2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

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

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

=>\(\sqrt{3x}=2\)

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

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

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

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

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

\(10x^2-9x-8x\sqrt{2x^2-3x+1}+3=0\)

Đặt \(a=\sqrt{2x^2-3x+1}\ge0\) thì:

\(4x^2+3a^2-8ax=0\)

\(\Leftrightarrow\left(2x-a\right)\left(2x-3a\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{a}{2}\\x=\dfrac{3a}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{2x^2-3x+1}}{2}\\x=\dfrac{3\sqrt{2x^2-3x+1}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\sqrt{2x^2-3x+1}\\2x=3\sqrt{2x^2-3x+1}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}4x^2=2x^2-3x+1\\4x^2=9\left(2x^2-3x+1\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x^2+3x-1=0\\\left(3-2x\right)\left(7x-3\right)=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\x=\dfrac{3}{2}\\x=\dfrac{\sqrt{17}}{4}-\dfrac{3}{4}\end{matrix}\right.\)

ĐKXĐ : \(x\ge-\dfrac{3}{8}\)

Ta có : \(\sqrt{8x+3}=9x^2+10x+\dfrac{9}{4}\)

\(\Leftrightarrow36x^2+40x+9-4\sqrt{8x+3}=0\)

\(\Leftrightarrow\left(36x^2+48x+16\right)-8x-3-4\sqrt{8x+3}-4=0\)

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

\(\Leftrightarrow\left(6x+\sqrt{8x+3}+6\right).\left(6x+2-\sqrt{8x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{8x+3}=-6x-6\left(1\right)\\\sqrt{8x+3}=-6x-2\left(2\right)\end{matrix}\right.\)

Giải (1) ta có :

(1) <=> \(8x+3=\left(-6x-6\right)^2\) (với \(-6x-6\ge0\Leftrightarrow x\le-1\))

\(\Leftrightarrow36x^2+64x+33=0\)

\(\Leftrightarrow\left(6x+\dfrac{16}{3}\right)^2+\dfrac{41}{9}=0\)

\(\Leftrightarrow x\in\varnothing\) => (1) vô nghiệm

Giải (2) ta có 

(2) <=> \(8x+3=\left(-6x-2\right)^2\) (với \(x\le-\dfrac{1}{3}\)) (*)

\(\Leftrightarrow36x^2+16x+1=0\)

\(\Leftrightarrow x=\dfrac{-4\pm\sqrt{7}}{18}\) 

Kết hợp (*) và ĐKXĐ ta được \(x=\dfrac{-4+\sqrt{7}}{18}\) là nghiệm phương trình

Câu 1:

PT \(\Leftrightarrow x^2+3x+8=(x+5)\sqrt{x^2+x+2}\)

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

Đặt \(\sqrt{x^2+x+2}=a; x+5=b(a\geq 0)\)

\(PT\Leftrightarrow a^2+2b-4=ba\)

\(\Leftrightarrow (a^2-4)-b(a-2)=0\)

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

Nếu \(a=2\Rightarrow x^2+x+2=a^2=4\)

\(\Leftrightarrow x^2+x-2=0\Leftrightarrow (x-1)(x+2)=0\Rightarrow x=1; x=-2\) (đều thỏa mãn)

Nếu \(a+2=b\Leftrightarrow \sqrt{x^2+x+2}+2=x+5\)

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

\(\Rightarrow \left\{\begin{matrix} x+3\geq 0\\ x^2+x+2=(x+3)^2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+3\geq 0\\ 5x+7=0\end{matrix}\right.\Rightarrow x=\frac{-7}{5}\) (thỏa mãn)

Vậy..........

Câu 2:

ĐKXĐ: \(x\geq 1\) hoặc \(x\leq \frac{1}{2}\)

\(10x^2-9x-8x\sqrt{2x^2-3x+1}+3=0\)

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

Đặt \(\sqrt{2x^2-3x+1}=a(a\geq 0)\)

Khi đó PT \(\Leftrightarrow 3a^2-8xa+4x^2=0\)

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

Nếu \(a=\sqrt{2x^2-3x+1}=2x\Rightarrow \left\{\begin{matrix} x\geq 0\\ 2x^2-3x+1=4x^2\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ 2x^2+3x-1=0\end{matrix}\right.\Rightarrow x=\frac{-3+\sqrt{17}}{4}\) (t/m)

Nếu \(3a=3\sqrt{2x^2-3x+1}=2x\Rightarrow \left\{\begin{matrix} x\geq 0\\ 9(2x^2-3x+1)=4x^2\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ 14x^2-27x+9=0\end{matrix}\right.\Rightarrow x=\frac{3}{2}; x=\frac{3}{7}\) (t/m)

Vậy...........

