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19 tháng 8 2021
\(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)+\left(2x-y\right)^2=\left(2x-y\right)\left(2+2x-y\right)\)
\(B=9x^2-\left(y^2-4y+4\right)=9x^2-\left(y-2\right)^2=\left(3x-y+2\right)\left(3x+y-2\right)\)
\(C=-25x^2+y^2-6y+9=\left(y^2-6y+9\right)-25x^2=\left(y-3\right)^2-\left(5x\right)^2=\left(y-3-5x\right)\left(y-3+5x\right)\)\(D=x^2-4x-y^2-8y-12=\left(x^2-4x+4\right)-\left(y^2+8y+16\right)=\left(x-2\right)^2-\left(y+4\right)^2=\left(x-2-y-4\right)\left(x-2+y+4\right)=\left(x-y-6\right)\left(x+y+2\right)\)
26 tháng 8 2021
`a)x^2-2x+2+4y^2+4y`
`=x^2-2x+1+4y^2+4y+1`
`=(x-1)^2+(2y+1)^2`
`b)4x^2+y^2+12x+4y+13`
`=4x^2+12x+9+y^2+4y+4`
`=(2x+3)^2+(y+2)^2`
`c)x^2+17+4y^2+8x+4y`
`=x^2+8x+16+4y^2+4y+1`
`=(x+4)^2+(2y+1)^2`
`d)4x^2-12xy+y^2-4y+13`
`=4x^2-12x+9+y^2-4y+4`
`=(2x-3)^2+(y-2)^2`
26 tháng 8 2021
a) \(x^2-2x+2+4y^2+4y=\left(x-1\right)^2+\left(2y+1\right)^2\)
b) \(4x^2+y^2+12x+4y+13=\left(2x+3\right)^2+\left(y+2\right)^2\)
c) \(x^2+17+4y^2+8x+4y=\left(x+4\right)^2+\left(2y+1\right)^2\)
d) \(4x^2-12x+y^2-4y+13=\left(2x-3\right)^2+\left(y-2\right)^2\)
8 tháng 12 2023
10: \(x\left(x-y\right)+x^2-y^2\)
\(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+x+y\right)\)
\(=\left(x-y\right)\left(2x+y\right)\)
11: \(x^2-y^2+10x-10y\)
\(=\left(x^2-y^2\right)+\left(10x-10y\right)\)
\(=\left(x-y\right)\left(x+y\right)+10\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+10\right)\)
12: \(x^2-y^2+20x+20y\)
\(=\left(x^2-y^2\right)+\left(20x+20y\right)\)
\(=\left(x-y\right)\left(x+y\right)+20\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+20\right)\)
13: \(4x^2-9y^2-4x-6y\)
\(=\left(4x^2-9y^2\right)-\left(4x+6y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)-2\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x-3y-2\right)\)
14: \(x^3-y^3+7x^2-7y^2\)
\(=\left(x^3-y^3\right)+\left(7x^2-7y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+7\cdot\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+7\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2+7x+7y\right)\)
15: \(x^3+4x-\left(y^3+4y\right)\)
\(=x^3-y^3+4x-4y\)
\(=\left(x^3-y^3\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2+4\right)\)
16: \(x^3+y^3+2x+2y\)
\(=\left(x^3+y^3\right)+\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+2\right)\)
17: \(x^3-y^3-2x^2y+2xy^2\)
\(=\left(x^3-y^3\right)-\left(2x^2y-2xy^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)-2xy\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2-2xy\right)\)
\(=\left(x-y\right)\left(x^2-xy+y^2\right)\)
18: \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x^2-4x+4\right)-y^2\right]\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
TH
1
H9
HT.Phong (9A5)
CTVHS
24 tháng 10 2023
a) \(A=4x^2-4x+23\)
\(A=4x^2-4x+1+22\)
\(A=\left(2x-1\right)^2+22\)
Mà: \(\left(2x-1\right)^2\ge0\forall x\)
\(\Rightarrow A=\left(2x-1\right)^2+22\ge22\forall x\)
Dấu "=" xảy ra:
\(2x-1=0\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy: \(A_{min}=22\Leftrightarrow x=\dfrac{1}{2}\)
b) \(B=25x^2+y^2+10x-4y+2\)
\(B=25x^2+10x+1+y^2-4y+4-3\)
\(B=\left(5x+1\right)^2+\left(y-2\right)^2-3\)
