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Ta có: x2 – x – 12 = x2 – x – 16 + 4
= (x2 – 16) – (x – 4)
= (x – 4).(x + 4) – (x – 4)
= (x – 4).(x + 4 – 1)
= (x – 4).(x + 3)
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
1
a) x^2+2x-5 b) x^2+x+7 9 (dư 8)
2
x=2; x = -(3*căn bậc hai(7)*i+1)/2;x = (3*căn bậc hai(7)*i-1)/2;
3
a=2
Bài 1 :
b, Ta có : \(4x^2-25-\left(2x-5\right)\left(2x+7\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)\)
\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)
\(=-2\left(2x-5\right)\)
c, Ta có : \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=x\left(x+3\right)\left(x-2\right)\)
Bài 2 :
a, Để \(x^3+3x^2+3x-2⋮x+1\)
<=> \(x^3+1+3x^2+3x-3⋮x+1\)
<=> \(\left(x+1\right)^3-3⋮x+1\)
Ta thấy : \(\left(x+1\right)^3⋮x+1\)
<=> \(-3⋮x+1\)
<=> \(x+1\inƯ_{\left(3\right)}\)
<=> \(x+1=\left\{1,-1,3,-3\right\}\)
<=> \(x=\left\{0,-2,2,-4\right\}\)
Vậy ...
b, Để \(2x^2+x-7⋮x-2\)
<=> \(2x^2-8x+8+9x-15⋮x-2\)
<=> \(2\left(x-2\right)^2+9x-15⋮x-2\)
Ta thấy : \(2\left(x-2\right)^2⋮x-2\)
<=> \(9x-15⋮x-2\)
<=> \(9x-18+3⋮x-2\)
Ta thấy : \(8\left(x-2\right)⋮x-2\)
<=> \(3⋮x-2\)
<=> \(x-2\inƯ_{\left(3\right)}\)
<=> \(x-2=\left\{1,-1,3,-3\right\}\)
<=> \(x=\left\{3,1,5,-1\right\}\)
Vậy ...
1. \(x^3-x^2+x-1=(x^3-x^2)+(x-1)\)
\(=x^2(x-1)+(x-1)=(x^2+1)(x-1)\)
2. \(6x^2y-2xy^2+3x-y=2xy(3x-y)+(3x-y)\)
\(=(3x-y)(2xy+1)\)
3. \(4x^2+1\) thì còn cái gì để phân tích hả bạn? Hay ý bạn là \(4x^4+1\)?
\(4x^4+1=(2x^2)^2+1=(2x^2)^2+1+4x^2-4x^2\)
\(=(2x^2+1)^2-(2x)^2=(2x^2+1-2x)(2x^2+1+2x)\)
4. \(x^2-9x+8=(x^2-x)-(8x-8)\)
\(=x(x-1)-8(x-1)=(x-1)(x-8)\)
5. \(x^3-2x^2y+3xy^2=x(x^2-2xy+3y^2)\)
6. \(x^2-6x+y-y^2\) (sai đề)
7. \(x^2-xy-2x+2y=(x^2-xy)-(2x-2y)\)
\(=x(x-y)-2(x-y)=(x-y)(x-2)\)
Ta có :
\(\left(3x^{n-1}y^6-5x^{n+1}y^4\right):2x^3y^n=\frac{3}{2}x^{n-4}y^{6-n}-\frac{5}{2}x^{n-2}y^{4-n}\)
Để A chia hết cho B thì tất cả số mũ của phần biến phải không âm
\(n-4\ge0\)\(\Leftrightarrow\)\(n\ge4\)
\(6-n\ge0\)\(\Leftrightarrow\)\(n\le6\)
\(n-2\ge0\)\(\Leftrightarrow\)\(n\ge2\)
\(4-n\ge0\)\(\Leftrightarrow\)\(n\le4\)
Từ những dữ kiện trên \(\Rightarrow\)\(4\le n\le4\)\(\Rightarrow\)\(n=4\)
Vậy \(n=4\)
Chúc bạn học tốt ~
\(\left(3x^{n-1}y^6-5x^{n+1}y^4\right):2x^3y^n=\frac{3}{2}x^{n-4}y^{6-n}-\frac{5}{2}x^{n-2}y^{4-n}\)
Để \(\left(3x^{n-1}y^6-5x^{n+1}y^4\right)⋮2x^3y^n\) thì các số mũ của phần biến phải không âm, do đó :
\(n-4\ge0\)\(\Leftrightarrow\)\(n\ge4\)
\(6-n\ge0\)\(\Leftrightarrow\)\(n\le6\)
\(n-2\ge0\)\(\Leftrightarrow\)\(n\ge2\)
\(4-n\ge0\)\(\Leftrightarrow\)\(n\le4\)
\(\Rightarrow\)\(4\le n\le4\)\(\Rightarrow\)\(n=4\)
\(\left(7x^{n-1}y^5-5x^3y^4\right):5x^2y^n=\frac{7}{5}x^{n-3}y^{5-n}-xy^{4-n}\)
Để \(\left(7x^{n-1}y^5-5x^3y^4\right)⋮5x^2y^n\) thì các số mũ của phần biến phải không âm, do đó :
\(n-3\ge0\)\(\Leftrightarrow\)\(n\ge3\)
\(5-n\ge0\)\(\Leftrightarrow\)\(n\le5\)
\(4-n\ge0\)\(\Leftrightarrow\)\(n\le4\)
\(\Rightarrow\)\(3\le n\le4\)\(\Rightarrow\)\(n\in\left\{3;4\right\}\)
Chúc bạn học tốt ~
\(a.4x^3-8x^2+4xy^3=4x\left(x^2-8x+y^3\right)\)
\(b.x^2+2xy+y^2-36=\left(x+y\right)^2-36=\left(x+y-6\right)\left(x+y+6\right)\) \(c.x^2-2xy+y^2-25=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\) \(d.x^2-5x+2xy-5y+y^2=\left(x+y\right)^2-5\left(x+y\right)=\left(x+y\right)\left(x+y-5\right)\) \(e.49+2xy-x^2-y^2=-\left(x^2-2xy+y^2-49\right)=-\left[\left(x-y\right)^2-49\right]=-\left(x-y-7\right)\left(x-y+7\right)\) \(f.3x^2-6x+3-3y^2=3\left(x^2-2x-y^2+1\right)\)
\(g.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)\left(x+1\right)\)
\(h,\) giống câu f.
