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\(\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}\)
\(=\dfrac{4\left(x-3\right)+x+7}{x^2-9}\)
\(=\dfrac{4x-12+x+7}{x^2-9}\)
\(=\dfrac{5x-5}{x^2-9}\)
Bài 1:
\(a,\left(x+4\right)\left(x+3\right)-7x=x^2+4x+3x+12-7x=x^2+12\\
b,\left(x+4\right)^2+x-16=x^2+8x+16+x-16=x^2+9x\\
c,\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}=\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{4x-12+x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{5x-5}{\left(x+3\right)\left(x-3\right)}\)
Bài 2:
\(7a-7b=7\left(a-b\right)\\
b,x^2-8x+16=\left(x-4\right)^2\\
c,ax-ay+3x-3y=a\left(x-y\right)+3\left(x-y\right)=\left(a+3\right)\left(x-y\right)\\
d,x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2 :
f(x) có bậc 3 chia cho đa thức \(x^2-x-2\) có bậc 2 sẽ được thương có bậc 1
Gọi thương của phép chia f(x) cho \(x^2-x-2\) là \(cx+d\)
\(\left(cx+d\right)\left(x^2-x-2\right)=f\left(x\right)\)
hay \(cx^3-cx^2-2cx+dx^2-dx-2d=x^3+ax+b\)
\(\Rightarrow cx^3+\left(d-c\right)x^2-\left(2c+d\right)x-2d=x^3+ax+b\)
\(\Rightarrow\left\{{}\begin{matrix}cx^3=x^3\\\left(d-c\right)x^2=0\\-\left(2c+d\right)x=ax\\-2d=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}c=1\\d-1=0\\a=-2.1-d\\-2d=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}c=1\\d=1\\a=-3\\b=-2\end{matrix}\right.\)
a) 5x(10x+7)-25x(2x-3) = 40
50x2+35x-50x2+75x = 40
(50x2-50x2)+(35x+75x) = 40
110x = 40
x =\(\dfrac{40}{110}\)=\(\dfrac{4}{11}\)
b)
Bài 2:
a) x + 2a.(x-y) - y
= 2a.(x-y) + (x-y)
= (x-y).(2a+1)
b) 5a2 - 5ax - 7a + 7x
= 5a.(a-x) - 7.(a-x)
= (a-x).(5a-7)
Bài 1:
a) 5x.(10x+7) - 25x.(2x-3) = 40
50x2 + 35x - 50x2 + 75x = 40
110x = 40
x = 4/11
b) (3x+2).(x-2) - (x-1).(x-3) = 4
3x2 - 6x + 2x - 4 - x2 + 3x + x - 3 = 4
2x2 - 7 = 4
...
bn tự làm tiếp nha
2\
a3+4a2-7a-10
= a3-2a2+6a2-12a+5a-10
=a2(a-2) +6a(a-2) +5(a-2)
= (a-2)(a2+6a+5)
= (a-2)(a+1)(a+5)
4\
(a2+a)2+4(a2+a)-12
= (a2+a)2+4(a2+a)+4-16
= (a2+a+2)2-16
= (a2+a+6)(a2+a-2)
5/
(x2+x+1)(x2+x+2)-12
đặt x2+x+1=a
⇒ a(a+1)-12
= a2+a-12
= a2-3a+4a-12
= a(a-3)+4(a-3)
= (a-3)(a+4)
⇒ (x2+x-2)(x2+x+5)
6\
x8+x+1
= x8+x7+x6-x7-x6-x5+x5+x4+x3-x4-x3-x2+x2+x+1
= x6(x2+x+1) - x5(x2+x+1) +x3(x2+x+1)-x2(x2+x+1)+(x2+x+1)
= (x2+x+1)(x6-x5+x3+x2+1)
7\
x10+x5+1
= x10+x9+x8-x9-x8-x7+x7+x6+x5-x6-x5-x4+x5+x4+x3-x3-x2-x+x2+x+1
= x8(x2+x+1)-x7(x2+x+1)+x5(x2+x+1)-x4(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1)
= (x2+x+1)(x8-x7+x5-x4+x3-x+1)
4. Đặt t= a^2 +a
Suy ra t^2 +4t - 12 = (t-2)(t+6) = (a^2+a-2) (a^2+a +6) = (a-1)(a+2)(a^2+a+6)
5. Đặt t = x^2 +x+1
Ta có: t(t+1) -12
= t^2 +t-12
= (t-3)(t+4)
= ( x^2 +x -2 ) (x^2+x+5)
= (x-1) ( x+2) (x^2+x+5)
6. x^8 + x^7 + x^6 - x^7- x^6 - x^5 + x^5+ x^4 + x^3- x^4- x^3- x^2 + x^2 + x +1
= (x^2 +x+1) ( x^6 - x^5 +x^3 -x^2 +1)
7. x^10 + x^9 +x^8 - x^9- x^8- x^7 +x^7+x^6+x^5 - x^6-x^5 - x^4 + x^5+ x^4 + x^3 - x^3 - x^2 - x + x^2 + x +1
= (x^2 + x + 1) ( x^8 -x^7 + x^5 - x^4 + x^3 -x + 1)
a3 - 7a - 6
= a3 - a - 6a - 6
= a ( a2 - 1 ) - 6 ( a + 1 )
= a ( a - 1 ) ( a + 1 ) - 6 ( a + 1 )
= ( a + 1 ) [ ( a ( a - 1 ) - 6 ]
= ( a + 1 ) ( a2 - a - 6 )
= ( a + 1 ) ( a2 + 2a - 3a - 6 )
= ( a + 1 ) ( a + 2 ) ( a - 3 )
1.
$a^3-7a-6=a^3-a-(6a+6)=a(a^2-1)-6(a+1)$
$=a(a-1)(a+1)-6(a+1)=(a+1)(a^2-a-6)$
$=(a+1)(a^2+2a-3a-6)$
$=(a+1)[a(a+2)-3(a+2)]=(a+1)(a+2)(a-3)$
2.
\(a^3+4a^2-7a-10=a^3+a^2+(3a^2+3a)-(10a+10)\)
\(=a^2(a+1)+3a(a+1)-10(a+1)=(a+1)(a^2+3a-10)\)
\(=(a+1)[a(a-2)+5(a-2)]=(a+1)(a-2)(a+5)\)
3.
\(a(b+c)^2+b(c+a)^2+c(a+b)^2-4abc\)
\(=a(b^2+c^2+2bc)+b(c^2+a^2+2ac)+c(a^2+b^2+2ab)-4abc\)
\(=ab(a+b)+bc(b+c)+ca(c+a)+2abc\)
\(=ab(a+b+c)+bc(a+b+c)+ac(a+c)\)
\(=(a+b+c)(ab+bc)+ac(a+c)=(ab+b^2+bc)(a+c)+ac(a+c)\)
\(=(a+c)(ab+b^2+bc+ac)=(a+c)(a+b)(b+c)\)