a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

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

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

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

\(\Leftrightarrow\left(3x-1\right)^3=1\)

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

https://hoc24.vn/hoi-dap/question/655171.html

Lần sau ghi cho rõ đề

a) \(27x^3+27x^2+9x+1=64\)

\(\Rightarrow27x^3+27x^2+9x-63=0\)

\(\Rightarrow27x^3-27x^2+54x^2-54x+63x-63=0\)

\(\Rightarrow27x^2\left(x-1\right)+54x\left(x-1\right)+63\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(27x^2+54x+63\right)=0\)

\(\Rightarrow\left(x-1\right).9\left(3x^2+6x+7\right)=0\)

\(\Rightarrow\left(x-1\right)\left(3x^2+6x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x^2+6x+7=0\end{matrix}\right.\)

Mà ta có:

\(3x^2+6x+7\)

\(=3\left(x^2+2x+\dfrac{7}{3}\right)\)

\(=3\left(x^2+2x+1-1+\dfrac{7}{3}\right)\)

\(=3\left(x+1\right)^2+4\)

Vì \(3\left(x+1\right)^2\ge0\) với mọi x

\(\Rightarrow3\left(x+1\right)^2+4\ge4\)

\(\Rightarrow3x^2+6x+7\) vô nghiệm

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

b) \(\left(x-2\right)^3-x^2\left(x-6\right)=4\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+6x^2=4\)

\(\Rightarrow12x-8=4\)

\(\Rightarrow12x=12\)

\(\Rightarrow x=1\)

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

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

\(\Rightarrow x^3-3x^2+3x-1-x^3-9+3x^2-12=2\)

\(\Rightarrow3x-22=2\)

\(\Rightarrow3x=24\)

\(\Rightarrow x=8\)

\(\frac{2}{5}x\left(y-1\right)-\frac{2}{5}y\left(y-1\right)\)

\(=\left(y-1\right)\left[\left(\frac{2}{5}x-\frac{2}{5}y\right)\right]\)

\(=\left(y-1\right)\frac{2}{5}\left(x-y\right)\)

\(\frac{1}{25}x^2-64y^2\)

\(=\left(\frac{1}{5}x\right)^2-8^2\)

\(=\left(\frac{1}{5}x+8\right)\left(\frac{1}{5}x-8\right)\)

\(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3\)

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

\(8x^3+12x^2y+6xy+y^3\)

\(=2^3+3.4x^2y+3.2x.y^2+y^3\)

\(=\left(2+y\right)^3\)

a: \(=\dfrac{\left(x^4-y^4\right)^2}{x^2+y^2}=\left(x^2-y^2\right)^2\cdot\left(x^2+y^2\right)\)

b: \(=\dfrac{\left(4x+3\right)\left(16x^2-12x+9\right)}{16x^2-12x+9}=4x+3\)

1.

\(\frac{x^2+2x+5}{x+4}-\left(x-3\right)\ge0\)

\(\Leftrightarrow\frac{x^2+2x+5-\left(x-3\right)\left(x+4\right)}{x+4}\ge0\)

\(\Leftrightarrow\frac{x+17}{x+4}\ge0\Rightarrow\left[{}\begin{matrix}x>-4\\x\le-12\end{matrix}\right.\)

2.

\(\frac{x^2-3x-1}{2-x}+x>0\)

\(\Leftrightarrow\frac{x^2-3x-1+x\left(2-x\right)}{2-x}>0\)

\(\Leftrightarrow\frac{-x-1}{2-x}>0\Rightarrow\left[{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)

3.

\(\frac{3x-47}{3x-1}-\frac{4x-47}{2x-1}>0\)

\(\Leftrightarrow\frac{\left(3x-47\right)\left(2x-1\right)-\left(4x-47\right)\left(3x-1\right)}{\left(3x-1\right)\left(2x-1\right)}>0\)

\(\Leftrightarrow\frac{-6x\left(x-8\right)}{\left(3x-1\right)\left(2x-1\right)}>0\Rightarrow\left[{}\begin{matrix}0< x< \frac{1}{3}\\\frac{1}{2}< x< 8\end{matrix}\right.\)

4.

\(\frac{x\left(x+2\right)+9}{x+2}-4\ge0\)

\(\Leftrightarrow\frac{x^2+2x+9-4\left(x+2\right)}{x+2}\ge0\)

\(\Leftrightarrow\frac{x^2-2x+1}{x+2}\ge0\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{x+2}\ge0\Rightarrow x>-2\)

5.

\(\frac{\left(x-1\right)^3\left(x+2\right)^4\left(x+6\right)}{\left(x-7\right)^3\left(x-2\right)^2}\le0\Rightarrow\left[{}\begin{matrix}x\le-6\\1\le x< 2\\2< x< 7\\x=-2\end{matrix}\right.\)

6. Xem lại đề