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a)\(\frac{x^2+5x+4}{x^2-1}=\frac{A}{x^2-2x+1}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}=\frac{A}{\left(x-1\right)^2}\)
\(\Leftrightarrow\frac{x+4}{x-1}=\frac{A}{\left(x-1\right)^2}\). Nhân 2 vế ở tử với x-1 ta có:
\(x+4=\frac{A}{x-1}\Leftrightarrow A=\left(x-1\right)\left(x+4\right)=x^2+3x-4\)
b)\(\frac{x^2-3x}{2x^2-7x+3}=\frac{x^2+4x}{A}\)
\(\Leftrightarrow\frac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}=\frac{x\left(x+4\right)}{A}\)
\(\Leftrightarrow\frac{x}{2x-1}=\frac{x\left(x+4\right)}{A}\).Nhân 2 vế ở mẫu với x ta có:
\(2x-1=\frac{x+4}{A}\)\(\Leftrightarrow\left(2x-1\right)\left(x+4\right)=A\Leftrightarrow A=2x^2+7x-4\)
Bài 7:(Sbt/25) Dùng tính chất cơ bản của phân thức hoặc quy tắc đổi dấu để biến mỗi cặp phân thức sau thành một cặp phân thức bằng nó và có cùng mẫu thức :
a. \(\dfrac{3x}{x-5}\) và \(\dfrac{7x+2}{5-x}\)
Ta có:
\(\dfrac{3x}{x-5}=\dfrac{-\left(3x\right)}{-\left(x-5\right)}=\dfrac{-3x}{5-x}\)
\(\dfrac{7x+2}{5-x}\)
Vậy .....
b.\(\dfrac{4x}{x+1}\) và \(\dfrac{3x}{x-1}\)
Ta có:
\(\dfrac{4x}{x+1}=\dfrac{4x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x^2-4x}{x^2-1}\)
\(\dfrac{3x}{x-1}=\dfrac{3x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x^2+3x}{x^2-1}\)
Vậy ..........
c. \(\dfrac{2}{x^2+8x+16}\) và \(\dfrac{x-4}{2x+8}\)
Ta có:
\(\dfrac{2}{x^2+8x+16}=\dfrac{4}{2\left(x+4\right)^2}\)
\(\dfrac{x-4}{2x+8}=\dfrac{\left(x-4\right)\left(x+4\right)}{2\left(x+4\right)\left(x+4\right)}=\dfrac{x^2-16}{2\left(x+4\right)^2}\)
Vậy .........
d. \(\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) và \(\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
Ta có:
\(\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x\left(x-2\right)}{\left(x+1\right)\left(x-3\right)\left(x-2\right)}=\dfrac{2x^2-4x}{\left(x+1\right)\left(x-2\right)\left(x-3\right)}\)
\(\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}=\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x+1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{x^2-9}{\left(x+1\right)\left(x-2\right)\left(x-3\right)}\)
Vậy .........
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
a ) \(\dfrac{x^2+3x+2}{3x+6}=\dfrac{\left(x+1\right)\left(x+2\right)}{3\left(x+2\right)}=\dfrac{x+1}{3}\) (1)
\(\dfrac{2x^2+x-1}{6x-3}=\dfrac{\left(2x-1\right)\left(x+1\right)}{3\left(2x-1\right)}=\dfrac{x+1}{3}\) (2)
Từ (1) ; (2) \(\Rightarrow\dfrac{x^2+3x+2}{3x+6}=\dfrac{2x^2+x-1}{6x-3}\) (đpcm)
b ) \(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=\dfrac{5}{x+1}\) (3)
\(\dfrac{5x^2-5x+5}{x^3+1}=\dfrac{5\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5}{x+1}\) (4)
Từ (3) và (4) \(\Rightarrow\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5x^2-5x+5}{x^3+1}\) (đpcm)
