Tìm x ,y
4x^2-12xy=-9y^2
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XO
12 tháng 8 2023
13x2 + 9y2 - 30x + 12xy + 25 = 0
<=> (9y2 + 12xy + 4y2) + (9x2 - 30x + 25) = 0
<=> (3y + 2x)2 + (3x - 5)2 = 0
Dễ thấy \(\left(3y+2x\right)^2\ge0;\left(3x-5\right)^2\ge0\forall x,y\)
nên \(\left(3y+2x\right)^2+\left(3x-5\right)^2\ge0\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}3y+2x=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{10}{9}\\x=\dfrac{5}{3}\end{matrix}\right.\)
25 tháng 7 2021
a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)
25 tháng 7 2021
c) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-25\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
29 tháng 10 2021
\(1,=8xy+14y^2-4xz-7yz\\ 2,=y\left(4x^2-12x+9\right)=y\left(2x-3\right)^2\\ 3,\Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
29 tháng 10 2021
Câu 1: \(\left(2y-z\right)\left(4x+7y\right)=8xy-4xz+14y^2-7yz\)
câu 2: \(4x^2y-12xy+9y=y\left(4x^2-12x+9\right)\)
câu 3: \(\left(x-2\right)\left(x+3\right)+x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\\ \Leftrightarrow2\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
19 tháng 7 2016
a)\(\left(3x\right)^2-2×3x+1^2=\left(3x-1\right)^2\)
b)\(\left(2x\right)^2+2×2x+1^2=\left(2x+1\right)^2\)
c)\(\left(2x\right)^2+2×2x×3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
d)\(-\left(4x^2-12xy+9y^2\right)=-\left[\left(2x\right)^2-2×2x×3y+\left(3x\right)^2\right]=-\left[\left(2x-3y\right)^2\right]\)
\(4x^2-12xy=-9y^2\)
\(\Leftrightarrow4x^2-12xy+9y^2=0\)
\(\Leftrightarrow\left(2x\right)^2-2xy+\left(3y\right)^2=0\)
\(\Leftrightarrow\left(2x-3y\right)^2=0\)
\(\Leftrightarrow2x-3y=0\)
\(\Leftrightarrow2x=3y\)
\(\Leftrightarrow\frac{x}{y}=\frac{3}{2}\)