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a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2
b: =x^3+3x^2-2x-3x^2-9x+6
=x^3-11x+6
c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)
\(=2x^2-3x-1+\dfrac{5}{2x+1}\)
a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)
\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)
\(=2x^5-16x^3-2x^5-x^3\)
\(=-17x^3\)
b) \(\left(x+3\right)\left(x^2+3x-2\right)\)
\(=x^3+3x^2-2x+3x^2+9x-6\)
\(=x^3+6x^2+7x-6\)
c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)
\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)
Bài 1:
a. $2x^3+3x^2-2x=2x(x^2+3x-2)=2x[(x^2-2x)+(x-2)]$
$=2x[x(x-2)+(x-2)]=2x(x-2)(x+1)$
b.
$(x+1)(x+2)(x+3)(x+4)-24$
$=[(x+1)(x+4)][(x+2)(x+3)]-24$
$=(x^2+5x+4)(x^2+5x+6)-24$
$=a(a+2)-24$ (đặt $x^2+5x+4=a$)
$=a^2+2a-24=(a^2-4a)+(6a-24)$
$=a(a-4)+6(a-4)=(a-4)(a+6)=(x^2+5x)(x^2+5x+10)$
$=x(x+5)(x^2+5x+10)$
Bài 2:
a. ĐKXĐ: $x\neq 3; 4$
\(A=\frac{2x+1-(x+3)(x-3)+(2x-1)(x-4)}{(x-3)(x-4)}\\ =\frac{2x+1-(x^2-9)+(2x^2-9x+4)}{(x-3)(x-4)}\\ =\frac{x^2-7x+14}{(x-3)(x-4)}\)
b. $x^2+20=9x$
$\Leftrightarrow x^2-9x+20=0$
$\Leftrightarrow (x-4)(x-5)=0$
$\Rightarrow x=5$ (do $x\neq 4$)
Khi đó: $A=\frac{5^2-7.5+14}{(5-4)(5-3)}=2$
a: =(x-3)(2x+5)
b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)
=>(x-2)(5-x)=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
a) (x + 2)(x2 + 3x + 1)
= x.x2 + x.3x + x.1 + 2.x2 + 2.3x + 2.1
= x3 + 3x2 + x + 2x2 + 6x + 2
= x3 + 5x2 + 7x + 2
b) (2x3 + 10x2 + 9x + 4) : (x + 4)
= (2x3 + 8x2 + 2x2 + 8x + x + 4) : (x + 4)
= [(2x3 + 8x2) + (2x2 + 8x) + (x + 4)] : (x + 4)
= [2x2(x + 4) + 2x(x + 4) + (x + 4)] : (x + 4)
= (x + 4)(2x2 + 2x + 1) : (x + 4)
= 2x2 + 2x + 1
\(a,x^2\left(x-2x^3\right)\)
\(=x^3-2x^5\)
\(b,\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
\(c,\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
\(d,\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2+3x-6-6x^3-x^2+2x\)
\(=17x^2+5x-6-6x^3-x^2\)
\(e,\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
\(f,\left(xy-2\right)\left(x^3-2x-6\right)\)
\(=x^4y-2x^2y-6xy-2x^3+4x-12\)
\(g,\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)
\(=20x^5-4x^4+8x^3-12x^2-5x^4+x^3-2x^2+3x+10x^3-2x^2+4x-6\)
\(=20x^5-9x^4+19x^3-16x^2+7x-6\)
a. x2(x−2x3)= x3-2x5
b. (x−2)(x−x2+4)= x2-x3+4x-2x+2x2-8= -x3+3x2+2x-8
c. (x2−1)(x2+2x)= x4+2x3-x2-2x
d. (2x−1)(3x+2)(3−x) = (6x2+x-2)(3-x)=18x2-6x3+3x-x2-6+2x =-6x3+17x2+5x-6
