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1: \(=\left(x-1\right)^2\)
2: \(x\in\left\{0;20\right\}\)
Câu 13:
\(1,=\left(x-1\right)^2\\ 2,\Leftrightarrow x\left(x-20\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=20\end{matrix}\right.\\ 3,\text{Đề lỗi}\)
Câu 14:
\(1,ĐK:x\ne-2\\ 2,=\dfrac{\left(x+2\right)^2}{x+2}=x+2\\ 3,\Leftrightarrow x+2=0\Leftrightarrow x=-2\left(ktm\right)\Leftrightarrow x\in\varnothing\)
Câu 16:
\(A=x^2-4x+4+20=\left(x-2\right)^2+20\ge20\)
Dấu \("="\Leftrightarrow x=2\)
a: \(=\dfrac{3b+4a}{6ab}\)
b: \(=\dfrac{x^2-2x+1-x^2-2x-1}{x^2-1}=\dfrac{-4x}{x^2-1}\)
c: \(=\dfrac{xz+yz-xy-xz}{xyz}=\dfrac{yz-xy}{xyz}=\dfrac{z-x}{xz}\)
d: \(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
e: \(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)
\(\left[\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2\right]:\left(x^2-6x+9\right)=\left[\left(3-x\right)^5-7\left(3-x\right)^4-4\left(3-x\right)^2\right]:\left(3-x\right)^2=\left(3-x\right)^2\left[\left(3-x\right)^3-7\left(3-x\right)^2-4\right]:\left(3-x\right)^2=\left(3-x\right)^3-7\left(3-x\right)^2-4=27-27x+9x^2-x^3-63+42x-7x^2-4=-x^3+2x^2+15x-40\)
\(\dfrac{\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{x^2-6x+9}\)
\(=\dfrac{-\left(x-3\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{\left(x-3\right)^2}\)
\(=-\left(x-3\right)^3-7\left(x-3\right)^2-4\)
a: \(=\dfrac{x\left(x^2+x-2\right)}{x+2}=\dfrac{x\left(x+2\right)\left(x-1\right)}{x+2}=x^2-x\)
b: \(=\dfrac{x^3-3x^2+2x+24}{x+2}=\dfrac{x^3+2x^2-5x^2-10x+12x+24}{x+2}=x^2-5x+12\)
