\(\sqrt{16x}=8\Leftrightarrow16x=64\Leftrightarrow x=4\)

Ta có: \(\sqrt{16x}=8\)

\(\Leftrightarrow16x=64\)

hay x=4

c: Ta có: \(\sqrt{2x}=\sqrt{5}\)

\(\Leftrightarrow2x=5\)

hay \(x=\dfrac{5}{2}\)

d: Ta có: \(\sqrt{3x-1}=4\)

\(\Leftrightarrow3x-1=16\)

\(\Leftrightarrow3x=17\)

hay \(x=\dfrac{17}{3}\)​

\(\dfrac{180}{x-4}-\dfrac{180}{x}=\dfrac{1}{2}\)

\(\Leftrightarrow\) \(\dfrac{2x\cdot180}{2x\left(x-4\right)}-\dfrac{2\cdot180\cdot\left(x-4\right)}{2x\left(x-4\right)}=0\)

\(\Leftrightarrow\) \(\dfrac{360x-360x+1440-x^2+4x}{2x\left(x-4\right)}=0\)

\(\Leftrightarrow\) \(\dfrac{-x^2+4x+1440}{2x\left(x-4\right)}=0\)

\(\Leftrightarrow-x^2+4x+1440=0\)

\(\Leftrightarrow-x^2+40x-36x+1440=0\)

\(\Leftrightarrow-x\cdot\left(x-40\right)\cdot\left(-36\right)\cdot\left(x-40\right)=0\)

\(\Leftrightarrow\left(x-40\right)\cdot\left(x-36\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-40=0\\x+36=0\end{matrix}\right.\)

 \(x-40=0\)

  \(x=0+40\)

 \(x=40\)

\(x+36=0\)

   \(x=0-36\)

   \(x=-36\)

\(\Leftrightarrow\left[{}\begin{matrix}x=40\\x=-36\end{matrix}\right.\)

\(180\left(\dfrac{1}{x-4}-\dfrac{1}{x}\right)=\dfrac{1}{2}\)

\(\dfrac{1}{x-4}-\dfrac{1}{x}=\dfrac{1}{360}\left(đk:x\ne0,4\right)\)

\(\dfrac{x-x+4}{x\left(x-4\right)}=\dfrac{1}{360}\)

\(\dfrac{4}{x\left(x-4\right)}=\dfrac{1}{360}\)

\(x^2-4x=1440\)

\(x^2-4x+4=1444\)

\(\left(x-2\right)^2=1444=38^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=38\\x-2=-38\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=40\\x=-36\end{matrix}\right.\)

         (12x-1)(6x-1)(4x-1)(3x-1)=5

<=>(12x-1)(12x-2)(12x-3)(12x-4)=40

<=>[(12x-1)(12x-4)] [(12x-2)(12x-3)] =40

<=>(144x^2 - 60x + 4) (144x^2 - 60x + 6) =40

đặt 144x^2 - 60x +4 = t  =>144x^2 - 60x +6 = t+2

ta có phương trình:

        t ( t+2 ) =40

<=> t^2 + 2t -40 =0

<=> (t+1)^2 -39 =0

<=> t+1=\(\sqrt{39}\)      hoặc t+1=\(-\sqrt{39}\)    <=> x=\(\sqrt{39}\) -1 hoặc x=\(-\sqrt{39}\) -1

tick nha

Ta có: \(\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}-\dfrac{2x}{x-1}\)

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

Suy ra: \(2x^2-2x-5x+5-2x^2-6x=4\)

\(\Leftrightarrow13x=-1\)

hay \(x=-\dfrac{1}{13}\)

Ta có: \(\sqrt{4x^2-4x+9}=3\)

\(\Leftrightarrow4x^2-4x=0\)

\(\Leftrightarrow4x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(ĐK:x\ne\pm\dfrac{3}{2}\\ PT\Leftrightarrow2x+3+2x-3=2x+4\\ \Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)

\(\dfrac{1}{2x-3}+\dfrac{1}{2x+3}=\dfrac{2x+4}{4x^2-9}\)

\(\dfrac{2x+3+2x-3}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{2x+4}{4x^2-9}\)

\(\dfrac{4x}{4x^2-9}=\dfrac{2x+4}{4x^2-9}\Rightarrow4x=2x+4\)

\(\Rightarrow2x=4\Rightarrow x=2\)

  \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

Đặt \(x^2+x-1=a\)

Ta có : \(x^2+x-1=\left(x+\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

\(\Rightarrow a\ge-\frac{5}{4}\)

Ta có pt : \(\left(a+1\right)\left(a-1\right)=24\)

\(\Leftrightarrow a^2-1=24\)

\(\Leftrightarrow a^2=25\)

\(\Leftrightarrow a=5\left(Do\text{ }a\ge-\frac{5}{4}\right)\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)