\(i.x^3-2x^2y+xy^2-64x=x\left(x^2-2xy+y^2-64\right)=x\left[\left(x-y\right)^2-64\right]=x\left(x-y-8\right)\left(x-y+8\right)\) \(k.3x+3y-x^2-2xy-y^2=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
1.
a. x2 - 2x + 1 = 0
x2 - 2x*1 + 12 = 0
(x-1)2 = 0
............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)
1, Tìm x biết:
a, x2 - 2x +1 = 0
(x-1)2 = 0
x-1 = 0
x = 1. Vậy ...
b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30
25x2 +10x + 1 - (25x2 -9) = 30
25x2 +10x + 1 - 25x2 +9 = 30
10x + 10 =30
10(x+1) = 30
x+1 =3
x = 2. vậy ...
c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5
(x3 - 1) - x(x2 -4) = 5
x3 - 1 - x3 + 4x = 5
4x - 1 = 5
4x = 6
x = \(\dfrac{3}{2}\) .vậy ...
d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15
x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15
x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15
24x + 25 = 15
24x = -10
x = \(\dfrac{-5}{12}\) vậy ...
a )
Để A \(⋮\) B thì \(x^n\ge x^3\) \(\Rightarrow n\ge3\)
Để M \(⋮\) N thì \(y^n\ge y^2\Rightarrow n\ge2\)
a, A= 5\(x^ny^3\)
B= 4\(x^3y\)
=> A\(⋮\)B -> n \(\ge\)3
b, làm tương tự như trên
1, \(\left(2x^4-5x^2y^2+3xy^3\right)\left(5x^3+x^2y-y^3\right)\)
\(=10x^7-25x^5y^2+15x^4y^3+2x^6y-5x^4y^3+5x^2y^5+3xy^6\)
2, a, \(4-2x+5x^2-4x^2\&5x-3+x^2\)
Sắp xếp: \(4-2x+5x^2-4x^2=5x^2-4x^2-2x+4=x^2-2x+4\)
\(5x-3+x^2=x^2+5x-3\)
- \(\left(x^2-2x+4\right)\left(x^2+5x-3\right)=x^4+3x^3-9x^2-14x-12\)
b, Làm tương tự câu a
1 ) \(\left(2x^4-5x^2y^2+3xy^3\right)\left(5x^3+x^2y-y^3\right)\)
\(=2x^4\left(5x^3+x^2y-y^3\right)-5x^2y^2\left(5x^3+x^2y-y^3\right)+3xy^3\left(5x^3+x^2y-y^3\right)\)\(=10x^7+2x^6y-2x^4y-25x^5y^2-5x^4y^3+5x^2y^5+15x^4y^3+3x^3y^4-3xy^6\)2 ) a ) \(4-2x+5x^2-4x^2=x^2-2x+4\)
\(5x-3+x^2=x^2+5x-3\)
\(\left(x^2-2x+4\right)\left(x^2+5x-3\right)\)
\(=x^4-2x^3+4x^2+5x^3-10x^2+20x-3x^2+6x-12\)
\(=x^4+3x^3-9x^2+26x-12\)
b ) \(10-x^4+3x-4x^2=-x^4-4x^2+3x+10\)
\(2x+x^3-1=x^3+2x-1\)
\(\left(-x^4-4x^2+3x+10\right)\left(x^3+2x-1\right)\)
\(=-x^4\left(x^3+2x-1\right)-4x^2\left(x^3+2x-1\right)+3x\left(x^3+2x-1\right)+10\left(x^3+2x-1\right)\)\(=-x^7-2x^5+x^4-4x^5-8x^3+4x^2+3x^4+6x^2-3x+10x^3+20x-10\)\(=-x^7-\left(2x^5+4x^5\right)+\left(3x^4+x^4\right)+\left(10x^3-8x^3\right)+\left(4x^2+6x^2\right)+\left(20x-3x\right)-10\)\(=-x^7-6x^5+4x^4+2x^3+10x^2+17x-10\)
Bài 5.5:
\(\left(2x-3\right)\left(x+1\right)+\left(4x^3-6x^2-6x\right):\left(-2x\right)=18\)
\(\Leftrightarrow\left(2x^2+2x-3x-3\right)+2x\cdot\left(2x^2-3x-3\right):\left(-2x\right)=18\)
\(\Leftrightarrow2x^2-x-3-2x^2+3x+3=18\)
\(\Leftrightarrow2x=18\)
\(\Leftrightarrow x=\dfrac{18}{2}\)
\(\Leftrightarrow x=9\)