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4:
\(P=\left(x+4\right)\left(x^2-4x+16\right)-\left(64-x^3\right)\)
\(=x^3+64-64+x^3=2x^3\)
Khi x=-20 thì \(P=2\cdot\left(-20\right)^3=-16000\)
=>Chọn C
2: Đề khó hiểu quá bạn ơi
\(1,\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\left(B\right)\\ 2,Giống.1\\ 3,=5x^2\left(B\right)\\ 4,x^3=x\Leftrightarrow x^3-x=0\\ \Leftrightarrow x\left(x^2-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\\ \Leftrightarrow A\)
\(a,\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x^2+4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2+2x+x+2-\left(x^2-2x-x+2\right)=2x^2+4\)
\(\Leftrightarrow x^2+3x+2-x^2+2x+x-2=2x^2+4\)
\(\Leftrightarrow6x=2x^2+4\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)
\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)
\(\Leftrightarrow3x^2-13x+4=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{3}\end{cases}}}\)
= 0.
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