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a: Sửa đề: \(2A+\left(2x^2+y^2\right)=6x^2+5y^2-2x^2y^2\)
=>\(2A=6x^2+5y^2-2x^2y^2-2x^2-y^2\)
=>\(2A=4x^2+4y^2-2x^2y^2\)
=>\(A=2x^2+2y^2-x^2y^2\)
b: \(2A-\left(xy+3x^2-2y^2\right)=x^2-8y+xy\)
=>\(2A=x^2-8y+xy+xy+3x^2-2y^2\)
=>\(2A=4x^2+2xy-8y-2y^2\)
=>\(A=2x^2+xy-4y-y^2\)
c: Sửa đề: \(A+\left(3x^2y-2xy^2\right)=2x^2y+4xy^3\)
=>\(A=2x^2y+4xy^3-3x^2y+2xy^2\)
=>\(A=-x^2y+4xy^3+2xy^2\)
( - 12 x 4 y + 4 x 3 – 8 x 2 y 2 ) : ( - 4 x ) 2 = ( - 12 x 4 y ) : ( - 4 x 2 ) + ( 4 x 3 ) : ( - 4 x 2 ) – ( 8 x 2 y 2 ) : ( - 4 x 2 ) = 3 x 2 y – x + 2 y 2
Đáp án cần chọn là: D
\(a,=5\left(x-2y\right)\\ b,=3xy\left(x-2y\right)\\ c,=\left(x-y\right)\left(x+3\right)\\ d,=\left(x-1\right)\left(2x-4x^2\right)=2x\left(1-2x\right)\left(x-1\right)\\ e,=\left(x-2y\right)^2\\ f,=\left(3x-4y\right)\left(3x+4y\right)\\ g,=\left(x-3\right)\left(x^2+3x+9\right)\)
a. 5x - 10y
= 5(x - 2y)
b. 3x2y - 6xy2
= 3xy(x - 2y)
c. x(x - y) - 3(y - x)
= x(x - y) + 3(x - y)
= (x + 3)(x - y)
d. 2x(x - 1) + 4x2(1 - x)
= 2x(x - 1) - 4x2(x - 1)
= (2x - 4x2)(x - 1)
= 2x(1 - 2x)(x - 1)
e. x2 - 4xy + 4y2
= (x - 2y)2
f. 9x2 - 16y2
= (3x - 4y)(3x + 4y)
g. x3 - 27
= (x - 3)(x2 + 3x + 9)
a)
\(=\left(x+2y\right)\left(x^2-xy+y^2\right)-3xy\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x^2-xy+y^2-3xy\right)\)
\(=\left(x+2y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+2\right)\left(x-2\right)^2\)
b)
\(3x\left(2x-1\right)\left(2x+1\right)=0\)
3x=0 =>x=0
hoặc 2x-1=0 => 2x=1=>x=1/2
hoặc 2x+1=0=>2x=-1=>x=-1/2
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
A
đáp án A đúng nha