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Ta co:
\(\dfrac{1}{x^2-4}=\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\dfrac{1}{\left(x-2\right)\left(x+2\right)}=\dfrac{a}{x-2}+\dfrac{b}{x+2}\)
\(\Rightarrow\dfrac{a\left(x+2\right)+b\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{ax+2a+bx-2b}{\left(x-2\right)\left(x+2\right)}\)
Ta có: \(\dfrac{1}{x^2-4}=\dfrac{a}{x-2}+\dfrac{b}{x+2}\Rightarrow\dfrac{1}{x^2-4}=\dfrac{ax+2a+bx-2b}{x^2-4}\)
\(\Rightarrow ax+2a+bx-2b=1\)
\(\Rightarrow x\left(a+b\right)+\left(2a-2b\right)=0x+1\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=0\\2a-2b=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{4}\\b=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy: \(a=\dfrac{1}{4};b=-\dfrac{1}{4}\).
a) \(\dfrac{\left(x+2\right)P}{x-2}=\dfrac{\left(x-1\right)Q}{x^2-4}\)
\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2\left(x-2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2P=\left(x-1\right)Q\)
\(\Leftrightarrow P=x-1\)
\(Q=\left(x+2\right)^2=x^2+4x+4\)
b)\(\dfrac{\left(x+2\right)P}{x^2-1}=\dfrac{\left(x-2\right)Q}{x^2-2x+1}\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)P=\left(x+1\right)\left(x-1\right)\left(x-2\right)Q\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)P=\left(x+1\right)\left(x-2\right)Q\)
\(\Leftrightarrow P=\left(x+1\right)\left(x-2\right)=x^2-x-2\)
\(Q=\left(x-1\right)\left(x+2\right)=x^2+x-2\)
a)2(x-y)/(-3)(x-y)=-2/3
b)8-x^3=(2-x)(x^2+2x+4) => Vế phải =(2-x)/x=(x-2)/-x
c)y^2-x^2=(y+x)(y-x) bạn đổi dấu rồi rút gọn là được,cũng tương tự như trên ý
a. \(x^2y^3.35xy=5.7x^3y^4\)
\(\Leftrightarrow35x^3y^4=35x^3y^4\Rightarrowđpcm\)
\(b.x^2\left(x+2\right).\left(x+2\right)=x\left(x+2\right)^2.x\)
\(\Leftrightarrow x^2\left(x+2\right)^2=x^2\left(x+2\right)^2\Rightarrowđpcm\)
\(c.\left(3-x\right)\left(9-x^2\right)=\left(3+x\right)\left(x^2-6x+9\right)\)
\(\Leftrightarrow\left(3-x\right)\left(3-x\right)\left(3+x\right)=\left(3+x\right)\left(3-x\right)^2\)
\(\Leftrightarrow\left(3-x\right)^2\left(3+x\right)=\left(3-x\right)^2\left(3+x\right)\)
\(\Rightarrowđpcm\)
\(d.5\left(x^3-4x\right)=\left(10-5x\right)\left(-x^2-2x\right)\)
\(\Leftrightarrow5x^3-20x=5x^3-20x\Rightarrowđpcm\)
a)\(\dfrac{x+5}{3x-2}=\dfrac{x\left(x+5\right)}{x\left(3x-2\right)}\) b)\(\dfrac{2x-1}{4}=\dfrac{\left(2x-1\right)\left(2x+1\right)}{8x+4}\) c)\(\dfrac{2x\left(x-2\right)}{x^2-4x+4}=\dfrac{2x}{x-2}\) d) \(\dfrac{5x^2+10x}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x}{x-2}\)
Bài 3: (SBT/24):
a. \(\dfrac{5x+3}{x-2}\)=\(\dfrac{5x^2+13x+6}{x^2-4}\)
(5x+3) . (x2-4) = 5x3-20x+3x3-12
(x-2) . (5x2+13x+6) = 5x3+13x2+6x-10x2-26x-12 = 5x3-20x+3x2-12
=> (5x+3) (x2-4) = (x-2) (5x2+13x+6)
Vậy \(\dfrac{5x+3}{x-2}\)=\(\dfrac{5x^2+13x+6}{x^2-4}\)(đẳng thức đúng)
b. \(\dfrac{x+1}{x+3}\)=\(\dfrac{x^2+3}{x^2+6x+9}\)
(x+1) . (x2+6x+9) = x3+6x2+9x+x2+6x+9 = x3+7x2+15x+9
(x+3) . (x2+3) = x3+3x+3x2+9
=> (x+1) (x2+6x+9) ≠ (x+3) (x2+3)
Vậy \(\dfrac{x+1}{x+3}\)≠\(\dfrac{x^2+3}{x^2+6x+9}\)(đẳng thức sai)
Chữa lại: \(\dfrac{x+1}{x+3}\)=\(\dfrac{x^2+3}{x^{2_{ }}+6x+9}\)
c. \(\dfrac{x^2-2}{x^2-1}\)=\(\dfrac{x+2}{x+1}\)
(x2-2) . (x+1) = x3+x2-2x-2
(x2-1) . (x+2) = x3+2x2-x-2
=> (x2-2) (x+1) ≠ (x2-1) (x+2)
Vậy \(\dfrac{x^2-2}{x^2-1}\)≠\(\dfrac{x+2}{x+1}\)(đẳng thức sai)
Chữa lại: \(\dfrac{x^2+x-2}{x^2-1}\)=\(\dfrac{x+2}{x+1}\)
d. \(\dfrac{2x^2-5x+3}{x^2+3x-4}\)=\(\dfrac{2x^2-x-3}{x^2+5x+4}\)
(2x2-5x+3) . (x2+5x+4) = 2x4+10x3+8x2-5x3-25x2-20x+3x2+15x+12
= 2x4+5x3-14x2-5x+12
(x2+3x-4) . (2x2-x-3) = 2x4-x3-3x2+6x3-3x2-9x-8x2+4x+12
= 2x4+5x3-14x2-5x+12
=> (2x2-5x+3) (x2+5x+4) = (x2+3x-4) (2x2-x-3)
Vậy \(\dfrac{2x^2-5x+3}{x^2+3x-4}\)=\(\dfrac{2x^2-x-3}{x^2+5x+4}\)
(1/2x^2-1/3y^2)(1/2x^2+1/3y^2)
=(1/2x^2)^2-(1/3y^2)^2
=1/4x^4-1/9y^4
=>a=1/4