3:

ĐK: \(x\geq -\sqrt[3]{3}\)

Đặt \(\sqrt{x^3+3}=a(a\geq 0)\)

PT \(\Leftrightarrow (x^3+3)+6x^2-2x-(5x-1)\sqrt{x^3+3}=0\)

\(\Leftrightarrow a^2+6x^2-2x-(5x-1)a=0\)

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

Coi đây là pt bậc 2 ẩn $x$.

Ta thấy \(\Delta=(5a+2)^2-24(a^2+a)=(a-2)^2\)

\(\Rightarrow x=\frac{(5a+2)\pm \sqrt{\Delta}}{12}\Rightarrow x=\frac{a}{2}\) hoặc \(x=\frac{a+1}{3}\)

Nếu \(x=\frac{a}{2}=\frac{\sqrt{x^3+3}}{2}\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2=\frac{x^3+3}{4}\end{matrix}\right.\)

\( \Rightarrow \left\{\begin{matrix} x\geq 0\\ x^3-4x^2+3=0\end{matrix}\right. \Rightarrow \left\{\begin{matrix} x\geq 0\\ (x-1)(x^2-3x-3)=0\end{matrix}\right.\)

\( \Rightarrow \left[\begin{matrix} x=1\\ x=\frac{3+\sqrt{21}}{2}\end{matrix}\right.\) (t.m)

Nếu \(x=\frac{a+1}{3}\Rightarrow 3x-1=a=\sqrt{x^3+3}\)

\( \Rightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ (3x-1)^2=x^3+3\end{matrix}\right. \Rightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ x^3-9x^2+6x+2=0\end{matrix}\right.\)

\( \Rightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ (x-1)(x^2-8x-2)=0\end{matrix}\right.\Rightarrow x=1; x=4+3\sqrt{2}\)

Vậy \(x\in\left\{1; 4+3\sqrt{2}; \frac{3+\sqrt{21}}{2}\right\}\)

a: ĐKXĐ: x>=5

\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

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

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x>=1/2

\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)

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

=>\(5-\sqrt{2x-1}=0\)

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

=>2x-1=25

=>2x=26

=>x=13(nhận)

c: \(\sqrt{x^2-10x+25}=2\)

=>\(\sqrt{\left(x-5\right)^2}=2\)

=>\(\left|x-5\right|=2\)

=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

d: \(\sqrt{x^2-14x+49}-5=0\)

=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)

=>\(\sqrt{\left(x-7\right)^2}=5\)

=>|x-7|=5

=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)

\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)

\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a)ĐKXĐ:x\ge5\\ \sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=\dfrac{4}{2}\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2=2^2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=4+5\\ \Leftrightarrow x=9\left(tmđk\right)\)

Vậy \(S=\left\{9\right\}\)

\(b)ĐKXĐ:x\ge2\\ \sqrt{2x-1}-\sqrt{8x-4}+5=0\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{8x-4}=0-5\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow-\left(\sqrt{2x-1}\right)=\left(-5\right)^2\\ \Leftrightarrow-2x+1=-25\\ \Leftrightarrow-2x=\left(-25\right)-1\\ \Leftrightarrow-2x=-26\\ \Leftrightarrow x=\dfrac{-26}{-2}\\ \Leftrightarrow x=13\left(tmđk\right)\)

Vậy \(S=\left\{13\right\}\)

\(c)\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+5\\x=\left(-2\right)+5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{7;3\right\}\)

\(d)\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{x^2-14x+49}=0+5\\ \Leftrightarrow\sqrt{x^2-14x+49}=5\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5+7\\x=\left(-5\right)+7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{12;2\right\}.\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2\ge0\\3x^2-17x+4=\left(3x-2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\3x^2-17x+4=9x^2-12x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\6x^2+5x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\left[{}\begin{matrix}x=0< \dfrac{2}{3}\left(loại\right)\\x=-\dfrac{5}{6}< \dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

b.

ĐKXĐ: \(\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\)

Đặt \(\sqrt{x^2-5x+4}=t\ge0\Leftrightarrow x^2-5x=t^2-4\)

\(\Rightarrow2x^2-10x=2t^2-8\)

Phương trình trở thành:

\(2t^2-8-3t+6=0\)

\(\Leftrightarrow2t^2-3t-2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-\dfrac{1}{2}< 0\left(loại\right)\end{matrix}\right.\)

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

\(\Leftrightarrow x^2-5x=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

Nhìn không đủ chán rồi không dám động vào

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À do nãy máy lag sr :) Chứ bài đặt ẩn phụ mệt lắm :)