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a) \(\frac{4x+3}{x^2-5}=\frac{\left(4x+3\right).3x}{\left(x^2-5\right).3x}=\frac{12x^2+9x}{3x\left(x^2-5\right)}\)
b) \(\frac{8x^2-8x+2}{\left(4x-2\right)\left(15x-1\right)}=\frac{2\left(4x^2-4x+1\right)}{2\left(2x-1\right)\left(15x-1\right)}=\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(15x-1\right)}=\frac{2x-1}{15x-1}\)
\(\dfrac{8x^2-8x+2}{\left(4x-2\right)\left(15-x\right)}=\dfrac{2\left(4x^2-4x+1\right)}{2\left(2x-1\right)\left(15-x\right)}\\ =\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)\left(15-x\right)}=\dfrac{2x-1}{15-x}=\dfrac{1-2x}{x-15}\\ =\dfrac{A}{x-15}\)
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 .........
b)(x2+x+1)(x2+x+2)-12
Đặt t=x2+x+1
t(t+1)-12=t2+t-12
=(t-3)(t+4)=(x2+x+1-3)(x2+x+1+4)
=(x2+x-2)(x2+x+5)
=(x-1)(x+2)(x2+x+5)
c)(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=t2+8t+15
=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+15)
=(x2+8x+10)(x2+8x+22)
d)(x+2)(x+3)(x+4)(x+5)-24
=(x2+7x+10)(x2+7x+12)-24
Đặt t=x2+7x+10
t(t+2)-24=(t-4)(t+6)
=(x2+7x+10-4)(x2+7x+10+6)
=(x2+7x+6)(x2+7x+16)
=(x+1)(x+6)(x2+7x+16)
a/ Đặt x2 + 4x + 8 = a
Thì đa thức ban đầu thành
a2 + 3ax + 2x2 = (a2 + 2ax + x2) + (ax + x2)
= (a + x)2 + x(a + x) = (a + x)(a + 2x)
`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`
`<=>x ne 4,x me -1`
`b,ĐKXĐ:4x^2-25 ne 0`
`<=>(2x-5)(2x+5) ne 0`
`<=>x ne +-5/2`
`c,ĐKXĐ:8x^3+27 ne 0`
`<=>8x^3 ne -27`
`<=>2x ne -3`
`<=>x ne -3/2`
`d,2x+2 ne 0,4y^2-9 ne 0`
`<=>2x ne -2,(2y-3)(2y+3) ne 0`
`<=>x ne -1,y ne +-3/2`
b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)
d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)
\(a.\) \(ax^2-a^2x-x+a\)
\(=\left(ax^2-a^2x\right)-\left(x-a\right)\)
\(=ax\left(x-a\right)-\left(x-a\right)\)
\(=\left(ax-1\right)\left(x-a\right)\)
\(b.\) \(18x^3-12x^2+2x\)
\(=2x\left(9x^2-6x+1\right)\)
\(=2x\left(3x-1\right)^2\)
\(c.\) \(x^3-5x^2-4x+20\)
\(=\left(x^3-5x^2\right)-\left(4x-20\right)\)
\(=x^2\left(x-5\right)-4\left(x-5\right)\)
\(=\left(x^2-4\right)\left(x-5\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-5\right)\)
\(d.\) \(\left(x+7\right)\left(x+15\right)+15\)
\(=x^2+15x+7x+105+15\)
\(=x^2+22x+120\)
\(=\left(x+10\right)\left(x+12\right)\)
d: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{A}\)
hay A=x-2
a ) \(\frac{4x+3}{x^2-5},A=12x2+9x\)
Suy ra \(\frac{4x+3}{x^2-5}=\frac{\left(4x+3\right).3x}{\left(x^2-5\right).3x}=\frac{12x^2+9x}{3x^3-15x}\)
Chúc bạn học tốt !!!
