a, \(\sqrt{4x^2+20x+25}\) + \(\sqrt{x^2-8x+16}\) = \(\sqrt{x^2+18x+81}\)

⇔ 4x² + 20x + 25 + \(2\sqrt{\left(4x^2+20x+25\right)\left(x^2-8x+16\right)}\) = x² + 18x + 81

⇔ 4x² + 20x + 25 - x² - 18x - 81 + \(2\sqrt{\left(2x+5\right)^2.\left(x-4\right)^2}\) = 0

⇔ 3x² + 2x - 56 + 2.(2x + 5) . (x - 4) = 0

⇔ 3x² + 2x - 56 + (4x + 10) . (x - 4) = 0

⇔ 3x² + 2x - 56 + 4x² - 16x + 10x - 40 = 0

⇔ 7x² - 4x - 96 = 0

x₁ = 4 ( nhận )

x₂ = \(\frac{-24}{7}\) ( nhận )

Vậy: S = {4; \(\frac{-24}{7}\)}

\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\\ \Leftrightarrow-2\sqrt{x-1}=-2\Leftrightarrow\sqrt{x-1}=1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{1}{3}\sqrt{2x}-2\sqrt{2x}+3\sqrt{2x}=12\\ \Leftrightarrow\dfrac{4}{3}\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=9\\ \Leftrightarrow2x=81\Leftrightarrow x=\dfrac{81}{2}\left(tm\right)\)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(3\sqrt{18x}-5\sqrt{8x}+4\sqrt{50x}=38\)

\(\Leftrightarrow9\sqrt{2x}-10\sqrt{2x}+20\sqrt{2x}=38\)

\(\Leftrightarrow19\sqrt{2x}=38\)

\(\Leftrightarrow\sqrt{2x}=2\)

\(\Leftrightarrow2x=4\)

hay x=2(thỏa ĐK)

b) ĐKXĐ: \(x\ge0\)

Ta có: \(3\sqrt{12x}-2\sqrt{27x}+4\sqrt{3x}=8\)

\(\Leftrightarrow6\sqrt{3x}-6\sqrt{3x}+4\sqrt{3x}=8\)

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

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

c) ĐKXĐ: \(x\ge5\)

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

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\)

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

\(\Leftrightarrow x-5=4\)

hay x=9

a)

\(3.3\sqrt{2x}-5.2\sqrt{2x}+4.5.\sqrt{2x}=38\\ \Leftrightarrow19\sqrt{2x}=38\\ \Leftrightarrow\sqrt{2x}=2\\ \Leftrightarrow x=2\)

b)

\(3.2.\sqrt{3x}-2.3.\sqrt{3x}+4.\sqrt{3x}=8\\ \Leftrightarrow4\sqrt{3x}=8\\ \Leftrightarrow\sqrt{3x}=2\\\Leftrightarrow x=\dfrac{2^2}{3}=\dfrac{4}{3} \)

c)

\(\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\)

a: =>3x^2-3x-2x+2=0

=>(x-1)(3x-2)=0

=>x=2/3 hoặc x=1

b: =>2x^2=11

=>x^2=11/2

=>\(x=\pm\dfrac{\sqrt{22}}{2}\)

c: Δ=5^2-4*1*7=25-28=-3<0

=>PTVN

f: =>6x^4-6x^2-x^2+1=0

=>(x^2-1)(6x^2-1)=0

=>x^2=1 hoặc x^2=1/6

=>\(\left[{}\begin{matrix}x=\pm1\\x=\pm\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)

d: =>(5-2x)(5+2x)=0

=>x=5/2 hoặc x=-5/2

e: =>4x^2+4x+1=x^2-x+9 và x>=-1/2

=>3x^2+5x-8=0 và x>=-1/2

=>3x^2+8x-3x-8=0 và x>=-1/2

=>(3x+8)(x-1)=0 và x>=-1/2

=>x=1

x 2  + (x + 2)(11x – 7) = 4

⇔  x 2  – 4 + (x + 2)(11x – 7) = 0

⇔ (x + 2)(x – 2) + (x + 2)(11x – 7) = 0

⇔ (x + 2)[(x – 2) + (11x – 7)] = 0

⇔ (x + 2)(x – 2 + 11x – 7) = 0

⇔ (x + 2)(12x – 9) = 0 ⇔ x + 2 = 0 hoặc 12x – 9 = 0

       x + 2 = 0 ⇔ x = - 2

      12x – 9 = 0 ⇔ x = 0,75

Vậy phương trình có hai nghiệm x = - 2 hoặc x = 0,75

\(x^2+\left(x+2\right)\left(11x-7\right)=4\)

\(\Leftrightarrow x^2+\left(11x^2+15x-14\right)=4\)

\(\Leftrightarrow12x^2+15x-18=0\)

\(\Leftrightarrow12x^2+24x-9x-18=0\)

\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)

\(\Leftrightarrow\left(12x-9\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}12x-9=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-2\end{cases}}}\)

Vậy nghiệm của phương trình là : \(S=\left\{-2;\frac{3}{4}\right\}\)