a) Ta có: \(\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)=\left(4-x\right)^2\)

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

\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)-\left(x-4\right)\right]-2\left(3x-2\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-3-x+4\right)-6x+4=0\)

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

\(\Leftrightarrow-5x=0\)

mà -5<0

nên x=0

Vậy: x=0

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

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

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

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

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

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

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

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

(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:

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

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

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

\(\left(3x-5\right)\left(-2x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\-2x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=5\\-2x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-7}{2}\end{cases}}}\)

\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)

\(\Leftrightarrow9x^2-1=2x-3+6x^2-9x\)

\(\Leftrightarrow9x^2-1=-7x-3+6x^2\)

\(\Leftrightarrow9x^2-1+7x+3-6x^2=0\)

\(\Leftrightarrow3x^2+2+7x=0\)

\(\Leftrightarrow3x^2+6x+x+2=0\)

\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}\)

Xét \(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)

\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3+3x+3-6x+3x^2}\)

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

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

Thay vào ta tính được:

\(A=\left[f\left(\frac{1}{2020}\right)+f\left(\frac{2019}{2020}\right)\right]+...+\left[f\left(\frac{1009}{2020}\right)+f\left(\frac{1011}{2020}\right)\right]+f\left(\frac{1010}{2020}\right)\)

\(A=1+...+1+f\left(\frac{1010}{2020}\right)\) (với 1009 số 1)

\(A=1009+f\left(\frac{1}{2}\right)=1009+\frac{\left(\frac{1}{2}\right)^3}{1-3\cdot\frac{1}{2}+3\cdot\left(\frac{1}{2}\right)^2}\)

\(A=1009+\frac{1}{2}=\frac{2019}{2}\)

Vậy \(A=\frac{2019}{2}\)

Tks bạn nhé

hello ae xin chào

a)(x+3)³-x(3x+1)²+(2x+1)(4x²-2x+1-3x²)=54

\(\Rightarrow\)x³+9x²+27x+27-x(9x²+6x+1)+(2x+1)(x²-2x+1)=54

\(\Rightarrow\)x³+9x²+27x+27-9x³-6x²-x+2x³-4x²+2x+x²-2x+1=54

\(\Rightarrow\)-6x³+26x+28=54

\(\Rightarrow\)-6x³+26x=54-28

\(\Rightarrow\)-6x³+26x=26

\(\Rightarrow\)-6x³+26x-26=0

\(\Rightarrow\)-2(3x³+13x+14)

\(\left(x^2-3x+2\right)\sqrt{\dfrac{x+3}{x-1}}=-\dfrac{1}{2}x^3+\dfrac{15}{2}x-11\left(1\right)\)

Đk: \(\sqrt{\dfrac{x+3}{x-1}}\ge0\Leftrightarrow\left[{}\begin{matrix}x>1\\x\le-3\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow-2\left(x-1\right)\left(x-2\right)\sqrt{\dfrac{x+3}{x-1}}=x^3-15x+22\)

\(\Rightarrow-2\sqrt{\left(x-1\right)\left(x+3\right)}.\left(x-2\right)=\left(x-2\right)\left(x^2+2x-11\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\-2\sqrt{\left(x-1\right)\left(x+3\right)}=x^2+2x-11\left(2\right)\end{matrix}\right.\)

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

Đặt \(a=\sqrt{x^2+2x-3}\left(a\ge0\right)\). Từ phương trình (2) suy ra:

\(a^2+2a-8=0\Leftrightarrow\left[{}\begin{matrix}a=2\left(nhận\right)\\a=-4\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+2x-3}=2\Leftrightarrow x^2+2x-7=0\)

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

Thử lại ta có \(x=2\) và \(x=-1+2\sqrt{2}\) là 2 nghiệm của phương trình (1).

\(\Leftrightarrow2\left(x^2-3x+2\right)\cdot\sqrt{\dfrac{x+3}{x-1}}=-x^3+15x-22\)

\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)\cdot\dfrac{\sqrt{\left(x+3\right)\left(x-1\right)}}{x-1}=-x^3+2x^2-2x^2+4x+11x-22\)

\(\Leftrightarrow2\left(x-2\right)\sqrt{\left(x+3\right)\left(x-1\right)}=\left(x-2\right)\left(-x^2-2x+11\right)\)

\(\Leftrightarrow\left(x-2\right)\left(\sqrt{4\left(x^2+2x-3\right)}+x^2+2x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\left(1\right)\\2\sqrt{x^2+2x-3}+x^2+2x-11=0\left(2\right)\end{matrix}\right.\)

(1) =>x=2

(2): Đặt \(\sqrt{x^2+2x-3}=a\left(a>=0\right)\)

=>2a+a^2-8=0

=>(a+4)(a-2)=0

=>a=2

=>x^2+2x-3=4

=>x^2+2x-7=0

=>\(x=-1\pm2\sqrt{2}\)