*Cộng các phân thức sau:a) x^2/x+1 + 2x/x^2-1 + 1/1+x+1 b) 2x+y/2x^2-y + 8y/y^2-4x^2+2x-y/2x^2+xy
K
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Những câu hỏi liên quan
12 tháng 12 2020
a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)
\(=\frac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^3-x^2-2x+x-1-x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^3-2x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2021
a. \(y'=\dfrac{-1}{\left(x-1\right)}\)
b. \(y'=\dfrac{5}{\left(1-3x\right)^2}\)
c. \(y=\dfrac{\left(x+1\right)^2+1}{x+1}=x+1+\dfrac{1}{x+1}\Rightarrow y'=1-\dfrac{1}{\left(x+1\right)^2}=\dfrac{x^2+2x}{\left(x+1\right)^2}\)
d. \(y'=\dfrac{4x\left(x^2-2x-3\right)-2x^2\left(2x-2\right)}{\left(x^2-2x-3\right)^2}=\dfrac{-4x^2-12x}{\left(x^2-2x-3\right)^2}\)
e. \(y'=1+\dfrac{2}{\left(x-1\right)^2}=\dfrac{x^2-2x+3}{\left(x-1\right)^2}\)
g. \(y'=\dfrac{\left(4x-4\right)\left(2x+1\right)-2\left(2x^2-4x+5\right)}{\left(2x+1\right)^2}=\dfrac{4x^2+4x-14}{\left(2x+1\right)^2}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2021
2.
a. \(y'=4\left(x^2+x+1\right)^3.\left(x^2+x+1\right)'=4\left(x^2+x+1\right)^3\left(2x+1\right)\)
b. \(y'=5\left(1-2x^2\right)^4.\left(1-2x^2\right)'=-20x\left(1-2x^2\right)^4\)
c. \(y'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{2x+1}{x-1}\right)'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{-3}{\left(x-1\right)^2}\right)=\dfrac{-9\left(2x+1\right)^2}{\left(x-1\right)^4}\)
d. \(y'=\dfrac{2\left(x+1\right)\left(x-1\right)^3-3\left(x-1\right)^2\left(x+1\right)^2}{\left(x-1\right)^6}=\dfrac{-x^2-6x-5}{\left(x-1\right)^4}\)
e. \(y'=-\dfrac{\left[\left(x^2-2x+5\right)^2\right]'}{\left(x^2-2x+5\right)^4}=-\dfrac{2\left(x^2-2x+5\right)\left(2x-2\right)}{\left(x^2-2x+5\right)^4}=-\dfrac{4\left(x-1\right)}{\left(x^2-2x+5\right)^3}\)
f. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)
C
1
24 tháng 12 2020
a, \(\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}=\frac{x}{y\left(x-y\right)}+\frac{2x-y}{x\left(y-x\right)}\)
\(=\frac{x^2}{xy\left(x-y\right)}-\frac{2xy-y^2}{xy\left(x-y\right)}=\frac{\left(x-y\right)^2}{xy\left(x-y\right)}=\frac{x-y}{xy}\)
b, \(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x^2}{x^2-1}=\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x-1+x+1+2x^2}{\left(x-1\right)\left(x+1\right)}=\frac{2x+2x^2}{\left(x-1\right)\left(x+1\right)}=\frac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{2x}{x-1}\)
23 tháng 10 2023
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: \(\dfrac{1}{x^2+xy}+\dfrac{2}{y^2-x^2}+\dfrac{1}{xy-x^2}\)
\(=\dfrac{1}{x\left(x+y\right)}-\dfrac{2}{\left(x-y\right)\left(x+y\right)}-\dfrac{1}{x\left(x-y\right)}\)
\(=\dfrac{x-y-2x-x-y}{x\left(x-y\right)\left(x+y\right)}=\dfrac{-2x-2y}{x\left(x-y\right)\left(x+y\right)}\)
\(=-\dfrac{2}{x\left(x-y\right)}\)
18 tháng 9 2023
a) \(\dfrac{x^3-1}{x^2+x+1}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)
b) \(\dfrac{x^2+2xy+y^2}{2x^2+xy-y^2}\)
\(=\dfrac{\left(x+y\right)^2}{x^2+xy+x^2-y^2}=\dfrac{\left(x+y\right)^2}{x\left(x+y\right)+\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{\left(2x-y\right)\left(x+y\right)}=\dfrac{x+y}{\left(2x-y\right)}\)
c) \(\dfrac{ax^4-a^4x}{a^2+ax+x^2}\)
\(=\dfrac{ax\left(x^3-a^3\right)}{a^2+ax+x^2}\)
\(=\dfrac{ax\left(x-a\right)\left(a^2+ax+x^2\right)}{a^2+ax+x^2}\)
\(=ax\left(x-a\right)\)
4 tháng 9 2021
\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow x=-2\)
\(B=x^2+10x-1=\left(x+5\right)^2-26\ge-26\)
Dấu \("="\Leftrightarrow x=-5\)
\(C=5-4x+4x^2=\left(2x-1\right)^2+4\ge4\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)
\(D=x^2+y^2-2x+6y-3=\left(x-1\right)^2+\left(y+3\right)^2-13\ge-13\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
\(E=2x^2+y^2+2xy+2x+3=\left(x+y\right)^2+\left(x+1\right)^2+2\ge2\)
Dấu \("="\Leftrightarrow x=-y=-1\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
