Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = ( x + y )3 + 2x2 + 4xy + 2y2
A = 53 + 2 ( x2 + 2xy + y2 )
A = 125 + 2 ( x + y )2
A = 125 + 2 . 52
A = 125 + 2 . 25
A = 125 + 50 = 175
\(A=\left(x+y\right)^3+2x^2+4xy+2y^2\)
\(=\left(x+y\right)^3+2\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3+2\left(x+y\right)^2\)
\(=7^3+2\cdot7^2=441\)
a) \(A=x^2+2xy+y^2-4x-4y+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
b) \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37=100\)
c) \(C=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10=25\)
a) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
\(\left(3x+4y\right)^2-4\left(x-2y\right)^2=\left(3x+4y\right)^2-\left(2x-4y\right)^2=5x\left(x+8y\right)\)
\(25\left(x-y\right)\left(x+y\right)-\left(5x-2\right)^2=25x^2-25y^2-25x^2+20x-4=-25y^2+20x-4\)
\(4x^2+12x+2018=4x^2+12x+9+2009=\left(2x+3\right)^2+2009\ge0+2009=2009\Rightarrow GTNNla:2009\Leftrightarrow2x+3=0\Leftrightarrow x=\frac{-3}{2}\)
\(5x^2-4xy+y^2-6x+13=\left(4x^2-4xy+y^2\right)+\left(x^2-6x+9\right)+4=\left(2x-y\right)^2+\left(x-3\right)^2+4\ge4\Rightarrow GTNNla:4\Leftrightarrow\left\{{}\begin{matrix}y=2x\\x=3\end{matrix}\right.\Leftrightarrow x=3;y=6\)
bài 1
a) \(7x\left(5x-1\right)+5x-1=\left(5x-1\right)\left(7x+1\right)\)
b) \(4xy-4x^2-y^2+25=25-\left(4x^2-4xy+y^2\right)\)
\(=5^2-\left(2x-y\right)=\left(5-2x+y\right)\left(5+2x-y\right)\)
c) \(2x^2-2y+xy-4x=\left(2x^2+xy\right)-\left(2y+4x\right)\)
\(=x^2\left(2x+y\right)-2\left(2x+y\right)=\left(2x+y\right)\left(x^2-2\right)\)
d) \(3x^2-7x+2=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
bài 2
a) * Rút gọn:
\(Q=3\left(2x-1\right)^2+2\left(2x+3\right)\left(x-1\right)-\left(x-3\right)\left(x+3\right)\)
\(Q=\left[3\left(4x^2-4x+1\right)\right]+\left[2\left(2x^2-2x+3x-3\right)\right]-\left(x^2-9\right)\)
\(Q=\left(12x^2-12x+3\right)+\left(4x^2-4x+6x-6\right)-\left(x^2-9\right)\)
\(Q=12x^2-12x+3+4x^2-4x+6x-6-x^2+9\)
\(Q=15x^2-10x+6=5x\left(3x-2\right)+6\)
Thế x = 2 vào biểu thức Q ta được:
\(Q=5\cdot2\left(3\cdot2-2\right)+6=46\)
b) \(Q=5x\left(3x-2\right)+6=6\)
\(\Leftrightarrow5x\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
1)
\(a,\) \(A=4x^2+4x+11\)
\(=\left(4x^2+4x+1\right)+10=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy : min \(A=10\Leftrightarrow x=-\frac{1}{2}\)
b) \(C=x^2-2x+y^2-4y+7\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow x=1,y=2\)
Vậy : \(minC=2\Leftrightarrow x=1,y=2\)
2,
a) \(A=5-8x-x^2\)
\(=-\left(x^2+8x+16\right)+21=-\left(x+4\right)^2+21\le21\)
Dấu "=" xảy ra \(\Leftrightarrow x=-4\)
b) \(B=5-x^2+2x-4y^2-4y\)
\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\)
Dấu "=" xảy ra \(\Leftrightarrow x=1,y=-\frac{1}{2}\)
