Thực hiên phép chia sau : ( x^2 + 4x - 7 ) : ( x - 2...
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| Thực hiên phép chia sau : ( x^2 + 4x - 7 ) : ( x - 2 ) |
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1
Những câu hỏi liên quan
H
2 tháng 1 2023
\(a,đk:x\ne0;4;1\)
\(\dfrac{x-1}{x^2-5x+4}-\dfrac{4}{x^2-4x}\\ =\dfrac{x-1}{\left(x-1\right)\left(x-4\right)}-\dfrac{4}{x\left(x-4\right)}\\ =\dfrac{x\left(x-1\right)}{x\left(x-1\right)\left(x-4\right)}-\dfrac{4\left(x-1\right)}{x\left(x-1\right)\left(x-4\right)}\\ =\dfrac{x^2-x-4x+4}{x\left(x-1\right)\left(x-4\right)}\\ =\dfrac{x^2-5x+4}{x.\left(x-1\right)\left(x-4\right)}=\dfrac{\left(x-1\right)\left(x-4\right)}{x.\left(x-1\right)\left(x-4\right)}=\dfrac{1}{x}\)
\(đk:x\ne-2;1\)
\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{x\left(7x-7\right)}{\left(x+2\right)\left(7x-7\right)}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{7x^2-7x+7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{7x^2-16}{\left(x+2\right)\left(7x-7\right)}\)
NH
2 tháng 1 2023
a)
\(\dfrac{x-1}{x^2-5x+4}-\dfrac{4}{x^2-4x}\) \(ĐKXĐ:x\ne0;x\ne4;x\ne1\)
\(=\dfrac{x-1}{x^2-4x-x+4}-\dfrac{4}{x\left(x-4\right)}\)
\(=\dfrac{x-1}{x\left(x-4\right)-\left(x-4\right)}-\dfrac{4}{x\left(x-4\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x-4\right)}-\dfrac{4}{x\left(x-4\right)}\)
\(=\dfrac{x^2-x}{x\left(x-1\right)\left(x-4\right)}-\dfrac{4\left(x-1\right)}{x\left(x-1\right)\left(x-4\right)}\)
\(=\dfrac{x^2-x-4x+4}{x\left(x-1\right)\left(x-4\right)}\)
\(=\dfrac{x\left(x-1\right)-4\left(x-1\right)}{x\left(x-1\right)\left(x-4\right)}\)
\(=\dfrac{\left(x-1\right)\left(x-4\right)}{x\left(x-1\right)\left(x-4\right)}\\ =\dfrac{1}{x}\)
b)
\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\) \(ĐKXĐ:x\ne-2;x\ne1\)
\(=\dfrac{x\left(7x-7\right)}{\left(x+2\right)\left(7x-7\right)}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\)
\(=\dfrac{7x^2-7x+7x-16}{\left(x+2\right)\left(7x-7\right)}\)
\(=\dfrac{7x^2-16}{\left(x+2\right)\left(7x-7\right)}\)
HT
1
AH
Akai Haruma
Giáo viên
14 tháng 11 2021
Lời giải:
\(17x^2-2x^3-3x^4-4x-5=-3x^4-2x^3+17x^2-4x-5\)
\(=-3x^2(x^2+x-5)+x^3+2x^2-4x-5\)
\(=-3x^2(x^2+x-5)+x(x^2+x-5)+(x^2+x-5)\)
\(=(x^2+x-5)(-3x^2+x+1)\)
Do đó: $(17x^2-2x^3-3x^4-4x-5):(x^2+x-5)=(-3x^2+x+1)$
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
23 tháng 9 2023
\(a,\dfrac{x^2+2}{x^3+1}-\dfrac{1}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ =\dfrac{x^2+2-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\\ c,\dfrac{1}{2-2x}-\dfrac{3}{2+2x}+\dfrac{2x}{x^2-1}\\ =\dfrac{-1}{2\left(x-1\right)}-\dfrac{3}{2\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm1\right)\\ =\dfrac{-1\left(x+1\right)-3\left(x-1\right)+2x.2}{2\left(x+1\right)\left(x-1\right)}\\ =\dfrac{-x-1-3x+3+4x}{2\left(x+1\right)\left(x-1\right)}=\dfrac{2}{2\left(x+1\right)\left(x-1\right)}=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
4 tháng 12 2016
a)\(dk,x\ne7;x\ne0\)
\(\frac{4x+13}{5x\left(x-7\right)}-\frac{x-48}{5x\left(7-x\right)}=\frac{4x+13}{5x\left(x-7\right)}+\frac{x-48}{5x\left(x-7\right)}=\frac{\left(4x+13\right)+\left(x-48\right)}{5x\left(x-7\right)}\\ \)
\(=\frac{5x-35}{5x\left(x-7\right)}=\frac{5\left(x-7\right)}{5x\left(x-7\right)}=\frac{1}{x}\)
b)
\(\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{1-\left(5x\right)^2}=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{\left(1-5x\right)\left(1+5x\right)}\)
\(\frac{1+5x}{x\left(1-5x\right)\left(1+5x\right)}+\frac{x\left(25x-15\right)}{x\left(1-5x\right)\left(1+5x\right)}=\frac{25x^2-15x+5x+1}{x\left(1-5x\right)\left(1+5x\right)}=\frac{25x^2-10x+1}{x\left(1-5x\right)\left(1+5x\right)}\)
HQ
Hà Quang Minh
Giáo viên
20 tháng 9 2023
\(\frac{{9{x^2} + 5x + x}}{{3x}} = \frac{{9{x^2} + 6x}}{{3x}} = \frac{{9{x^2}}}{{3x}} + \frac{{6x}}{{3x}} = 3x + 2\)
\(\frac{{2{x^2} - 3x - 2}}{{2 - x}} = \frac{{2{x^2} - 3x - 2}}{{ - x + 2}} = - 2x - 1\)
PT
1
CM
15 tháng 1 2018
2 x 3 x 4 x 8 x 50 x 25 x 125
= 3 x 2 x 4 x 50 x 8 x 25 x 125
= 3 x (2 x 50) x (4 x 25) x (8 x 125)
= 30 000 000.
BM
1
10 tháng 12 2021
a) (x + 3y) (2x2y - 6xy2)
= (x + 3y) + 2xy (x - 3y)
= 2xy [(x + 3y) (x - 3y)]
= 2xy (x2 - 3y2)
b) (6x5y2 - 9x4y3 + 15x3y4) : 3x3y2
= (6x5y2 : 3x3y2) + (-9x4y3 : 3x3y2) + (15x3y4 : 3x3y2)
= [(6 : 3) (x5 : x3) (y2 : y2)] + [(-9 : 3) (x4 : x3) (y3 : y2)] + [(15 : 3) (x3 : x3) (y4 : y2)]
= 2x2 + (-3xy) + 5y2
= 2x2 - 3xy + 5y2
EC
5 tháng 9 2016
a)\(\left(\frac{1}{2}x-1\right)\left(2x-3\right)=x^2-\frac{3}{2}x-2x+3=x^2-\frac{7}{2}x+3\)
b)\(\left(x-7\right)\left(x-5\right)=x^2-5x-7x+5=x^2-12x+5\)
c)\(\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(4x-1\right)=\left(x^2-\frac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\frac{1}{4}\)
x^2+4x-7=(x-2)(x+6)+5
=> x^2+4x-7 chia x-2 được thương x+6 dư 5