Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,\(8x^2-8xy+2x=2x\left(4x-8y+1\right)\)
b,\(\left(x^2+2x\right)\left(x^2+4x+3\right)-24=x\left(x+2\right)\left(x+1\right)\left(x+3\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)-24=\left(t+1\right)\left(t-1\right)-24=t^2-5^2=\left(t+5\right)\left(t-5\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+3x-4\right)\)( đặt t = x2 + 3x + 1 )
a) \(A=x-x^2=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Vậy Max A = \(\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)
b) \(B=2x-2x^2=2\left(x-x^2\right)=-2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\le\frac{1}{2}\)
Vậy Max B = \(\frac{1}{2}\Leftrightarrow x=\frac{1}{2}\)
a, \(\frac{x+1}{2x+6}=\frac{x+1}{2\left(x+3\right)}\)
b, \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)
c, \(\frac{x-x-2xy+x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x-2xy}{x+2y}+\frac{4xy}{\left(2y-x\right)\left(x+2y\right)}\)
\(=\frac{\left(x-2xy\right)\left(2y-x\right)}{\left(x+2y\right)\left(2y-x\right)}+\frac{4xy}{\left(2y-x\right)\left(x+2y\right)}=\frac{2xy-x^2+4xy^2+2x^2y}{\left(2y-x\right)\left(x+2y\right)}\)
A=(x4−2x3−3x2)−(2x3−4x2−6x)−(3x2−6x−9)
=x2(x2−2x−3)−2x(x2−2x−3)−3(x2−2x−3)
=(x2−2x−3)(x2−2x−3)
=(x2−2x−3)2
⇒ A là SCP với mọi x nguyên
chúc học tốt!
a ) \(A=x-x^2=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Vậy MAX \(A=\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)
b) \(B=2x-2x^2=2\left(x-x^2\right)=-2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\le\frac{1}{2}\)
Vậy MAX \(B=\frac{1}{2}\Leftrightarrow x=\frac{1}{2}\)
a)(2x2+1)(3x3-2x2+3
= 6x5-4x4+6x2+3x3-2x2+3
= 6x5-4x4+3x3+4x2+3
b)(-3x+1)(4x4-x³+x)
= -12x5+3x4-3x2+4x4-x³+x
= -12x5+7x4-x3-3x2+x
\(8x^5+4x^3-8x^5+16x^3=20\)
\(20x^3=20\)
\(x^3=1\)
\(=>x=1\)