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Bài 1: (Sgk/36):
a. \(\dfrac{5y}{7}\)=\(\dfrac{20xy}{28x}\) vì
5y . 28x = 140xy
7 . 20xy = 140xy
=> 5y . 28x = 7 . 20xy
Vậy \(\dfrac{5y}{7}\)=\(\dfrac{20xy}{28x}\)
b. \(\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}\)=\(\dfrac{3x}{2}\) vì
3x . 2(x+5) = 6x2+30x
2 . 3x(x+5) = 6x2+30x
=> 3x . 2(x+5) = 2 . 3x(x+5)
Vậy \(\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}\)=\(\dfrac{3x}{2}\)
c. \(\dfrac{x+2}{x-1}\)=\(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}\) vì
(x+2) (x2-1) = (x+2) (x-1) (x-1)
=> (x+2) (x2-1) = (x-1) (x+2) (x+1)
Vậy \(\dfrac{x+2}{x-1}\)=\(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}\)
d. \(\dfrac{x^2-x-2}{x+1}\)=\(\dfrac{x^2-3x+2}{x-1}\)
(x-1) (x2-x-2) = x3-2x2-x+2
(x+1) (x2-3x+2) = x3-2x2-x+2
=> (x-1) (x2-x-2) = (x2-3x+2) (x+1)
Vậy \(\dfrac{x^2-x-2}{x+1}\)=\(\dfrac{x^2-3x+2}{x-1}\)
a. \(x^2y^3.35xy=5.7x^3y^4\)
\(\Leftrightarrow35x^3y^4=35x^3y^4\Rightarrowđpcm\)
\(b.x^2\left(x+2\right).\left(x+2\right)=x\left(x+2\right)^2.x\)
\(\Leftrightarrow x^2\left(x+2\right)^2=x^2\left(x+2\right)^2\Rightarrowđpcm\)
\(c.\left(3-x\right)\left(9-x^2\right)=\left(3+x\right)\left(x^2-6x+9\right)\)
\(\Leftrightarrow\left(3-x\right)\left(3-x\right)\left(3+x\right)=\left(3+x\right)\left(3-x\right)^2\)
\(\Leftrightarrow\left(3-x\right)^2\left(3+x\right)=\left(3-x\right)^2\left(3+x\right)\)
\(\Rightarrowđpcm\)
\(d.5\left(x^3-4x\right)=\left(10-5x\right)\left(-x^2-2x\right)\)
\(\Leftrightarrow5x^3-20x=5x^3-20x\Rightarrowđpcm\)
Xin được mạn phép chữa đề.
\(\text{c) }\dfrac{x+2}{x+1}=\dfrac{\left(x+2\right)\left(x-1\right)}{x^2-1}\)
\(\text{Ta có : }\dfrac{\left(x+2\right)\left(x-1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+2}{x+1}\left(đpcm\right)\)
Vậy.......................
c) x+2x+1=(x+2)(x−1)x2−1c) x+2x+1=(x+2)(x−1)x2−1
Ta có : (x+2)(x−1)x2−1=(x+2)(x−1)(x−1)(x+1)=x+2x+1(đpcm)
Vậy
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 .........
từ vế trái ta có
\(\frac{x.x\left(x+3\right)}{x.\left(x+3\right)\left(x+3\right)}\)
Rút gọn đi x và (x+3) còn
\(\frac{x}{x+3}\)
từ đó suy ra cái bên trên đó .
Xét VT, ta có: \(\frac{x^2\left(x+3\right)}{x\left(x+3\right)^2}=\frac{x}{x+3}\)= VP
Vậy ...
1. Đ
\(\dfrac{\left(x-8\right)^3}{2\left(8-x\right)}=\dfrac{\left(8-x\right)^3:\left(8-x\right)}{2\left(8-x\right):\left(8-x\right)}=\dfrac{\left(8-x\right)^2}{2}\)
2. ta có:\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
MTC: \(x^3-1\)
NTP:
NTP:\(x-1\)
\(\dfrac{3x}{x^3-1}\) giữ nguyên
\(\dfrac{x-1}{x^2+x+1}=\dfrac{\left(x-1\right)^2}{x^3-1}\) ta nhân cho x-1
bây giờ mới xong đó
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) => \(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5x^2-5x+5}{x^3+1}\)
\(\dfrac{x+2}{x-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}\)
\(\Rightarrow\left(x+2\right)\left(x^2-1\right)=\left(x-1\right)\left(x+2\right)\left(x+1\right)\)
\(\Rightarrow\left(x+2\right)\left(x^2-1\right)=\left(x+2\right)\left(x^2-1\right)\)
-> đpcm.