e. (x+3)(x2+3x−5)= x3+3x2-5x+3x2+9x-15= x3+6x2+4x-15
f. (xy−2)(x3−2x−6)= x4y-2x2y-6xy-2x3+4x+12
g. (5x3−x2+2x−3)(4x2−x+2)= 20x5-9x4+19x3-12x2+7x-6
a) \(x-2=\left(x-2\right)^2\)
\(\left(x-2\right)^2-\left(x-2\right)=0\)
\(\left(x-2\right)\left(x-2-1\right)=0\)
\(\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow x-2=0\) hoặc \(x-3=0\)
*) \(x-2=0\)
\(x=2\)
*) \(x-3=0\)
\(x=3\)
Vậy \(x=2;x=3\)
b) \(x+5=2\left(x+5\right)^2\)
\(2\left(x+5\right)^2-\left(x+5\right)=0\)
\(\left(x+5\right)\left[2\left(x+5\right)-1\right]=0\)
\(\left(x+5\right)\left(2x+10-1\right)=0\)
\(\left(x+5\right)\left(2x+9\right)=0\)
\(\Rightarrow x+5=0\) hoặc \(2x+9=0\)
*) \(x+5=0\)
\(x=-5\)
*) \(2x+9=0\)
\(2x=-9\)
\(x=-\dfrac{9}{2}\)
Vậy \(x=-5;x=-\dfrac{9}{2}\)
c) \(\left(x^2+1\right)\left(2x-1\right)+2x=1\)
\(\left(x^2+1\right)\left(2x-1\right)+2x-1=0\)
\(\left(x^2+1\right)\left(2x-1\right)+\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(x^2+1+1\right)=0\)
\(\left(2x-1\right)\left(x^2+2\right)=0\)
\(\Rightarrow2x-1=0\) hoặc \(x^2+2=0\)
*) \(2x-1=0\)
\(2x=1\)
\(x=\dfrac{1}{2}\)
*) \(x^2+2=0\)
\(x^2=-2\) (vô lí)
Vậy \(x=\dfrac{1}{2}\)
d) Sửa đề:
\(\left(x^2+3\right)\left(x+1\right)+x=-1\)
\(\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\)
\(\left(x+1\right)\left(x^2+3+1\right)=0\)
\(\left(x+1\right)\left(x^2+4\right)=0\)
\(\Rightarrow x+1=0\) hoặc \(x^2+4=0\)
*) \(x+1=0\)
\(x=-1\)
*) \(x^2+4=0\)
\(x^2=-4\) (vô lí)
Vậy \(x=-1\)
a) (x2 – 2x+ 1)(x – 1)
= x2 . x + x2.(-1) + (-2x). x + (-2x). (-1) + 1 . x + 1 . (-1)
= x3 - x2 - 2x2 + 2x + x – 1
= x3 - 3x2 + 3x – 1
b) (x3 – 2x2 + x -1)(5 – x)
= x3 . 5 + x3 . (-x) + (-2 x2) . 5 + (-2x2)(-x) + x . 5 + x(-x) + (-1) . 5 + (-1) . (-x)
= 5 x3 – x4 – 10x2 + 2x3 +5x – x2 – 5 + x
= - x4 + 7x3 – 11x2+ 6x - 5.
Suy ra kết quả của phép nhan:
(x3 – 2x2 + x -1)(x - 5) = (x3 – 2x2 + x -1)(-(5 - x))
= - (x3 – 2x2 + x -1)(5 – x)
= - (- x4 + 7x3 – 11x2+ 6x -5)
= x4 - 7x3 + 11x2- 6x + 5
a) (x2 – 2x+ 1)(x – 1)
= x2 . x + x2.(-1) + (-2x). x + (-2x). (-1) + 1 . x + 1 . (-1)
= x3 - x2 - 2x2 + 2x + x – 1
= x3 - 3x2 + 3x – 1
b) (x3 – 2x2 + x -1)(5 – x)
= x3 . 5 + x3 . (-x) + (-2 x2) . 5 + (-2x2)(-x) + x . 5 + x(-x) + (-1) . 5 + (-1) . (-x)
= 5 x3 – x4 – 10x2 + 2x3 +5x – x2 – 5 + x
= - x4 + 7x3 – 11x2+ 6x - 5.
Suy ra kết quả của phép nhan:
(x3 – 2x2 + x -1)(x - 5) = (x3 – 2x2 + x -1)(-(5 - x))
= - (x3 – 2x2 + x -1)(5 – x)
= - (- x4 + 7x3 – 11x2+ 6x -5)
= x4 - 7x3 + 11x2- 6x + 5