Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2x^4 + x^3 - x^2 - 4x - 2 2x^2 - x - 2 x^2 + x + 1 2x^4 - x^3 - 2x^2 2x^3 + x^2 - 4x 2x^3 - x^2 - 2x 2x^2 - 2x - 2 2x^2 - x - 2 -x
\(\left(2x^4+x^3-x^2-4x-2\right):\left(2x^2-x-2\right)=x^2+x+1-\frac{x}{2x^2-x-2}\)
làm như bình thường là đc:
a) ( x + 2 )^3 - x^2 . ( x+ 6 ) - 8
= ( x^3 + 3.x^2.2 +3.x.2^2 + 2^3 ) - ( x^3 + 6x^2 ) - 8
= x^3 + 6x^2 + 12x + 8 - x^3 - 6x^2 - 8
=12x
b) ( x - 2 ) . ( x^2 + 2x + 4 ) - ( x^3 + 2 )
= x^3 - 2^3 - ( x^3 + 2 )
= x^3 - 8 - x^3 - 2
= -6
\(\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)\)
\(=\left(x^3-8\right)\left(x^3+8\right)\)
\(=\left(x^3\right)^2-8^2\)
\(=x^6-64\)
a) 2(x-1)2 - 4(x+3)2 + 2x(x-5)
= 2(x2 -2x +1)- 4(x2 + 6x +9) + 2x2 -10x
= 2x2 - 4x + 2 -4x2 - 24x - 36 + 2x2 - 10x
= (2x2 + 2x2 - 4x2) - (4x + 24x+10x) +(2-36)
= -38x-34
b) 2(2x+5)2 -3(4x+1)(1-4x)
= 2(4x2 + 20x + 25) + 3(4x+1)(4x-1)
= 8x2 +40x + 50 + 3(16x2 -1)
= 8x2 + 40x + 50 + 48x2 - 3
=56x2 +40x + 47
a, \(2\left(x-1\right)^2-4\left(x+3\right)^2+2x\left(x-5\right)\)
\(=2\left(x^2-2x+1\right)-4\left(x^2+6x+9\right)+2x\left(x-5\right)\)
\(=2x^2-4x+2-4x^2-24x-36+2x^2-10=-28x-44\)
b, \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
\(=2\left(4x^2+20x+25\right)-3\left(1-16x^2\right)\)
\(=8x^2+40x+50-3+48x^2=56x^2+40x+47\)
\(\left(x-2\right)\left(x^2-2x+1\right)\left(x+2\right)\left(x^2+2x+4\right)\)
\(=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]\left[\left(x+2\right)\left(x^2-2x+4\right)\right]\)
\(=\left(x^3-8\right)\left(x^3+8\right)\)
\(=x^6-64\)
\(\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)\left(x+2\right)\left(x^2-2x+4\right)\)
\(=\left(x^3-8\right)\left(x^3+8\right)\)
\(=x^6+64\)
\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}\)( ĐKXĐ : \(x\ne1\))
\(=\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)
\(=\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x\left(x+5\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-x^2+5x-4}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x^2+5x}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-x^2+5x-4-\left(x^2+5x\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-x^2+5x-4-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-2\left(x^2+2\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-2}{x\left(x-1\right)}=\frac{-2}{x\left(x-1\right)}\)
Đang đánh máy thì bấm gửi -..-
2x^4 + 2x^3 + 3x^2 - 5x - 20 x^2 + x + 4 2x^2 - 5 2x^4 + 2x^3 + 8x^2 -5x^2 - 5x - 20 -5x^2 - 5x - 20 0
Vậy \(\left(2x^4+2x^3+3x^2-5x-20\right):\left(x^2+x+4\right)=2x^2-5\)
\(\dfrac{5x+2}{x^2-4}+\dfrac{x-5}{x-2}=\dfrac{5x+2+x^2-3x-10}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+4}{x+2}\\ \left(x+4\right)^2-\left(x+3\right)\left(x-2\right)=-13\\ \Leftrightarrow x^2+8x+16-x^2+x+6=-13\\ \Leftrightarrow9x=-13-22=-35\\ \Leftrightarrow x=-\dfrac{35}{9}\)
\(\left(2+x\right)\left(x^2-4\right)-\left(x-2\right)\left(x^2+2x+4\right)=\left(x+2\right)^2\left(x-2\right)-\left(x-2\right)\left(x^2+2x+4\right)=\left(x-2\right)\left[\left(x+2\right)^2-x^2-2x-4\right]=\left(x-2\right)\left(x^2+4x+4-x^2-2x-4\right)=\left(x-2\right)2x=2x^2-4x\)
\(\left(x+2\right)\left(x^2-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-4x+2x^2-8-x^3+8\)
\(=2x^2-4x\)