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a. \(\left(20x^4y-25x^2y^2-3x^2y\right):5x^2y\)
\(=4x^2-5y-\frac{3}{5}\)
b. \(\left(15xy^2+17xy^3+18y^2\right):6y^2\)
\(=\frac{5}{2}x+\frac{17}{6}xy+3\)
c. \(\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(y-x\right)^2\)
\(=\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(x-y\right)^2\)
\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)
d. \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left(y-x\right)\)
\(=\left(y-x\right)^2:\left(y-x\right)\)
\(=y-x\)
\(a,3x^3y^3-15x^2y^2=3x^2y^2\left(xy-5\right)\)
\(b,5x^3y^2-25x^2y^3+40xy^4\)
\(=5xy^2\left(x^2-5xy+8y^2\right)\)
\(c,-4x^3y^2+6x^2y^2-8x^4y^3\)
\(=-2x^2y^2\left(2x-3+4x^2y\right)\)
\(d,a^3x^2y-\frac{5}{2}a^3x^4+\frac{2}{3}a^4x^2y\)
\(=a^3x^2\left(y-\frac{5}{2}x^2+\frac{2}{3}ay\right)\)
\(e,a\left(x+1\right)-b\left(x+1\right)=\left(x+1\right)\left(a-b\right)\)
\(f,2x\left(x-5y\right)+8y\left(5y-x\right)\)
\(=2x\left(x-5y\right)-8y\left(x-5y\right)=\left(x-5y\right)\left(2x-8y\right)\)
\(g,a\left(x^2+1\right)+b\left(-1-x^2\right)-c\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(a-b-c\right)\)
\(h,9\left(x-y\right)^2-27\left(y-x\right)^3\)
\(=9\left(x-y\right)^2+27\left(x-y\right)^3\)
\(=9\left(x-y\right)^2\left(1+3x-3y\right)\)
a,3x3y3−15x2y2=3x2y2(xy−5)a,3x3y3−15x2y2=3x2y2(xy−5)
b,5x3y2−25x2y3+40xy4b,5x3y2−25x2y3+40xy4
=5xy2(x2−5xy+8y2)=5xy2(x2−5xy+8y2)
c,−4x3y2+6x2y2−8x4y3c,−4x3y2+6x2y2−8x4y3
=−2x2y2(2x−3+4x2y)=−2x2y2(2x−3+4x2y)
d,a3x2y−52a3x4+23a4x2yd,a3x2y−52a3x4+23a4x2y
=a3x2(y−52x2+23ay)=a3x2(y−52x2+23ay)
e,a(x+1)−b(x+1)=(x+1)(a−b)e,a(x+1)−b(x+1)=(x+1)(a−b)
f,2x(x−5y)+8y(5y−x)f,2x(x−5y)+8y(5y−x)
=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)
g,a(x2+1)+b(−1−x2)−c(x2+1)g,a(x2+1)+b(−1−x2)−c(x2+1)
=(x2+1)(a−b−c)=(x2+1)(a−b−c)
h,9(x−y)2−27(y−x)3h,9(x−y)2−27(y−x)3
=9(x−y)2+27(x−y)3
\(A\)xác định \(\Leftrightarrow x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\ne0\)
\(\Leftrightarrow x^2y^2+1+x^2-x^2y-y+y^2\ne0\)
\(\Leftrightarrow\left(x^2y^2+y^2\right)+\left(x^2+1\right)-\left(x^2y+y\right)\ne0\)
\(\Leftrightarrow y^2\left(x^2+1\right)+\left(x^2+1\right)-y\left(x^2+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left(y^2-y+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\)
Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(y-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall y\end{cases}}\)\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\forall x;y\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\forall x;y\)
\(\Leftrightarrow A\ne0\forall x;y\)
\(A=5x^2+2y^2+2xy-26x-16y+54\) \(=2\left(y^2+y\left(x-8\right)+\dfrac{\left(x-8\right)^2}{2}\right)-\dfrac{\left(x-8\right)^2}{2}+5x^2-26x+54\)
\(=2\left(y+\dfrac{x-8}{2}\right)^2+\dfrac{9}{2}x^2-18x+22\)
\(=2\left(y+\dfrac{x-8}{2}\right)^2+\dfrac{9}{2}\left(x-2\right)^2+4\ge4\)
Dấu '' = '' xảy ra khi: \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+\dfrac{x-8}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy: Min A = 4 tại \(x=2;y=3.\)
\(A=5x^2+2y^2+2xy-26x-16y+54\)
\(2A=4y^2+10x^2+4xy-52x-32y+108\)
\(2A=4y^2+4xy-32y-52x+10x^2+108\)
\(2A=\left(2y\right)^2+4y\left(x-8\right)+x^2-16x+64+9x^2-36x+44\)
\(2A=\left(2y\right)^2+2.2y\left(x-8\right)+\left(x-8\right)^2+\)\(9\left(x^2-4x+4\right)+8\)
\(2A=\left(2y+x-8\right)^2+9\left(x-2\right)^2+8\ge8\)
\(=>A\ge4\)
Để A nhỏ nhất thì \(x-2=0=>x=2;2y+x-8=0< =>2y-6=0=>y=3\)
Vậy ..................
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