đặt \(\hept{\begin{cases}\sqrt[3]{3x-2}=a\\\sqrt{6-5x}=b\ge0\end{cases}}\) ta sẽ có hệ sau \(\hept{\begin{cases}3a+4b=10\\5a^3+3b^2=8\end{cases}}\)

rút thế \(b=\frac{10-3a}{4}\)xuống phương trình dưới ta có\

\(5a^3+3\left(\frac{10-3a}{4}\right)^2=8\) hay 

\(80a^3+27a^2-180a+172=0\Leftrightarrow\left(a+2\right)\left(80a^2-133a+86\right)=0\Leftrightarrow a=-2\)

hay \(\sqrt[3]{3x-2}=-2\Leftrightarrow x=-2\) thay lại thỏa mãn

vậy phương trình có nghiệm duy nhất x=-2

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

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

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

\(\Leftrightarrow\frac{3x^2-5x-2}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-4}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x^2-3x-6}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x^2-3x+2}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)

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

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

Dễ thấy: \(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\) vô nghiệm

\(\Rightarrow x-2=0\Rightarrow x=2\)

sao cái từ "dễ thấy" nó khó thấy quá v 

haha :v

\(\text{x ∈ ∅}\)

Sr tui bj cuồng liên hợp làm mãi cách này có lố ko nhỉ :v

Đk:\(x\ge\frac{8}{3}\)

\(pt\Leftrightarrow4x-2-8-\left(3\sqrt{5x-6}-9\right)=\sqrt{3x-8}-1\)

\(\Leftrightarrow4x-2-10-\frac{9\left(5x-6\right)-81}{3\sqrt{5x-6}+9}=\frac{3x-8-1}{\sqrt{3x-8}+1}\)

\(\Leftrightarrow4\left(x-3\right)-\frac{45\left(x-3\right)}{3\sqrt{5x-6}+9}-\frac{3\left(x-3\right)}{\sqrt{3x-8}+1}=0\)

\(\Leftrightarrow\left(x-3\right)\left(4-\frac{45}{3\sqrt{5x-6}+9}-\frac{3}{\sqrt{3x-8}+1}\right)=0\)

Dễ thấy: \(4-\frac{45}{3\sqrt{5x-6}+9}-\frac{3}{\sqrt{3x-8}+1}>0\forall x\ge\frac{8}{3}\)

\(\Rightarrow x-3=0\Rightarrow x=3\)

cảm ơn bạn nhiều lắm :v mà cô bọn tui bắt làm bài này theo cách tổng bình phương :v hiccc

\(2\left(2x-1\right)-3\sqrt{5x-6}=\sqrt{3x-8}\left(x\ge\frac{8}{3}\right)\)

PT đã cho tương đương với \(2\left(2x-1\right)=\sqrt{3x-8}+3\sqrt{5x-6}\)

\(\Leftrightarrow\sqrt{3x-8}+3\sqrt{5x-6}-4x+2=0\)

\(\Leftrightarrow2\sqrt{3x-8}+2\cdot3\sqrt{5x-6}-8x+4=0\)

\(\Leftrightarrow-2\sqrt{3x-8}-2\cdot3\sqrt{5x-6}+8x-4=0\)

\(\Leftrightarrow\left(3x-8\right)-2\sqrt{3x-8}+1+\left(5x-6\right)-2\cdot3\sqrt{5x-6}+9=0\)

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

(*) chỉ thỏa mãn khi \(\hept{\begin{cases}\sqrt{3x-8}-1=0\\\sqrt{5x-8}-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{3x-8}=1\\\sqrt{5x-8}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}3x-8=1\\5x-8=9\end{cases}\Leftrightarrow}x=3\left(tm\right)}\)

Vậy x=3

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

=>\(\sqrt{3x+1}-2+2\sqrt{x+3}-4=3\sqrt{5x-1}-6\)

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

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

=>\(\left(x-1\right)\left(\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{2}{\sqrt{x+3}+2}-\dfrac{15}{\sqrt{5x-1}+2}\right)=0\)

=>x-1=0

=>x=1

ĐKXĐ: \(x\ge\dfrac{22}{27}\)

Đặt \(\sqrt{3x-2}=y\ge\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x^2=3\sqrt{3y-2}-2\\y^2=3x-2\end{matrix}\right.\)

Đặt \(\sqrt{3y-2}=z\ge0\Rightarrow\left\{{}\begin{matrix}x^2=3z-2\\z^2=3y-2\end{matrix}\right.\)

Ta nhận được 1 hệ hoán vị vòng quanh rất cơ bản: \(\left\{{}\begin{matrix}x^2=3z-2\\y^2=3x-2\\z^2=3y-2\end{matrix}\right.\)

Các hàm \(f\left(t\right)=t^2\) và \(g\left(t\right)=3t-2\) cùng đơn điệu tăng trên \(\left(0;+\infty\right)\) nên hệ trên có nghiệm duy nhất \(x=y=z\)

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

Đk: `x >= 0`.

`<=> sqrtx + sqrt(x+3) + 2sqrt(x(x+3)) - (3x+9) + 5x = 0`

Đặt `sqrt x = a, sqrt(x+3) = b`

`<=> a + b + 2ab - 3b^2 + 5a^2 = 0`

`<=> (a+b)(5a+1-3b) = 0`

`<=> a = -b` hoặc `5a + 1 = 3b`.

Đến đây bạn biến đổi ẩn rồi tự giải tiếp ha. 

ĐKXĐ:

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

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

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

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