K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
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\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
LD
1
8 tháng 11 2021
a: \(=x\left(x^2+4x+4-z^2\right)\)
\(=x\left(x+2-z\right)\left(x+2+z\right)\)
T
23 tháng 8 2023
\(a,A=y^2-\dfrac{1}{2}y+\dfrac{1}{16}\)
\(=y^2-2.y.\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\)
\(=\left(y-\dfrac{1}{4}\right)^2\)
Với \(y=100,25\), ta được:
\(A=\left(100,25-\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{401}{4}-\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{400}{4}\right)^2=100^2=10000\)
\(------\)
\(b,B=4x^2-9y^2-6y-1\)
\(=\left(2x\right)^2-\left[\left(3y\right)^2+2.3y.1+1\right]\)
\(=\left(2x\right)^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
Với \(x=23;y=1\), ta được:
\(B=\left(2.23-3.1-1\right)\left(2.23+3.1+1\right)\)
\(=\left(46-4\right)\left(46+4\right)\)
\(=42.50=2100\)
MN
3 tháng 7 2021
\(4x^2-4x+1+9y^2-6y+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(3y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\3y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
3 tháng 7 2021
Ta có:4x2-4x+9y2-6y+2=0
<=>(4x2-4x+1)+(9y2-6y+1)=0
<=> (2x-1)2+(3y-1)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\3y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
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: \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x^2-4x+4\right)-y^2\right]\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
2
24 tháng 8 2021
a) A = x2 - 4y2 + 2x + 4y = (x-2y)(x+2y)+2(x+2y)=(x+2y)(x-2y+2)
b) A = 4x2 - 9y2 - 4x - 6y=(2x-3y)(2x+3y)-2(2x+3y)=(2x+3y)(2x-3y-2)
c) A = 3x2 - 3xy - 5x + 5y=3x(x-y)-5(x-y)=(x-y)(3x-5)
24 tháng 8 2021
a) \(A=x^2-4y^2+2x+4y=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)=\left(x+2y\right)\left(x-2y+2\right)\)
b) \(A=4x^2-9y^2-4x-6y=\left(2x-3y\right)\left(2x+3y\right)-2\left(2x+3y\right)=\left(2x+3y\right)\left(2x-3y-2\right)\)
c) \(A=3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
18 tháng 7 2023
a)x2-6x+9
=x2-2.x.3+32
=(x-3)2
b)4x2+4x+1
=(2x)2+2.2x.1+12
=(2x+1)2
c)4x2+12xy+9y2
=(2x)2+2.2x.3y+(3y)2
=(2x+3y)2
d)4x4-4x2+4
=(2x2)2-2.2x2.2+22
=(2x2-2)2
H9
HT.Phong (9A5)
CTVHS
4 tháng 9 2023
a) \(2x^2+5x+2\)
\(=2x^2+4x+x+2\)
\(=2x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(2x+1\right)\)
b) \(4x^2-4x-9y^2+12y-3\)
\(=\left(4x^2-4x+1\right)-\left(9y^2-12y+4\right)\)
\(=\left(2x-1\right)^2-\left(3y-2\right)^2\)
\(=\left(2x-1+3y-2\right)\left(2x-1-3y+2\right)\)
\(=\left(2x+3y-3\right)\left(2x-3y+1\right)\)
c) \(x^4-2x^3-4x^2+4x-3\)
\(=x^4+x^3-x^2+x-3x^2-3x+3x-3\)
\(=\left(x^4+x^3-x^2+x\right)-\left(3x^2+3x-3x+3\right)\)
\(=x\left(x^3+x^2-x+1\right)-3\left(x^3+x^2-x+1\right)\)
\(=\left(x^3+x^2-x+1\right)\left(x-3\right)\)
d) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)=\left(2x-3y\right)\left(2x+3y+2\right)\)