a) \(\dfrac{x^2+3x+2}{3x+6}=\dfrac{x^2+x+2x+2}{3\left(x+2\right)}=\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{3\left(x+2\right)}=\dfrac{x\left(x+1\right)+2\left(x+1\right)}{3\left(x+2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{3\left(x+2\right)}=\dfrac{x+1}{3}\left(1\right)\) \(\dfrac{2x^2+x-1}{6x-3}=\dfrac{2x^2+2x-x-1}{3\left(2x-1\right)}=\dfrac{2x\left(x+1\right)-\left(x+1\right)}{3\left(2x-1\right)}=\dfrac{\left(2x-1\right)\left(x+1\right)}{3\left(2x-1\right)}=\dfrac{x+1}{3}\left(2\right)\) Từ (1)và (2)=> \(\dfrac{x^2+3x+2}{3x+6}=\dfrac{2x^2+x-1}{6x-3}\) b)\(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5\left(3x-2\right)}{3x\left(x+1\right)-2\left(x+1\right)}=\dfrac{5\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=\dfrac{5}{x+1}\left(3\right)\) \(\dfrac{5x^2-5x+5}{x^3+1}=\dfrac{5\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5}{x+1}\left(4\right)\) Từ (3) và (4) => \(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5x^2-5x+5}{x^3+1}\)
a) A= \(\frac{3x^2+5x-2}{3x^2-7x+2}=0\)
\(ĐK:3x^2-7x+2\ne0\)
\(\Leftrightarrow\orbr{\begin{cases}x\ne\frac{1}{3}\\x\ne2\end{cases}\left(^∗\right)}\)
=> 3x2 + 5x + 2 =0
<=> 3x2 + 3x + 2x +2 = 0
<=> 3x .( x + 1 ) + 2 .( x + 1 ) =0
<=> ( x + 1 )(3x + 2 ) =0
<=> \(\orbr{\begin{cases}x+1=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{-2}{3}\left(t/m\left(^∗\right)\right)\end{cases}}}\)
Vậy x = -2/3
b) \(B=\frac{2x^2+10x+12}{x^3-4x}=0\left(ĐK:x\ne0;x^2\ne4\Leftrightarrow x\ne0;x\ne\pm2\right)\)
<=> 2x2+ 10x + 12 = 0
<=> x2 + 5x+ 6 =0
<=> ( x + 2 ) ( x + 3 ) =0\(\Leftrightarrow\orbr{\begin{cases}x=-2\left(L\right)\\x=-3\left(t/m\right)\end{cases}}\)
Vậy x = -3
c)\(C=\frac{x^3+x^2-x-1}{x^3+2x-5}=0\) \(ĐK:x^3+2x-5\ne0\left(^∗\right)\)
<=> x3 + x2 -x -1 =0
<=> ( x - 1 )(x2 + 2x + 1 )
<=> ( x-1 ) (x+1)2 = 0
<=> \(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(t/m\left(^∗\right)\right)\\x=-1\left(t/m\left(^∗\right)\right)\end{cases}}}\)
Vậy x = { 1 ; -1 }
a) A = \(\frac{3x^2+5x-2}{3x^2-7x+2}=0\) (ĐKXĐ: x khác 1/3, x khác 2)
<=> 3x^2 + 5x - 2 = 0
<=> (3x - 1)(x + 2) = 0
<=> 3x - 1 = 0 hoặc x + 2 = 0
<=> 3x = 1 hoặc x = -2
<=> x = 1/3 (ktm) hoặc x = -2 (tm)
=> x = -2
b) B = \(\frac{2x^2+10x+12}{x^3-4x}=0\) (ĐKXĐ: x khác 0, x khác +-2)
<=> \(\frac{2\left(x^2+5x+6\right)}{x\left(x^2-4\right)}=0\)
<=> \(\frac{2\left(x+2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
<=> \(\frac{2\left(x+3\right)}{x\left(x-2\right)}=0\)
<=> 2(x + 3) = 0
<=> x + 3 = 0
<=> x = -3
c) C = \(\frac{x^3+x^2-x-1}{x^3+2x-5}=0\) (ĐKXĐ: x khác x^3 + 2x - 5)
<=> \(\frac{x^2\left(x+1\right)-\left(x+1\right)}{x^3+2x-5}=0\)
<=> \(\frac{\left(x+1\right)\left(x^2-1\right)}{x^3+2x-5}=0\)
<=> \(\frac{\left(x+1\right)\left(x-1\right)\left(x+1\right)}{x^3+2x-5}=0\)
<=> (x + 1)(x - 1) = 0
<=> x + 1 = 0 hoặc x - 1 = 0
<=> x = -1 hoặc x = 1
