LT
3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
R
2
5 tháng 1 2022
\(\dfrac{5x+2}{x^2-4}+\dfrac{x-5}{x-2}=\dfrac{5x+2+x^2-3x-10}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+4}{x+2}\\ \left(x+4\right)^2-\left(x+3\right)\left(x-2\right)=-13\\ \Leftrightarrow x^2+8x+16-x^2+x+6=-13\\ \Leftrightarrow9x=-13-22=-35\\ \Leftrightarrow x=-\dfrac{35}{9}\)
10 tháng 1 2023
a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)
b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)
c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)
d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)
CV
7 tháng 12 2021
Đáp án:
a.3x³−5x²+7xa.3x³−5x²+7x
b.−4x²y−10x²y+2xyb.−4x²y−10x²y+2xy
c.−x³+2x²+29x+20c.−x³+2x²+29x+20
d.2x⁴−3x³+2x²+3x−4d.2x⁴−3x³+2x²+3x−4
e.x²−4y²e.x²−4y²
h.2x²−6x+13h.2x²−6x+13
g.3xy⁴−12y²+2x²yg.3xy⁴−12y²+2x²y
f.−2x²y³+y−3f.−2x²y³+y−3
Giải thích các bước giải:
a.3x.(x²−5x+7)a.3x.(x²−5x+7)
=3x³−5x²+7x=3x³−5x²+7x
b.−2xy.(2x³+5x−1)b.−2xy.(2x³+5x−1)
=−4x⁴y−10xy²+2xy=−4x⁴y−10xy²+2xy
c.(x+4).(−x²+6x+5)c.(x+4).(−x²+6x+5)
=−x³+6x²+5x−4x²+24x+20=−x³+6x²+5x−4x²+24x+20
=−x³+2x²+29x+20=−x³+2x²+29x+20
d.(x²−1).(2x²−3x+4)d.(x²−1).(2x²−3x+4)
=2x⁴−3x³+4x²−2x²+3x−4=2x⁴−3x³+4x²−2x²+3x−4
=2x⁴−3x³+2x2+3x−4=2x⁴−3x³+2x2+3x−4
e.(x+2y).(x−2y)e.(x+2y).(x−2y)
=x²−(2y)²=x²−(2y)²
=x²−4y²=x²−4y²
h.(3x−1)²−7(x²+2)h.(3x−1)²−7(x²+2)
=9x²−6x+1−7x²−14=9x²−6x+1−7x²−14
=2x²−6x+13=2x²−6x+13
g.(6x²g.(6x²y⁵−xy³+4x³y²):2xy−xy³+4x³y²):2xy
=3xy⁴−12y²+2x²y=3xy⁴−12y²+2x²y
f.(−12x³y⁴+6xy²−18xy):6xyf.(−12x³y⁴+6xy²−18xy):6xy
=−2x³y³+y−3
MN
3 tháng 3 2020
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
\(=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3x-9-x^2}{3x\left(x+3\right)}\)
\(=\frac{\left(9+x^2-3x\right)\left(x+3\right)3x}{x\left(x-3\right)\left(x+3\right)\left(3x-9-x^2\right)}\)
\(=\frac{-3}{x-3}\)
AH
Akai Haruma
Giáo viên
6 tháng 7 2024
1.
$2x^3-21x^2+67x-60=2x^2(x-5)-11x(x-5)+12(x-5)$
$=(x-5)(2x^2-11x+12)$
$\Rightarrow (2x^3-21x^2+67x-60):(x-5)=2x^2-11x+12$
AH
Akai Haruma
Giáo viên
6 tháng 7 2024
2.
$x^4+2x^3+x-25=x^2(x^2+5)+2x(x^2+5)-5x^2-9x-25$
$=x^2(x^2+5)+2x(x^2+5)-5(x^2+5)-9x=(x^2+5)(x^2+2x-5)-9x$
$\Rightarrow (x^4+2x^3+x-25):(x^2+5)=x^2+2x-5$ và dư $-9x$
17 tháng 12 2022
a: \(=\dfrac{3x+6-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+10}{x^2-4}\)
b: \(=\dfrac{10x+15+4x-6+2x+5}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x+14}{\left(2x-3\right)\left(2x+3\right)}\)
22 tháng 6 2019
a) (x4 + 2x3 + x -25):(x2 +5)
= x2 +2x - 5 ( dư - 9x )
b) (27x3 - 8) : (6x + 9x2 + 4)
= 3x - 2
cau hoi nay de lam ma
\(\frac{4}{x-3}+\frac{5}{x+3}-\frac{13-9x^2}{x^2-9}\)
ĐKXĐ : \(x\ne\pm3\)
\(=\frac{4}{x-3}+\frac{5}{x+3}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{5\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}+\frac{5x-15}{\left(x+3\right)\left(x-3\right)}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4x+12+5x-15-13+9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{9x^2+9x-16}{\left(x+3\right)\left(x-3\right)}=\frac{9x^2+9x-16}{x^2-9}\)