Mà: \(\left\{{}\begin{matrix}\left(5x+1\right)^2\ge0\forall x\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow B=\left(5x+1\right)^2+\left(y-2\right)^2-3\ge-3\forall x,y\)
Dấu "=" xảy ra:
\(\left\{{}\begin{matrix}5x+1=0\\y-2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5x=-1\\y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\y=2\end{matrix}\right.\)
Vậy: \(B_{min}=-3\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\y=2\end{matrix}\right.\)
29 tháng 6 2021
b) Ta có: \(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)
c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
29 tháng 6 2021
\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)
\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)
\(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1=\left(x+1\right)^2+\left(y-2\right)^2+1\)
\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2
\(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)
dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)
\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)
=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2-1-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2\left(2-x\right)-1\ge-1\)
dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
15 tháng 10 2023
\(A=\left(2x+y\right)^2+\left(2x-y\right)^2+\left(4x^2-y^2\right)+3y\\ =\left(4x^2+4xy+y^2\right)+\left(4x^2-4xy+y^2\right)+\left(4x^2-y^2\right)+3y\\ =4x^2+4x^2+4x^2+4xy-4xy+y^2+y^2-y^2+3y=12x^2+3y-y^2\\ B=\left(x-2\right)\left(x+2\right)-\left(x+2\right)^2\\ =\left(x+2\right)\left(x-2-x-2\right)=-4\left(x+2\right)=-4x-8\\ C=\left(3x-4y\right)^2+\left(3x-4y\right)^2\\ =\left(9x^2-24xy+16y^2\right)+\left(9x^2-24xy+16y^2\right)\\ =18x^2-48xy+32y^2\)
AH
Akai Haruma
Giáo viên
11 tháng 1 2021
Lời giải:
a)
$A=4x^2+4x+11=(4x^2+4x+1)+10=(2x+1)^2+10\geq 10$
Vậy $A_{\min}=10$. Giá trị này đạt tại $(2x+1)^2=0$
$\Leftrightarrow x=-\frac{1}{2}$
b)
$C=x^2-2x+y^2-4y+7=(x^2-2x+1)+(y^2-4y+4)+2$
$=(x-1)^2+(y-2)^2+2\geq 2$
Vậy $C_{\min}=2$. Giá trị này đạt tại $(x-1)^2=(y-2)^2=0$
$\Leftrightarrow x=1; y=2$
AH
Akai Haruma
Giáo viên
28 tháng 8 2021
Lời giải:
a. $x^2+y^2+4y+13-6x$
$=(x^2-6x+9)+(y^2+4y+4)$
$=(x-3)^2+(y+2)^2$
b.
$4x^2-4xy+1+2y^2-2y$
$=(4x^2-4xy+y^2)+(y^2-2y+1)$
$=(2x-y)^2+(y-1)^2$
c.
$x^2-2xy+2y^2+2y+1$
$=(x^2-2xy+y^2)+(y^2+2y+1)$
$=(x-y)^2+(y+1)^2$
NT
28 tháng 8 2021
a. \(x^2+y^2+4y+12-6x=\left(x^2-6x+9\right)+\left(y^2+4y+4\right)=\left(x-3\right)^2+\left(y+2\right)^2\)b. \(4x^2-4xy+1+2y^2-2y=\left(4x^2-4xy+y^2\right)+\left(y^2-2y+1\right)=\left(2x-y\right)^2+\left(y-1\right)^2\)c. \(x^2-2xy+2y^2+2y+1=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2022
\(A=\left(4x^2+4x+1\right)+10=\left(2x+1\right)^2+10\ge10\)
\(A_{min}=10\) khi \(2x+1=0\Rightarrow x=-\dfrac{1}{2}\)
\(B=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)
\(B_{min}=-36\) khi \(x^2+5x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(C=\left(x^2-2x+1\right)+\left(y^2-4x+4\right)+2=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)
\(C_{min}=2\) khi \(\left(x;y\right)=\left(1;2\right)\)
\(4x^2-y^2+4y-4\)
\(=\left(2x\right)^2-\left(y^2-2\cdot y\cdot x+2^2\right)\)
\(=\left(2x\right)^2-\left(y-2\right)^2\)
\(=\left(2x-y-2\right)\left(2x+y-2\right)\)
= 4x2 - ( y2 - 4y + 4 )
= ( 2x)2 - ( y2 - 2.2y + 22 )
= ( 2x)2 - ( y - 2 )2
= ( 2x - y + 2 ) ( 2x + y - 2)