phân tích đa thức thành nhân tử: (x+1)(x+2)(x+3)(x+4)-8
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\(\dfrac{3x^2+9x-3}{x^2+x-2}-\dfrac{x+1}{x+2}+\dfrac{x-2}{1-x}\left(ĐK:x\ne\left\{1;-2\right\}\right)\\ =\dfrac{3x^2+9x-3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+1}{x+2}-\dfrac{x-2}{x-1}\\ =\dfrac{3x^2+9x-3-\left(x+1\right)\left(x-1\right)-\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-1\right)}\\ =\dfrac{3x^2+9x-3-\left(x^2-1\right)-\left(x^2-4\right)}{\left(x+2\right)\left(x-1\right)}\)
\(=\dfrac{3x^2+9x-3-x^2+1-x^2+4}{\left(x+2\right)\left(x-1\right)}\\ =\dfrac{x^2+9x+2}{\left(x+2\right)\left(x-1\right)}\)
(a):
\(P=\dfrac{3x-12}{3x^2-3x-36}=\dfrac{3\left(x-4\right)}{3\left(x^2-x-12\right)}\\ =\dfrac{3\left(x-4\right)}{3\left(x-4\right)\left(x+3\right)}\\ =\dfrac{1}{x+3}\left(ĐK:x\ne\left\{4;-3\right\}\right)\)
(b):
\(x=\dfrac{1}{2}\left(TMDK\right)=>P=1:\left(\dfrac{1}{2}+3\right)=1:\dfrac{7}{2}=\dfrac{2}{7}\)
(c):
\(P=\dfrac{1}{x+3}\in Z=>1⋮\left(x+3\right)\\ =>x+3\inƯ\left(1\right)=\left\{\pm1\right\}\\ =>x\in\left\{-4;-2\right\}\left(TMDK\right)\)
\(P=\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\\ =2x^3-3x^2+4x+4x^2-6x+8-\left(2x^3+x^2-2x-1\right)\\ =2x^3+x^2-2x+8-2x^3-x^2+2x+1\\ =9\)
A = (1-3m)(9m2 +3m +1) - (6-26m3)
A = 9m2 + 3m + 1 - 27m3 - 9m2 - 3m - 6 + 26m3
A = -m3 - 5
với m = 5 ⇔A = -53 - 5 = -130
b, B = ( 2x - 3)2 + ( 2x + 1)2 - 2 (4x2 - 9)
B = ( 2x - 3 - 2x - 1)2 + 2(2x-3)(2x+1) - 2(4x2 - 9)
B = 42 + 8x2 + 4x - 12x - 6 - 8x2 + 18
B =- 8x + 16 + 18 - 6
B = -8x + 28
với x = 3 ⇔ B = -8. 3 + 28 = 4
a/
\(\widehat{B}+\widehat{C}=200^o\) (1)
\(\widehat{B}+\widehat{D}=180^o\) (2)
\(\widehat{C}+\widehat{D}=120^o\) (3)
Cộng 2 vế của (1) (2) (3)
\(\Rightarrow2\left(\widehat{B}+\widehat{C}+\widehat{D}\right)=500^o\Rightarrow\widehat{B}+\widehat{C}+\widehat{D}=250^o\)
\(\Rightarrow\widehat{A}=360^o-250^o=110^o\) (tổng góc trong 1 tứ giác bằng \(360^o\) )
Cộng 2 vế của (1) với (2)
\(\Rightarrow\widehat{B}+\widehat{B}+\widehat{C}+\widehat{D}=380^o\Rightarrow\widehat{B}=380^o-250^o=130^o\)
Cộng 2 vế của (2) với (3)
\(\Rightarrow\widehat{B}+\widehat{C}+\widehat{D}+\widehat{D}=300^o\Rightarrow\widehat{D}=300^o-250^o=50^o\)
\(\Rightarrow\widehat{C}=360^o-\widehat{A}-\widehat{B}-\widehat{D}=360^o-110^o-130^o-50^o=70^o\)
b/
Xét tg AIB có
\(\widehat{AIB}=180^o-\widehat{IAB}-\widehat{IBA}=180^o-\dfrac{\widehat{A}}{2}-\dfrac{\widehat{B}}{2}=\)
\(=180^o-\dfrac{110^o}{2}-\dfrac{130^o}{2}=60^o\)
\(\dfrac{\widehat{C}+\widehat{D}}{2}=\dfrac{70^o+50^o}{2}=60^o\)
\(\Rightarrow\widehat{AIB}=\dfrac{\widehat{C}+\widehat{D}}{2}=60^o\)
P = (x + 1)(x + 2)(x + 3)(x + 4) - 8
= [(x + 1)(x + 4)][(x + 2)(x + 3] - 8
= (x2 + 5x + 4)(x2 + 5x + 6) - 8
= (x2 + 5x + 5 - 1)(x2 + 5x + 5 + 1) - 8
= (x2 + 5x + 5)2 - 1 - 8 = (x2 + 5x + 5)2 - 9
= (x2 + 5x + 8)(x2 + 5x + 2)