ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\\frac{-1-\sqrt{5}}{4}\le x\le-\frac{1}{8}\end{matrix}\right.\)(Có thể chưa chính xác)

\(12x^2+16x+1=2\sqrt{24x^3+12x^2-6x}+4\sqrt{x^2-x}+4\sqrt{8x^3+9x^2+x}\)

Áp dụng AM-GM:

\(2\sqrt{24x^3+12x^2-6x}=2\sqrt{6x\left(4x^2+2x-1\right)}\le6x+\left(4x^2+2x-1\right)=4x^2+8x-1\left(1\right)\)

\(4\sqrt{x^2-x}=2\sqrt{1.\left(4x^2-4x\right)}\le4x^2-4x+1\left(2\right)\)

\(4\sqrt{8x^3+9x^2+x}=2\sqrt{\left(4x^2+4x\right)\left(8x+1\right)}\le\left(4x^2+4x\right)+\left(8x+1\right)=4x^2+12x+1\left(3\right)\)

Cộng \(\left(1\right),\left(2\right),\left(3\right)\), ta có: \(VP\le VT\)

Dấu ''='' xảy ra khi :

\(\left\{{}\begin{matrix}4x^2+2x-1=6x\\4x^2-4x=1\\4x^2+4x=8x+1\end{matrix}\right.\)\(\Rightarrow4x^2-4x-1=0\)

\(\Rightarrow x=\frac{1\pm\sqrt{2}}{2}\) (t/m ĐKXĐ)

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

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

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

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

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

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

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

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

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

 \(x-40=0\)

  \(x=0+40\)

 \(x=40\)

\(x+36=0\)

   \(x=0-36\)

   \(x=-36\)

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

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

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

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

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

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

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

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

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

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 \(x=-\dfrac{1}{13}\)

học ngu, dễ vậy

Quy đồng, khử mẫu

\(\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x-4}-\dfrac{12x-1}{4-4x}\) Đkxđ : x≠1,-1

\(\Leftrightarrow\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x-4}+\dfrac{12x-1}{4x-4}\)

\(\Leftrightarrow\dfrac{6}{x^2-1}+5=\dfrac{20x-2}{4x-4}\)

\(\Leftrightarrow\dfrac{5x^2+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{10x-1}{2\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{5x^2+1}{\left(x+1\right)\left(x-1\right)}-\dfrac{10x-1}{2\left(x-1\right)}=0\)

\(\Leftrightarrow\dfrac{1-9x}{2\left(x+1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow1-9x=0\)

\(\Leftrightarrow9x=1\)

\(\Leftrightarrow x=\dfrac{1}{9}\)

Vậy S=\(\left\{\dfrac{1}{9}\right\}\)

\(\sqrt{x^2-x+16}=4\)

\(\Rightarrow x^2-x+16=16\\ \Rightarrow x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Ta có: \(\sqrt{x^2-x+16}=4\)

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

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

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

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

<=> \(\frac{6\left(x+4\right)-30x+120}{30}=\frac{10x-15x+30}{30}\)

<=> 6x + 24 - 30x + 120 = -5x + 30

<=> -24x + 5x = 30 - 144

<=> -19x = -114

<=> x = 6

Vậy S = {6}

a: Khi m=1 thì (1) sẽ là 2x^2-3x-5=0

=>2x^2-5x+2x-5=0

=>(2x-5)(x+1)=0

=>x=5/2 hoặc x=-1

b: 2x1(2+x2)+4x2(1-x1)+8x1x2=2015

=>4x1+4x2+8x1x2=2015

=>4*(x1+x2)+8x1x2=2015

=>4*(2m+1)/2+8*(-m-4)/2=2015

=>4m+2-4m-16=2015

=>-14=2015(loại)

b: Ta có: \(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow x^2+3x+2-x^2+3x-2-2x^2-4=0\)

\(\Leftrightarrow-2x^2+6x-4=0\)

a=-2; b=6; c=-4

Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=1\left(nhận\right);x_2=\dfrac{c}{a}=2\left(loại\right)\)