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chán wá
Câu 4:
a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)
b: \(A=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x^2-25\right)}{2x\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
c: Để A=-3 thì x-1=-6
hay x=-5(loại)
Thì bạn học thuộc rồi cứ thay số vào rồi tính thôi. Tui thấy cũng khá dễ mà.
\(M=x^2+xy+y^2-3x-3y\)
\(\Rightarrow4M=4x^2+4xy+4y^2-12x-12y\)
\(=\left(x^2+4y^2+9+4xy-12y-6x\right)+\left(3x^2-6x+3\right)-12\)
\(=\left(x+2y-3\right)^2+3\left(x-1\right)^2-12\ge-12\)
\(\Rightarrow M\ge-3\)
\(\Rightarrow Min_M=-3\Leftrightarrow x=y=1\)
Ta có: A = \(x^2+xy+y^2-3x-3y\)
= \(\left(x^2+xy-3x\right)+y^2-3y\)
= \(\left[x^2+x\left(y-3\right)\right]+y^2-3y\)
= \(\left[x^2+2x\dfrac{\left(y-3\right)}{2}+\dfrac{\left(y-3\right)^2}{4}\right]-\dfrac{\left(y-3\right)^2}{4}+y^2-3y\)
= \(\left(x+\dfrac{y-3}{2}\right)^2-\dfrac{y^2-6y+9}{4}+\dfrac{4y^2}{4}-\dfrac{12y}{4}\)
= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{-y^2+6y-9+4y^2-12y}{4}\)
= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3y^2-6y-9}{4}\)
= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3\left(y^2-2y+1-4\right)}{4}\)
= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3\left(y-1\right)^2-12}{4}\)
= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3\left(y-1\right)^2}{4}-3\ge-3\)
Vậy Min A = -3 \(\Leftrightarrow\left\{{}\begin{matrix}y-1=0\\x+\dfrac{y-3}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
\(f\left(x\right)=x^3+5x^2+ax+b\)
\(f\left(-2\right)=0\Leftrightarrow12-2a+b=0\left(1\right)\)
\(f\left(3\right)=0\Leftrightarrow72+3a+b=0\left(2\right)\)
\(\left(2\right)-\left(1\right)=0\Leftrightarrow\left(72+3a+b\right)-\left(12-2a+b\right)=0\Leftrightarrow60+5a=0\Leftrightarrow5a=-60\Leftrightarrow a=-12\)
=4