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\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
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24 tháng 11 2018
\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)
\(=\frac{2x+5}{3x-1}\)
Còn bài b bạn tự làm nhé
24 tháng 11 2018
Điều kiện: \(x\ne\left\{-1;-2;-5\right\}\)
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\)
Điều kiện: \(x\ne\left\{3;\frac{1}{3}\right\}\)
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
\(=\frac{2x^2-x-15}{3x^2-10x+3}=\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)
\(=\frac{\left(2x+5\right)\left(x-3\right)}{\left(3x-1\right)\left(x-3\right)}=\frac{2x+5}{3x-1}\)
19 tháng 1 2017
f(0)=-4/10
a/b=-4/10=-2/5
f(1)=-6/26=-3/13=(a+1)/(b+1)
5a=-2b
a/-2=b/5=(a+b)/3
13a+13=-3b-3
15a=-6b
26a=-6b-6
11a=-6
a+b=-3/2.a=3/2.6/11=9/11
a+b=9/11
1
11 tháng 7 2017
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}=\frac{\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)+5\left(x+2\right)}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\Rightarrow a=-2;b=5\)
\(\Rightarrow\)\(a+b=-2+5=3\)
KN
9 tháng 11 2017
a) \(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
= \(\dfrac{\left(x^3+x^2\right)-\left(4x+4\right)}{\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)}\)
=\(\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
= \(\dfrac{\left(x^2-4\right)\left(x+1\right)}{\left(x^2+7x+10\right)\left(x+1\right)}\)
= \(\dfrac{\left(x+2\right)\left(x-2\right)\left(x+1\right)}{\left[\left(x^2+2x\right)+\left(5x+10\right)\right]\left(x+1\right)}\)
= \(\dfrac{\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)+5\left(x+2\right)}\)
= \(\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+5\right)\left(x+2\right)}\)
= \(\dfrac{x-2}{x+5}\)
b) \(\dfrac{x^3+3x^2-4}{x^3-3x+2}\)
= \(\dfrac{x^3-x^2+4x^2+4x-4x-4}{x^3-x-2x+2}\)
= \(\dfrac{\left(x^3-x^2\right)+\left(4x^2-4x\right)+\left(4x-4\right)}{\left(x^3-x\right)-\left(2x-2\right)}\)
= \(\dfrac{x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)}{x^2\left(x-1\right)-2\left(x-1\right)}\)
= \(\dfrac{\left(x^2+4x+4\right)\left(x-1\right)}{\left(x^2-2\right)\left(x-1\right)}\)
= \(\dfrac{\left(x+2\right)^2\left(x-1\right)}{\left(x^2-2\right)\left(x-1\right)}\)
= \(\dfrac{\left(x+2\right)^2}{x^2-2}\)
Nhớ tik nha...
HV
23 tháng 11 2017
a)
\(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
\(=\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+2x^2+6x^2+12x+5x+10}\)
\(=\dfrac{\left(x+1\right)\left(x^2-4\right)}{x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+6x+5\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left[x\left(x+5\right)+\left(x+5\right)\right]}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{x-2}{x+5}\)
b)
\(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)
\(=\dfrac{x^4+3x^3+x^2+3x^3+9x^2+3x-x^2-3x-1}{x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1}\)
\(=\dfrac{x^2\left(x^2+3x+1\right)+3x\left(x^2+3x+1\right)-\left(x^2+3x+1\right)}{x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)}\)
\(=\dfrac{\left(x^2+3x+1\right)\left(x^2+3x-1\right)}{\left(x^2+3x-1\right)\left(x^2+3x-1\right)}\)
\(=\dfrac{x^2+3x+1}{x^2+3x-1}\)
PJ
11 tháng 2 2017
Phân tích phương trình:
\(\frac{x^3+x^2-4\cdot x-4}{x^3+8\cdot x^2+17\cdot x+10}=\frac{x^2\cdot\left(x+1\right)-4\cdot\left(x+1\right)}{x^2\cdot\left(x+1\right)+7\cdot x\cdot\left(x+1\right)+10\cdot\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\cdot\left(x^2-4\right)}{\left(x+1\right)\cdot\left(x^2+7\cdot x+10\right)}\)
\(=\frac{\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x-2\right)}{\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+5\right)}=\frac{x-2}{x+5}\)
Vậy \(a=-2;b=5\)
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
\(=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+x^2+7x^2+7x+10x+10}\)
\(=\frac{\left(x^2-4\right)\left(x+1\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\frac{x^2-4}{x^2+7x+10}\)
\(=\frac{x^2-4}{x^2+5x+2x+10}\)
\(=\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+5\right)+2\left(x+5\right)}\)
\(=\frac{x-2}{x+5}\)