\(\left(1\right)\Leftrightarrow2x-3x^2+11-33x=6x-4-15x^2+10x\)

\(\Leftrightarrow12x^2-47x+15=0\)

\(\Delta=47^2-4.12.15=1489,\sqrt{\Delta}=\sqrt{1489}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{47+\sqrt{1489}}{24}\\x=\frac{47-\sqrt{1489}}{24}\end{cases}}\)

\(\left(2\right)\Leftrightarrow\frac{\left(x-3\right)^2-\left(x+3\right)^2}{x^2-9}=\frac{-5}{x^2-9}\)

\(\Leftrightarrow\left(x-3\right)^2-\left(x+3\right)^2=-5\)

\(\Leftrightarrow x^2-6x+9-x^2-6x-9=-5\)

\(\Leftrightarrow-12x=-5\Leftrightarrow x=\frac{5}{12}\)

(2-3x)(x+11)=(3x-2)(2-5x)

<=>(3x-2)(2-5x)-(2-3x)(x+11)=0

<=>(3x-2)(2-5x)+(3x-2)(x+11)=0

<=>(3x-2)[2-5x+x+11]=0

<=>(3x-2)(13-4x)=0

<=>\(\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}\)

\(\frac{x-3}{x+3}-\frac{x+3}{x-3}=-\frac{5}{x^2-9}\)

Đk:\(x\ne-3;x\ne3\)(*)

Với đk trên pt tương đương với:

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

\(x^2-6x+9-x^2-6x-9=-5.-12x=-5\)

\(x=\frac{15}{12}\left(tmđk\right)\)(*)

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)

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

\(=\frac{1}{x}-\frac{1}{x+3}=\frac{x+3}{x.\left(x+3\right)}-\frac{x}{x.\left(x+3\right)}\)

\(=\frac{3}{x.\left(x+3\right)}=\frac{3}{x^2+3x}\)

a/ \(\left(x-1\right)\left(x+2\right)+4\left(x-1\right)\sqrt{\frac{x+2}{x-1}}=12\)

Điều kiện: \(\left[\begin{matrix}x\le-2\\x>1\end{matrix}\right.\)

Xét \(x\le-2\) thì ta có

\(\left(x-1\right)\left(x+2\right)+4\left(x-1\right)\sqrt{\frac{x+2}{x-1}}=12\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-4\sqrt{\left(x-1\right)\left(x+2\right)}=12\)

Đặt \(\sqrt{\left(x-1\right)\left(x+2\right)}=a\left(a\ge0\right)\) thì pt thành

\(a^2-4a-12=0\)

\(\Leftrightarrow\left[\begin{matrix}a=-2\left(l\right)\\a=6\end{matrix}\right.\)

\(\Rightarrow\sqrt{\left(x-1\right)\left(x+2\right)}=6\)

\(\Leftrightarrow x^2+x-38=0\)

\(\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{2}+\frac{3\sqrt{17}}{2}\left(l\right)\\x=-\frac{1}{2}-\frac{3\sqrt{17}}{2}\end{matrix}\right.\)

Trường hợp x > 1 làm tương tự nhé

a)\(ĐKXĐ:\hept{\begin{cases}x>3\\x\le-1\end{cases}}\)
TH1: \(x-3>0\)
 \(\left(x-3\right)\left(x+1\right)+4.\frac{x-3}{\sqrt{x-3}}\sqrt{x+1}=-3\)
\(\left(x-3\right)\left(x+1\right)+4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
Đặt \(t=\sqrt{\left(x-3\right)\left(x+1\right)}\left(t\ge0\right)\)
Phương trình trở thành:
\(t^2+4t+3=0\Leftrightarrow\orbr{\begin{cases}t=-1\\t=-3\end{cases}}\)(ktm)=> Vô Nghiệm
TH2: \(x-3< 0\)
\(\left(x-3\right)\left(x+1\right)-4.\frac{3-x}{\sqrt{3-x}}\sqrt{-x-1}=-3\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)-4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
Tự làm tiếp nhé
 

b)Nhân chéo chuyển vế rút gọn ta được:
\(x^3-2x^2+3x-2=0\)
\(\Leftrightarrow x\left(x^2-2x+1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)^2+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x+2\right)=0\)
\(\Rightarrow x=1\)

máy tính sẵn sàng

cậu giỏi nhỉ

Nguyễn Huy Thắng á hả

Ta có:

\(3\left(x^4+\frac{1}{x^4}+1\right)\ge\left(x^2+\frac{1}{x^2}+1\right)^2\)

\(\Leftrightarrow3\left(x^4+\frac{1}{x^4}+1\right)^3\ge\left(x^2+\frac{1}{x^2}+1\right)^2\left(x^4+\frac{1}{x^4}+1\right)^2\)

\(\ge\left(x^3+\frac{1}{x^3}+1\right)^4\)

Dấu = xảy ra khi \(x=1\)