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a/
\(9x^2+25y^2+1+30xy-6x-10y+4y^2-20y+25=0\)
\(\Leftrightarrow\left(3x+5y-1\right)^2+\left(2y-5\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y-1=0\\2y-5=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\frac{23}{6}\\y=\frac{5}{2}\end{matrix}\right.\)
b/
\(4x^2+4y^2+8xy+x^2-2x+1+y^2+2y+1=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
c/
\(y^2-2y+1+2=\frac{6}{x^2+2x+1+3}\)
\(\Leftrightarrow\left(y-1\right)^2+2=\frac{6}{\left(x+1\right)^2+3}\)
Ta có \(VT=\left(y-1\right)^2+2\ge2\)
\(\left(x+1\right)^2+3\ge3\Rightarrow VP=\frac{6}{\left(x+1\right)^2+3}\le\frac{6}{3}=2\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}y-1=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
d/
\(\frac{-9x^2+18x-9-8}{x^2-2x+1+2}=y^2+4y+4-4\)
\(\Leftrightarrow\frac{-9\left(x-1\right)^2-8}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow\frac{-9\left(x-1\right)^2-18+10}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow-9+\frac{10}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow\frac{10}{\left(x-1\right)^2+2}=\left(y+2\right)^2+5\)
Ta có \(\left(x-1\right)^2+2\ge2\Rightarrow\frac{10}{\left(x-1\right)^2+2}\le\frac{10}{2}=5\Rightarrow VT\le5\)
\(\left(y+2\right)^2+5\ge5\Rightarrow VP\ge5\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
bài 1 : ta có : \(A=27x^3+27x^2y+9xy^2+y^3=\left(3x+y\right)^3\)
\(=\left(3.\left(-3\right)+5\right)^3=\left(-9+5\right)^3=\left(-4\right)^3=-64\)
bài 2 : a) ta có : \(12a^2-3ab+8ac-2bc=3a\left(4a-b\right)+2c\left(4a-b\right)\)
\(=\left(3a+2c\right)\left(4a-b\right)\) câu này mk sữa đề lại chút .
b) ta có : câu này đề sai rồi .
nếu phân tích ra nó sẽ thành : \(17x^2+34x-5=\left(17x+17-\sqrt{374}\right)\left(x+\dfrac{17+\sqrt{374}}{17}\right)\)
c) ta có : \(4x^4+81=\left(2x^2\right)^2+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-36x^2=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)
câu 3 : a) ta có : \(-3x^2+2x+1=0\Leftrightarrow-3x^2+3x-x+1=0\)
\(\Leftrightarrow-3x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(-3x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-1}{3};x=1\)
b) ta có : \(x\left(x-3\right)=2x-6=x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
vậy \(x=2;x=3\)
\(P=\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)+2xy}\)
\(=\frac{x^2+5x+y^2+5y+2xy-6}{x^2+6x+y^2+6y+2xy}\)
\(=\frac{\left(x+y\right)^2+5\left(x+y\right)-6}{\left(x+y\right)^2+6\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(x+y+5\right)-6}{\left(x+y\right)\left(x+y+6\right)}\)
\(=\frac{2005\times\left(2005+5\right)-6}{2005\times\left(2005+6\right)}\)
\(=\frac{2005\times2010-6}{2005\times2011}\)
\(=\frac{2004}{2005}\)
Bài 2:
a: \(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y-4\right)\left(x+y+3\right)\)
b: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)
c: \(4x^4-32x^2+1\)
\(=4x^4+4x^2+1-36x^2\)
\(=\left(2x^2+1\right)^2-36x^2\)
\(=\left(2x^2-6x+1\right)\left(2x^2+6x+1\right)\)
d: \(=\left(x^2+3\right)\left(x^4-3x^2+9\right)\)
Đề có đúng không vậy bạn. Có phải là \(\dfrac{-9x^2+18x-17}{x^2-2x+3}=y\left(y+4\right)\)