Lời giải:

ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow (3x^2-10x-25)=2(x+3)(\sqrt{2x-1}-3)$

$\Leftrightarrow (x-5)(3x+5)=2(x+3).\frac{2(x-5)}{\sqrt{2x-1}+3}$

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

Xét biểu thức trong ngoặc vuông:

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

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

Dễ thấy điều này vô lý vì với $x\geq \frac{1}{2}$ thì vế trái không âm còn vế phải âm.

Vậy $x-5=0\Leftrightarrow x=5$

ĐKXĐ: ...

\(\sqrt[3]{x^3+5x^2}-x-2+x+1-\sqrt{\dfrac{5x^2-2}{6}}=0\)

\(\Leftrightarrow\dfrac{x^3+5x^2-\left(x+2\right)^3}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{x^3+5x^2}+\sqrt[3]{\left(x^3+5x^2\right)^2}}+\dfrac{\left(x+1\right)^2-\dfrac{5x^2-2}{6}}{x+1+\sqrt{\dfrac{5x^2-2}{6}}}=0\)

\(\Leftrightarrow\left(x^2+12x+8\right)\left(\dfrac{1}{6\left(x+1\right)+\sqrt{6\left(5x^2-2\right)}}-\dfrac{1}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{x^3+5x^2}+\sqrt[3]{\left(x^3+5x^2\right)^2}}\right)=0\)

\(\Leftrightarrow x^2+12x+8=0\)

\(\Leftrightarrow\sqrt{14x+7}-7-\left(\sqrt{2x+3}-3\right)=\sqrt{5x+1}-4\)

\(\Leftrightarrow\dfrac{14x+7-49}{\sqrt{14x+7}+7}-\dfrac{2x+3-9}{\sqrt{2x+3}+3}=\dfrac{5x+1-16}{\sqrt{5x+1}+4}\)

\(\Leftrightarrow\dfrac{14x-42}{\sqrt{14x+7}+7}-\dfrac{2x-6}{\sqrt{2x+3}+3}=\dfrac{5x-15}{\sqrt{5x+1}+4}\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{14}{\sqrt{14x+7}}-\dfrac{2}{\sqrt{2x+3}+3}-\dfrac{5}{\sqrt{5x+1}+4}\right)=0\Leftrightarrow x=3\)

`x^2+\sqrt{2x+1}+sqrt{x-3}=5x`

Bài này dùng pp liên hợp với đk của x là `x>=3`

`pt<=>x^2-16+\sqrt{2x+1}-3+\sqrt{x-3}-1=5x-20`

`<=>(x-4)(x+4)+(2x-8)/(\sqrt{2x+1}+3)+(x-4)/(\sqrt{x-3}+1)=5(x-4)`

`<=>(x-4)(x+4+2/(\sqrt{2x+1}+3)+1/(\sqrt{x-3}+1)-5)=0`

`<=>(x-4)(x-1+2/(\sqrt{2x+1}+3)+1/(\sqrt{x-3}+1))=0`

Vì `x>=3=>x-1>=2>0`

Mà `2/(\sqrt{2x+1}+3)+1/(\sqrt{x-3}+1)>0`

`=>x-1+2/(\sqrt{2x+1}+3)+1/(\sqrt{x-3}+1)>0`

`=>x-4=0<=>x=4(tm)`

Vậy `S={4}`

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

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

\(\Leftrightarrow5x+3=3-\sqrt{2}\)

\(\Leftrightarrow5x=-\sqrt{2}\)

\(\Leftrightarrow x=\frac{-\sqrt{2}}{5}\)

Do \(\Delta=5^2+4\cdot3\cdot4=25+48=73>0\) nên PT có 2 nghiệm phân biệt.

Khi đó: \(\left\{{}\begin{matrix}x_1+x_2=\frac{-b}{a}=\frac{-\left(-5\right)}{3}=\frac{5}{3}\\x_1x_2=\frac{c}{a}=\frac{-4}{3}\end{matrix}\right.\)

Từ đây, ta suy ra:

\(A=x_1^3x_2+x_1x_2^3\\ =x_1x_2\left(x_1^2+x^2_2\right)\\ =x_1x_2\left(x_1^2+2x_1x_2+x^2_2-2x_1x_2\right)\\ =x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =\frac{-4}{3}\cdot\left[\left(\frac{5}{3}\right)^2-\frac{-4\cdot2}{3}\right]\\ =\frac{-4}{3}\cdot\frac{25-\left(-8\cdot3\right)}{9}\\ =\frac{-4}{3}\cdot\frac{25+24}{9}\\ =\frac{-4}{3}\cdot\frac{49}{9}=\frac{-196}{27}\)

Chúc bạn học tốt nha.

Ta có:

A = x₁x₂(x₁² + x₂²) = x₁x₂[(x₁ + x₂)² - 2x₁x₂]

Ta có: \(\Delta=\left(-5\right)^2-4.3.\left(-4\right)=25+48>0\)

Áp dụng định lý Vi-ét với phương trình 3x² - 5x - 4 ta có:
x₁ + x₂ = \(\frac{-\left(-5\right)}{3}=\frac{5}{3}\)
x₁x₂ = \(\frac{-4}{3}\)

Thay vào A ta được:

A = \(\frac{-4}{3}\left[\left(\frac{5}{3}\right)^2-2.\frac{-4}{3}\right]=\frac{-4}{3}.\left(\frac{25}{9}+\frac{8}{3}\right)=\frac{-4}{3}.\frac{49}{3}=\frac{-196}{3}\)

(P/s: CÓ thể SAI)

Lời giải:
ĐKXĐ:.........

PT \(\Leftrightarrow 3(x^2-x)+[(x+1)-\sqrt{3x+1}]+[(x+2)-\sqrt{5x+4}]=0\)

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

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

Dễ thấy với $x\geq \frac{-1}{3}$ thì biểu thức trong ngoặc vuông luôn dương 

$\Rightarrow x^2-x=0$

$\Leftrightarrow x(x-1)=0$

$\Rightarrow x=0$ hoặc $x=1$ (đều tm)

ĐKXĐ: \(\left\{{}\begin{matrix}x-2>=0\\4-x>=0\end{matrix}\right.\)

=>2<=x<=4

\(\sqrt{x-2}-\sqrt{4-x}=2x^2-5x-3\)

=>\(\sqrt{x-2}-1+1-\sqrt{4-x}=2x^2-6x+x-3\)

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

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

=>x-3=0

=>x=3(nhận)