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a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)
=3x+4
b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)
\(=\dfrac{x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=x+3\)
c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)
d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)
=7x+1
e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)
\(=\dfrac{x^2\left(2x+1\right)+3x\left(2x+1\right)+6\left(2x+1\right)}{2x+1}=x^2+3x+6\)
f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)
g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)
a, (3x+1)(7x+3)=(5x-7)(3x+1)
<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0
<=> (3x+1)(7x+3-5x+7)=0
<=> (3x+1)(2x+10)=0
<=> 2(3x+1)(x+5)=0
=> 3x+1=0 hoặc x+5=0
=> x= -1/3 hoặc x=-5
Vậy...
a) (3x - 2)(4x + 5) = 0
⇔ 3x - 2 = 0 hoặc 4x + 5 = 0
1) 3x - 2 = 0 ⇔ 3x = 2 ⇔ x = 2/3
2) 4x + 5 = 0 ⇔ 4x = -5 ⇔ x = -5/4
Vậy phương trình có tập nghiệm S = {2/3;−5/4}
b) (2,3x - 6,9)(0,1x + 2) = 0
⇔ 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
1) 2,3x - 6,9 = 0 ⇔ 2,3x = 6,9 ⇔ x = 3
2) 0,1x + 2 = 0 ⇔ 0,1x = -2 ⇔ x = -20.
Vậy phương trình có tập hợp nghiệm S = {3;-20}
c) (4x + 2)(x2 + 1) = 0 ⇔ 4x + 2 = 0 hoặc x2 + 1 = 0
1) 4x + 2 = 0 ⇔ 4x = -2 ⇔ x = −1/2
2) x2 + 1 = 0 ⇔ x2 = -1 (vô lí vì x2 ≥ 0)
Vậy phương trình có tập hợp nghiệm S = {−1/2}
d) (2x + 7)(x - 5)(5x + 1) = 0
⇔ 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
1) 2x + 7 = 0 ⇔ 2x = -7 ⇔ x = −7/2
2) x - 5 = 0 ⇔ x = 5
3) 5x + 1 = 0 ⇔ 5x = -1 ⇔ x = −1/5
Vậy phương trình có tập nghiệm là S = {−7/2;5;−1/5}
1) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2-12\)
\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)
\(=x^3.\left(x+2\right)+4x.\left(x+2\right)-5.\left(x+2\right)=\left(x+2\right)\left(x^3+4x-5\right)\)
\(=\left(x+2\right)\left(x^3-x^2+x^2-x+5x-5\right)=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)
2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)
Đặt \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)
\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)
Thay a , ta có :
\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)
\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
1)x2-8x-9
= x^2 - 9x +x -9
= x(x+1) - 9 (x+1)
= (x-9) (x+1)
2)x2+3x-18
3)x3-5x2+4x
=x^3 - 4x^2 - x^2 + 4x
= x^2 (x-1) - 4x(x-1)
= (x^2 - 4x) (x-1)
= x(x-4)(x-1)
4)x3-11x2+30x
5)x3-7x-6
6)x16-64
\(=\left(x^8\right)^2-8^2\)
\(=\left(x^8-8\right)\left(x^8+8\right)\)
7)x3-5x2+8x-4
8)x2-3x+2
= x^2 - 2x - x +2
= x(x-1) -2(x-1)
= (x-2)(x-1)
1) \(\left(x-9\right)\left(x+1\right)\) 2) \(\left(x-3\right)\left(x+6\right)\) 3) \(x\left(x-4\right)\left(x-1\right)\)
4) \(x\left(x-6\right)\left(x-5\right)\) 5)\(\left(x-3\right)\left(x+1\right)\left(x+2\right)\) 6) ........
7) \(\left(x-1\right)\left(x-2\right)\left(x-2\right)\) 8) \(\left(x-2\right)\left(x-1\right)\)
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
a) \(2x^2+5x-18\)
\(=2x^2-4x+9x-18\)
\(=2x\left(x-2\right)+9\left(x-2\right)\)
\(=\left(x-2\right)\left(2x+9\right)\)
b) \(4x^2-17x+15\)
\(=4x^2-12x-5x+15\)
\(=4x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(4x-5\right)\)
c) \(-8x^2+10x+7\)
\(=-8x^2-4x+14x+7\)
\(=-4x\left(2x+1\right)+7\left(2x+1\right)\)
\(=\left(2x+1\right)\left(-4x+7\right)\)
d) \(7x^2-30x+8\)
\(=7x^2-28x-2x+8\)
\(=7x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(7x-2\right)\)
e) \(-x^3+11x^2-30x\)
\(=x\left(-x^2+11x-30\right)\)
\(=x\left(-x^2+5x+6x-30\right)\)
\(=x\left[-x\left(x-5\right)+6\left(x-5\right)\right]\)
\(=x\left(x-5\right)\left(-x+6\right)\)
a) 2x\(^2\) + 5x - 18 = 2x\(^2\) + 9x - 4x - 18 = x(2x + 9) - 2(2x + 9) = (x-2)(2x-9)
b) 4x\(^2\) - 17x - 15 = 4x\(^2\) + 20x - 3x - 15 = 4x(x + 5 ) - 3(x + 5) = (4x - 3 )(x + 5)
c) -8x\(^2\) + 10x + 7 = -8x\(^2\) + 14x - 4x + 7 =-2x(4x - 7) - (4x - 7) = (-2x - 1)(4x - 7)
d) 7x\(^2\) - 30x + 8 = 7x\(^2\) + 2x + 28x + 8 = x(7x + 2) + 4(7x + 2) = (x + 4)(7x + 2)
e) - x\(^3\) + 11x\(^2\) - 30x = -x(x\(^2\) - 11x + 30) = -x(x\(^2\) - 5x - 6x + 30) = -x\(\left[x\left(x-5\right)-6\left(x-5\right)\right]\) = -x(x-6)(x-5)