\(\left(2x^2+2x+1\right):\left(x-3\right)dư4\)

\(\left(2x^3-3x...">

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a: \(x^3+x^2-2x+a⋮x+1\)

\(\Leftrightarrow x^3+x^2-2x-2+a+2⋮x+1\)

=>a+2=0

hay a=-2

b: \(2x^3-4x^2-3a⋮2x-3\)

\(\Leftrightarrow2x^3-3x^2-x^2+1.5x-1.5x+2.25-3a-2.25⋮2x-3\)=>-3a-2,25=0

=>-3a=2,25

hay a=-0,75

c: \(4x^4+3x^2-ax+3⋮x+3\)

\(\Leftrightarrow4x^4+12x^3-12x^3-36x^2+39x^2+117x-ax+3⋮x+3\)

\(\Leftrightarrow-ax+3⋮x+3\)

\(\Leftrightarrow-ax-3a+3+3a⋮x+3\)

=>3a+3=0

hay a=-1

giúp mk với tứ tư mk phải nộp rùi bài 1: a, \(2x\left(3x^2-5x+3\right)\) b, \(-2x\left(x^2+5x-3\right)\) c, \(\dfrac{-1}{2}x\left(2x^3-4x+3\right)\) bài 2: a,\(\left(2x-1\right).\left(x^2-5-4\right)\) b,\(-\left(5x-4\right).\left(2x+3\right)\) c,\(\left(2x-y\right).\left(4x^2-2xy+y^2\right)\) d,\(\left(3x-4\right).\left(x+4\right).\left(5-x\right).\left(2x^2+3x-1\right)\) e,\(7\left(x-4\right)-\left(7x+3\right).\left(2x^2-x+4\right)\) bài 3: c/m rằng gtri của...
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giúp mk với tứ tư mk phải nộp rùi

bài 1:

a, \(2x\left(3x^2-5x+3\right)\)

b, \(-2x\left(x^2+5x-3\right)\)

c, \(\dfrac{-1}{2}x\left(2x^3-4x+3\right)\)

bài 2:

a,\(\left(2x-1\right).\left(x^2-5-4\right)\)

b,\(-\left(5x-4\right).\left(2x+3\right)\)

c,\(\left(2x-y\right).\left(4x^2-2xy+y^2\right)\)

d,\(\left(3x-4\right).\left(x+4\right).\left(5-x\right).\left(2x^2+3x-1\right)\)

e,\(7\left(x-4\right)-\left(7x+3\right).\left(2x^2-x+4\right)\)

bài 3:

c/m rằng gtri của biểu thức ko phụ thuộc vào gtri của biến

a,\(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)

b,\(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)

bài 4 :tìm x biết

a, \(3x+2\left(5-x\right)=0\)

b,\(x\left(2x-1\right).\left(x+5\right)-\left(2x^2+1\right).\left(x+4,5\right)=3,5\)

c,\(3x^2-3x\left(x-2\right)=36\)

d,\(\left(3x^2-x+1\right).\left(x-1\right)+x^2.\left(4-3x\right)=\dfrac{5}{2}\)

4
11 tháng 12 2017

1,

a,\(2x\left(3x^2-5x+3\right)\)

\(=6x^3-10x^2+6x\)

b,\(-2x\left(x^2+5x-3\right)\)

\(=-2x^3-10x^2+6x\)

c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)

\(=-x^4+2x^2-\dfrac{3}{2}x\)

Bài 2:

a) \(\left(2x-1\right)\left(x^2-5-4\right)\)

\(=\left(2x-1\right)\left(x^2-9\right)\)

\(=2x^3-18x-x^2+9\)

b) \(-\left(5x-4\right)\left(2x+3\right)\)

\(=-\left(10x^2+15x-8x-12\right)\)

\(=-10x^2-7x+12\)

c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)

\(=8x^3-y^3\)

20 tháng 11 2017

\(\text{a) }3x^2y^2:x^2=3y^2\)

\(\text{b) }\left(x^5+4x^3-6x^2\right):4x^2\\ =\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)

\(\text{c) }\left(x^3-8\right):\left(x^2+2x+4\right)\\ =\left(x-2\right)\left(x^2+2x+4\right):\left(x^2+2x+4\right)\\ =x-2\)

\(\text{d) }\left(3x^2-6x\right):\left(2-x\right)\\ =3x\left(x-2\right):\left(2-x\right)\\ =-3x\left(2-x\right):\left(2-x\right)\\ =-3x\)

\(\text{e) }\left(x^3+2x^2-2x-1\right):\left(x^2+3x+1\right)\\ =\left(x^3+3x^2-x^2+x-3x-1\right):\left(x^2+3x+1\right)\\ =\left[\left(x^3+3x^2+x\right)-\left(x^2+3x+1\right)\right]:\left(x^2+3x+1\right)\\ =\left[x\left(x^2+3x+1\right)-\left(x^2+3x-1\right)\right]:\left(x^2+3x+1\right)\\ =\left(x-1\right)\left(x^2+3x+1\right):\left(x^2+3x+1\right)\\ =x-1\)

31 tháng 12 2018

a) 3x2y2 : x2 = 3y2

b)( x5 + 4x3 - 6x2 ) : 4x2

=\(\dfrac{1}{4}\)x3+ x - \(\dfrac{3}{2}\)

8 tháng 2 2018

a. \(9\left(x+2\right)-3\left(x+2\right)=0\)

\(\Leftrightarrow9x+18-3x-6=0\)

\(\Leftrightarrow6x+12=0\)

\(\Leftrightarrow x=-2\)

e. \(\left(2x-1\right)^2-45=0\)

\(\Leftrightarrow4x^2-2x+1-45=0\)

\(\Leftrightarrow4x^2-2x-44=0\)

Đến đó tự giải tiếp nha!

c. \(2\left(2x-5\right)-3x=0\)

\(\Leftrightarrow4x-10-3x=0\)

\(\Leftrightarrow x-10=0\)

\(\Leftrightarrow x=10\)

g. \(2x^2-6x=0\)

\(\Leftrightarrow2x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

20 tháng 2 2018

sao làm nhung cau de the

a: \(=2x^2-x+5\)

b: \(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)

c: \(=-x^3+\dfrac{3}{2}-2x\)

d: \(=-2x^2+4xy-6y^2\)

e: \(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)

16 tháng 6 2018

\(a,x^2\left(x-2x^3\right)\)

\(=x^3-2x^5\)

\(b,\left(x-2\right)\left(x-x^2+4\right)\)

\(=x^2-x^3+4x-2x+2x^2-8\)

\(=3x^2-x^3+2x-8\)

\(c,\left(x^2-1\right)\left(x^2+2x\right)\)

\(=x^4+2x^3-x^2-2x\)

\(d,\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)

\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)

\(=\left(6x^2+x-2\right)\left(3-x\right)\)

\(=18x^2+3x-6-6x^3-x^2+2x\)

\(=17x^2+5x-6-6x^3-x^2\)

\(e,\left(x+3\right)\left(x^2+3x-5\right)\)

\(=x^3+3x^2-5x+3x^2+9x-15\)

\(=x^3+6x^2+4x-15\)

\(f,\left(xy-2\right)\left(x^3-2x-6\right)\)

\(=x^4y-2x^2y-6xy-2x^3+4x-12\)

\(g,\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)

\(=20x^5-4x^4+8x^3-12x^2-5x^4+x^3-2x^2+3x+10x^3-2x^2+4x-6\)

\(=20x^5-9x^4+19x^3-16x^2+7x-6\)

16 tháng 6 2018

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