Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(x+5\right)^2-62\ge-62\)
\(B=\left(\frac{1}{2}x^2+1-\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
\(C=\left(x-3y+2\right)^2+\left(x-5\right)^2-9\ge-9\)
\(D=\left(x-y+1\right)^2+\left(y-4\right)^2\ge0\)
\(A=-\left(x-3\right)^2+12\le12\)
\(B=-2x^2-5x+3=-2\left(x+\frac{5}{4}\right)^2+\frac{49}{8}\le\frac{49}{8}\)
\(C=\frac{1}{\left(x-2\right)^2+5}\le\frac{1}{5}\)
a, \(\dfrac{x^4-4x^2+3}{x^4+6x^2-7}\)=\(\dfrac{x^4-4x^2+4-1}{x^4+7x^2-x^2-7}\)=\(\dfrac{\left(x^2-2\right)^2-1}{x^2\left(x^2+7\right)-\left(x^2+7\right)}\)
=\(\dfrac{\left(x^2-1\right)\left(x^2-3\right)}{\left(x^2-1\right)\left(x^2+7\right)}\)=\(\dfrac{x^2-3}{x^2+7}\)
b, \(\dfrac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)=\(\dfrac{x^3\left(x+1\right)-\left(x+1\right)}{\left(x^2+1\right)^2+x\left(x^2+1\right)}\)=\(\dfrac{\left(x+1\right)\left(x^3-1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}\)=\(\dfrac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}\)=\(\dfrac{x^2-1}{x^2+1}\)
c, \(\dfrac{x^3+3x^2-4}{x^3-3x+2}\)=\(\dfrac{x^3+3x^2-3-1}{x^3-3x-1+3}\)=\(\dfrac{\left(x-1\right)\left(x^2+x+1\right)+3\left(x^2-1\right)}{\left(x-1\right)\left(x^2+x+1\right)-3\left(x-1\right)}\)=\(\dfrac{\left(x-1\right)\left(x^2+x+1+3x+3\right)}{\left(x-1\right)\left(x^2+x+1-3\right)}\)=\(\dfrac{x^2+4x+4}{x^2+x-2}\)=\(\dfrac{\left(x+2\right)^2}{x^2+x-2}\)
Mong là mk làm đúng. Chúc bn may mắn!
a) (7x + 4)2 - (7x + 4)(7x - 4)
= 49x2 + 56x + 16 - 49x2 + 16
= 56x + 32
b) (x - 2y)3 - 6xy(x - 2y)
= x3 - 6x2y + 12xy2 - 8y3 - 6x2y + 12xy2
= x3 - 12x2y + 24xy2 - 8y3
c) (3x + y)(9x2 - 3xy + y2) - (3xy)3 - 27x2y
= 27x3 + y3 - (3xy)3 - 27x2y
d) 5(x + 3)(x - 3) + (2x + 3)2 + (x - 6)2
= 5x2 - 45 + 4x2 + 12x + 9 + x2 - 12x + 36
= 10x2
e) (2x + 3)2 + (2x - 3)2 - 2(4x2 - 9)
= (2x + 3)2 + (2x - 3)2 - 2(2x - 3)(2x + 3)
= (2x + 3 - 2x + 3)2
= 62 = 36
g) (x + 2)3 + (x - 2)3 + x3 - 3x(x - 2)(x + 2)
= (x+2+x-2)(x2 + 4x + 4 - x2 + 4 + x2 - 4x + 4) + x3 - 3x3 + 12x
= 2x(x2 + 8) + x3 - 3x3 + 12x
= 2x3 + 16x + x3 - 3x3 + 12x
= 28x
Nếu \(x=400\Rightarrow\left\{{}\begin{matrix}799=2x-1\\403=x+3\\1198=3x-2\\1203=3\left(x+1\right)\end{matrix}\right.\)( * )
Thay ( * ) vào R , ta được :
\(R=\left(2x-1\right)x^2-3x^4+\left(x+3\right)x+\left(3x-2\right)x^3-3\left(x+1\right)\)
\(=2x^3-x^2-3x^4+x^2+3x+3x^4-2x^3-3x-3\)
\(=\left(2x^3-2x^3\right)+\left(3x^4-3x^4\right)+\left(x^2-x^2\right)+\left(3x-3x\right)-3\)
\(=-3\)
Vậy \(R=-3\) tại \(x=400\)
Cảm ơn bạn nha!