4 tháng 9 2021
\(A=x^2+4x+5\)
\(=x^2+4x+4+1\)
\(=\left(x+2\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=-2
\(C=4x^2-4x+5\)
\(=4x^2-4x+1+4\)
\(=\left(2x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
KV
30 tháng 10 2023
a) 3x³ + 6x²y
= 3x².(x + 2y)
b) 2x³ - 6x²
= 2x².(x - 2)
c) 18x² - 20xy
= 2x.(9x - 10y)
d) xy + y² - x - y
= (xy + y²) - (x + y)
= y(x + y) - (x + y)
= (x + y)(y - 1)
e) (x²y² - 8)² - 1
= (x²y² - 8 - 1)(x²y² - 8 + 1)
= (x²y² - 9)(x²y² - 7)
= (xy - 3)(xy + 3)(x²y² - 7)
f) x² - 7x - 8
= x² - 8x + x - 8
= (x² - 8x) + (x - 8)
= x(x - 8) + (x - 8)
= (x - 8)(x + 1)
30 tháng 10 2023
a: \(3x^3+6x^2y\)
\(=3x^2\cdot x+3x^2\cdot2y=3x^2\left(x+2y\right)\)
b: \(2x^3-6x^2=2x^2\cdot x-2x^2\cdot3=2x^2\left(x-3\right)\)
c: \(18x^2-20xy=2x\cdot9x-2x\cdot10y=2x\left(9x-10y\right)\)
d: \(xy+y^2-x-y\)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(y-1\right)\)
e: \(\left(x^2y^2-8\right)^2-1\)
\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left(x^2y^2-7\right)\left(x^2y^2-9\right)\)
\(=\left(x^2y^2-7\right)\left(xy-3\right)\left(xy+3\right)\)
f: \(x^2-7x-8\)
\(=x^2-8x+x-8\)
\(=x\left(x-8\right)+\left(x-8\right)=\left(x-8\right)\left(x+1\right)\)
g: \(10x^2\left(2x-y\right)+6xy\left(y-2x\right)\)
\(=2x\cdot\left(2x-y\right)\cdot5x-2x\cdot\left(2x-y\right)\cdot3y\)
\(=2x\left(2x-y\right)\left(5x-3y\right)\)
h: \(x^2-2x+1-y^2\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-1-y\right)\left(x-1+y\right)\)
i: \(2x\left(x+2\right)+x^2\left(-x-2\right)\)
\(=2x\left(x+2\right)-x^2\left(x+2\right)\)
\(=\left(x+2\right)\left(2x-x^2\right)=x\cdot\left(x+2\right)\left(2-x\right)\)
k: \(-x^2+6x-9=-\left(x^2-6x+9\right)\)
\(=-\left(x^2-2\cdot x\cdot3+3^2\right)=-\left(x-3\right)^2\)
l: \(-2x^2+8xy-8y^2\)
\(=-2\left(x^2-4xy+4y^2\right)\)
\(=-2\left(x-2y\right)^2\)
m: \(3x^2+5x-3y^2-5y\)
\(=3\left(x^2-y^2\right)+5\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+3y+5\right)\)
PT
23 tháng 11 2017
a) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{8y}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-8y}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{\left(2x+y\right)\left(2x+y\right)-8yx+\left(2x-y\right)\left(2x-y\right)}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{8x^2-8xy+2y^2}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{2\left(4x^2-4xy+y^2\right)}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{2\left(2x-y\right)^2}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{2\left(2x-y\right)}{x\left(2x+y\right)}\)
b) \(\dfrac{1}{x^2+3x+2}+\dfrac{2x}{x^2+4x+3}+\dfrac{1}{x^2+5x+6}\)
\(=\dfrac{1}{x^2+x+2x+2}+\dfrac{2x}{x^2+x+3x+3}+\dfrac{1}{x^2+2x+3x+6}\)
\(=\dfrac{1}{x\left(x+1\right)\left(x+2\right)}+\dfrac{2x}{x\left(x+1\right)+3\left(x+1\right)}+\dfrac{1}{x\left(x+2\right)+2\left(x+2\right)}\)
\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{2x}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{x+3+2x\left(x+2\right)+x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{x+3+2x^2+4x+x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2x^2+6x+4}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2\left(x^2+3x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2}{x+3}\)
a) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{1+x+1}\) \(=\dfrac{x^2.\left(x-1\right)\left(x+2\right)}{\left(x+1\right).\left(x-1\right)\left(x+2\right)}+\dfrac{2x.\left(x+2\right)}{\left(x-1\right).\left(x+1\right).\left(x+2\right)}+\dfrac{\left(x-1\right).\left(x+1\right)}{\left(x-1\right)\left(x+1\right).\left(x+2\right)}\)
\(=\dfrac{x^2.\left(x-1\right).\left(x+2\right)+2x.\left(x+2\right)+\left(x-1\right)\left(x+1\right)}{\left(x+1\right).\left(x-1\right).\left(x+2\right)}\)
\(=\dfrac{x^4+x^3-2x^2+2x^2+4x+x^2-1}{\left(x-1\right)\left(x+1\right).\left(x+2\right)}\)
\(=\dfrac{x^4+x^3+x^2+4x-1}{\left(x^2-1\right).\left(x+2\right)}\)
\(=\dfrac{x^4+x^3+x^2+4x-1}{x^3+2x^2-x-2}\)