4x2-4xy-3y2-2x+3y
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5 tháng 10 2021
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
5 tháng 10 2021
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
8 tháng 1 2024
a: \(A=x^3y-12xy-x^2y\)
\(=xy\cdot x^2-xy\cdot12-xy\cdot x\)
\(=xy\left(x^2-x-12\right)\)
\(=xy\left(x^2-4x+3x-12\right)\)
\(=xy\left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=xy\left(x-4\right)\left(x+3\right)\)
c: \(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)
=(x+1)(x+4)(x+2)(x+3)-120
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)-96\)
\(=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
\(=\left(x^2+5x+16\right)\left(x+6\right)\left(x-1\right)\)
d: \(D=x^5-x^4+x^2-1\)
\(=\left(x^5-x^4\right)+\left(x^2-1\right)\)
\(=x^4\left(x-1\right)+\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^4+x+1\right)\)
PD
27 tháng 8 2021
l/ $6x^2(x-1)-9x(x-1)\\=(6x^2-9)(x-1)\\=3(2x^2-3)(x-1)\\=3(\sqrt2 x-\sqrt 3)(\sqrt 2 x+\sqrt 3)(x-1)$
m/ $4x^2(x-2)+9x(2-x)\\=4x^2(x-2)-9x(x-2)\\=(4x^2-9x)(x-2)\\=x(4x-9)(x-2)$
n/ $4x^2y-4xy+y\\=y(4x^2-4x+1)\\=y(2x-1)^2$
o/ $3x(2x-3y)-6(3y-2x)\\=3x(2x-3y)+6(2x-3y)\\=(3x+6)(2x-3y)\\=3(x+2)(2x-3y)$
p/ $4x^2(x-1)+(1-x)\\=4x^2(x-1)-(x-1)\\=(4x^2-1)(x-1)\\=(2x-1)(2x+1)(x-1)$
HV
27 tháng 8 2021
l)\(6x^2\left(x-1\right)-9x\left(x-1\right)=3x\left(x-1\right)\left(2x-3\right)\)
m) \(4x^2\left(x-2\right)+9x\left(2-x\right)=4x^2\left(x-2\right)-9x\left(x-2\right)=x\left(x-2\right)\left(4x-9\right)\)
n) \(4x^2y-4xy+y=y\left(4x^2-4x+1\right)=y\left(2x-1\right)^2\)
o) \(3x\left(2x-3y\right)-6\left(3y-2x\right)=3x\left(2x-3y\right)+6\left(2x-3y\right)=3\left(2x-3y\right)\left(x+2\right)\)
p) \(4x^2\left(x-1\right)+\left(1-x\right)=4x^2\left(x-1\right)-\left(x-1\right)=\left(4x^2-1\right)\left(x-1\right)=\left(2x-1\right)\left(2x+1\right)\left(x-1\right)\)
ND
1
27 tháng 7 2021
b) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\dfrac{\left(2x+y\right)^2}{2x+y}=2x+y\)
c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\dfrac{\left(3y-x\right)^2}{3y-x}=3y-x\)
13 tháng 1 2023
3x^2+3y^2+4xy-2x+2y+2=0
=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0
=>x=1 và y=-1
M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1
30 tháng 10 2021
a) \(5x+10y=5\left(x+2y\right)\)
b) \(3x^3-12x=3x\left(x^2-4\right)=3x\left(x-2\right)\left(x+2\right)\)
c) \(4x^2+9x-4xy-9y=4x\left(x-y\right)+9\left(x-y\right)=\left(x-y\right)\left(4x+9\right)\)
d) \(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
KB
30 tháng 9 2018
a ) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)\)
\(=\left(2x+y\right)^2:\left(2x+y\right)\)
\(=2x+y\)
b ) \(\left(27x^3+1\right):\left(3x+1\right)\)
\(=\left(3x+1\right)\left(9x^2-3x+1\right):\left(3x+1\right)\)
\(=9x^2-3x+1\)
c ) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)\)
\(=\left(x-3y\right)^2:\left(3y-x\right)\)
\(=\left(3y-x\right)^2:\left(3y-x\right)\)
\(=3y-x\)
d ) \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)
\(=\left(2x-1\right)\left(4x^2+2x+1\right):\left(4x^2+2x+1\right)\)
\(=2x-1\)
:D
2 tháng 10 2023
Bài \(3\)
\(A=\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)
\(=2x^2+3x-10x-15-\left(2x^2-6x\right)+x+7\)
\(=2x^2+3x-10x-15-2x^2+6x+x+7\)
\(=\left(2x^2-2x^2\right)+\left(3x-10x+6x+x\right)+\left(-15+7\right)\)
\(=-8\)
Vậy biểu thức không phụ thuộc vào biến
\(B=4\left(y-6\right)-y^2\left(2+3y\right)+y\left(5y-4\right)+3y^2\)
Đề như này à?
Bài \(4\)
\(a,4a^2-16b^2=4\left(a^2-4b^2\right)=4\left(a-2b\right)\left(a+2b\right)\)
\(b,4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x+1\right)^2\)
\(c,\) ?
\(d,\left(x-y\right)^2-\left(2x-y\right)^2\\ =\left[\left(x-y\right)-\left(2x-y\right)\right]\left[\left(x-y\right)+\left(2x-y\right)\right]\\ =\left(x-y-2x+y\right)\left(x-y+2x-y\right)\\ =\left(-x\right)\left(3x-2y\right)\)
\(e,8x^3-y^3=\left(2x\right)^3-y^3\\ =\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(i,3x+6y+\left(x+2y\right)\\ =3\left(x+2y\right)+\left(x+2y\right)\\ =4\left(x+2y\right)\)
\(j,ax-ay-x+y=\left(ãx-ay\right)-\left(x-y\right)\\ =a\left(x-y\right)-\left(x-y\right)=\left(x-y\right)\left(a-1\right)\)
`k,` `y` hay `y^2` ạ? vì nó mới phân tích được nhân tử.
đề là phân tích thành nhân tử thì làm thế này nha:
4x2-4xy-3y2-2x+3y
=4x2-4xy+y2-4y2-2x+3y
=(2x-y)2-(2y)2-(2x-3y)
=(2x-y-2y)(2x-y+2y)-(2x-3y)
=(2x-3y)(2x+y)-(2x-3y)
=(2x-3y)(2x+y-1)
Đúng thì tick nha