Thực hiện các phép tính:
3 x + 1 x - 1 2 - 1 x + 1 + x + 3 1 - x 2
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Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
a: \(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\)
\(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)
\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)
b: \(\dfrac{x+1}{2x-2}+\dfrac{x-1}{2x+2}+\dfrac{x^2}{1-x^2}\)
\(=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{x-1}{2\left(x+1\right)}-\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2+\left(x-1\right)^2-2x^2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x+1+x^2-2x+1-2x^2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x^2-1}\)
c: \(\dfrac{1}{x^2+xy}+\dfrac{2}{y^2-x^2}+\dfrac{1}{xy-x^2}\)
\(=\dfrac{1}{x\left(x+y\right)}-\dfrac{2}{\left(x-y\right)\left(x+y\right)}-\dfrac{1}{x\left(x-y\right)}\)
\(=\dfrac{x-y-2x-x-y}{x\left(x-y\right)\left(x+y\right)}=\dfrac{-2x-2y}{x\left(x-y\right)\left(x+y\right)}\)
\(=-\dfrac{2}{x\left(x-y\right)}\)
\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)
\(=\dfrac{x^2+\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}:\left(\dfrac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\right)\)
\(=\dfrac{x^2+x^2-1}{x\left(x+1\right)}.\dfrac{x\left(x+1\right)}{x^2-x^2+1}\)
\(=\dfrac{\left(2x^2-1\right)x\left(x+1\right)}{x\left(x+1\right)}=2x^2-1\)
b)B = x3 – 3x2 + 3x – 1 – 4x(x2 – 1) + 3(x2 – 1)
= x3 – 3x2 + 3x – 1 – 4x3 + 4 + 3x2 – 3 = -3x2 + 7x – 4
b: \(\dfrac{xy}{2x-y}-\dfrac{2x^2}{y-2x}=\dfrac{xy}{2x-y}+\dfrac{2x^2}{2x-y}=\dfrac{xy+2x^2}{2x-y}\)
b: \(\dfrac{3x^2-x}{x-1}+\dfrac{x+2}{1-x}+\dfrac{3-2x^2}{x-1}\)
\(=\dfrac{3x^2-x-x-2+3-2x^2}{x-1}\)
\(=\dfrac{x^2-2x+1}{x-1}=x-1\)
Ta có : 1x2x3x...x9-1x2x3x...x8-1x2x3x...x8^2
=1x2x3x...x8x(9-1-8)
=1x2x3x...x8x0
=0
Nhớ k cho mik nha !!!
3 x + 1 x - 1 2 - 1 x + 1 + x + 3 1 - x 2
(Cộng với phân thức đối và áp dụng quy tắc đổi dấu)
(Quy đồng với MTC = (x + 1)(x – 1)2)
(Tách 4x = x + 3x để phân tích tử thành nhân tử)