrt{x^2+5}+5>0$

$\Rightarrow x>0$

Lại có:

PT $\Leftrightarrow \sqrt{x^2+12}-4=3(x-2)+(\sqrt{x^2+5}-3)$

$\Leftrightarrow \frac{x^2-4}{\sqrt{x^2+12}+4}=3(x-2)+\frac{x^2-4}{\sqrt{x^2+5}+3}$

$\Leftrightarrow (x-2)\left[\frac{x+2}{\sqrt{x^2+12}+4}-3-\frac{x+2}{\sqrt{x^2+5}+3}\right]=0$

Với $x>0$, dễ thấy:
$\frac{x+2}{\sqrt{x^2+5}+3}+3>\frac{x+2}{\sqrt{x^2+12}+4}$ nên biểu thức trong ngoặc vuông âm.

Do đó $x-2=0\Leftrightarrow x=2$ (tm)

Bài 1:

Đặt $\sqrt[3]{3x-2}=a; \sqrt{6-5x}=b$ với $b\geq 0$. Khi đó pt trở thành:
\(\left\{\begin{matrix} 2a+3b=8\\ 5a^3+3b^2=8\end{matrix}\right.\Rightarrow 5a^3+3(\frac{8-2a}{3})^2=8\)

\(\Leftrightarrow 15a^3+(8-2a)^2=24\)

\(\Leftrightarrow 15a^3+4a^2-32a+40=0\)

\(\Leftrightarrow 15a^2(a+2)-26a(a+2)+20(a+2)=0\)

$\Leftrightarrow (a+2)(15a^2-26a+20)=0$

Dễ thấy $15a^2-26a+20>0$ nên $a+2=0$

$\Leftrightarrow a=-2$
$\Rightarrow b=4$

$\Rightarrow x=-2$

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

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

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

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

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

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

Pt \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}=0\) vô n_o

(vì \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}< \dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\forall x\ge2\Rightarrow VT< 0\))

=> x - 2 = 0

<=> x = 2 (nhận)

\(\sqrt{4x+1}-\sqrt{3x-2}=\dfrac{x+3}{5}\)

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

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

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

TH1:

x + 3 = 0

<=> x = - 3 (loại)

TH2:

\(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}=0\)

\(\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=5\)

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

\(\Leftrightarrow\dfrac{4x+1-9}{\sqrt{4x+1}+3}+\dfrac{3x-2-4}{\sqrt{3x-2}+2}=0\)

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

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

Pt \(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}>0\forall x\ge\dfrac{2}{3}\) => vô n_o

=> x - 2 = 0

<=> x = 2 (nhận)

~ ~ ~

Vậy x = 2

\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

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

\(\Leftrightarrow\sqrt{2\left(x+3\right)\left(x+1\right)}-\dfrac{\left(4x^2+8x+4\right)-\left(x^2-1\right)}{\sqrt{x^2-1}+2x+2}=0\)

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

\(\Leftrightarrow\sqrt{x+1}\left[2\sqrt{x+3}-\dfrac{\sqrt{x+1}\left(3x+5\right)}{\sqrt{x+1}\left(\sqrt{x-1}+2\sqrt{x+1}\right)}\right]=0\)

\(\Leftrightarrow\sqrt{x+1}\left[2\sqrt{x+3}-\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}\right]=0\)

TH1

x + 1 = 0

<=> x = - 1 (loại)

TH2

\(2\sqrt{x+3}-\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}=0\)

mà \(2\sqrt{x+3}=\dfrac{4x+12}{2\sqrt{x+3}}>\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}\forall x\ge1\)

=> VT > 0

=> vô n_o

~ ~ ~

Vậy pt vô n_o

a/ ĐKXĐ: ...

\(\Leftrightarrow x+8+\sqrt{x+8}-\left(x+8\right)=\sqrt{x}+\sqrt{x+3}\)

\(\Leftrightarrow\sqrt{x+8}=\sqrt{x}+\sqrt{x+3}\)

\(\Leftrightarrow x+8=2x+3+2\sqrt{x^2+3x}\)

\(\Leftrightarrow5-x=2\sqrt{x^2+3x}\) (\(x\le5\))

\(\Leftrightarrow x^2-10x+25=4\left(x^2+3x\right)\)

\(\Leftrightarrow...\)

b/ ĐKXĐ: \(2\le x\le5\)

\(\Leftrightarrow2\left(x-2\right)+\sqrt{2\left(x-2\right)}\left(\sqrt{5-x}-\sqrt{3x-3}\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{2x-4}+\sqrt{5-x}=\sqrt{3x-3}\left(1\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\sqrt{\left(2x-4\right)\left(5-x\right)}=x-2\)

\(\Leftrightarrow\left(2x-4\right)\left(5-x\right)=\left(x-2\right)^2\)

\(\Leftrightarrow...\)

c/ ĐKXĐ: \(x\le12\)

\(\Leftrightarrow\sqrt[3]{24+x}\sqrt{12-x}-6\sqrt{12-x}+12-x=0\)

\(\Leftrightarrow\sqrt{12-x}\left(\sqrt[3]{24+x}-6+\sqrt{12-x}\right)=0\)

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

Xét (1):

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{24+x}=a\\\sqrt{12-x}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^3+b^2=36\end{matrix}\right.\)

\(\Leftrightarrow a^3+\left(6-a\right)^2=36\)

\(\Leftrightarrow a^3+a^2-12a=0\)

\(\Leftrightarrow a\left(a^2+a-12\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{24+x}=0\\\sqrt[3]{24+x}=3\\\sqrt[3]{24+x}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}24+x=0\\24+x=27\\24+x=-64\end{matrix}\right.\)

d/ ĐKXĐ: \(x\le\frac{3}{2}\) ; \(x\ne\frac{3}{8};x\ne-\frac{13}{24}\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2\sqrt{3-2x}-3}-\frac{1}{3-2\sqrt[3]{5+3x}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\frac{1}{2\sqrt{3-2x}-3}=\frac{1}{3-2\sqrt[3]{5+3x}}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\sqrt{3-2x}-3=3-2\sqrt[3]{5+3x}\)

\(\Leftrightarrow\sqrt[3]{5+3x}+\sqrt{3-2x}=3\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{5+3x}=a\\\sqrt{3-2x}=b\ge0\end{matrix}\right.\) ta được:

\(\left\{{}\begin{matrix}a+b=3\\2a^3+3b^2=19\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\2a^3+3b^2=19\end{matrix}\right.\)

\(\Leftrightarrow2a^3+3\left(3-a\right)^2=19\)

\(\Leftrightarrow2a^3+3a^2-18a+8=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-4\\a=\frac{1}{2}\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{5+3x}=-4\\\sqrt[3]{5+3x}=\frac{1}{2}\\\sqrt[3]{5+3x}=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5+3x=-64\\5+3x=\frac{1}{8}\\5+3x=8\end{matrix}\right.\)