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3 tháng 3 2022
a: \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\)
\(=\dfrac{5xy+y^3-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}=\dfrac{x^3+y^3}{x^2y^2}\)
b: \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x^2-3x}\)
DL
23 tháng 7 2019
a) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=\left(x^2-1\right)\left[\left(x^2-1\right)^2-\left(x^4+x^2+1\right)\right]\)
\(=\left(x^2-1\right)\left(x^4-2x^2+1-x^4-x^2-1\right)=\left(x^2-1\right)\left(-3x^2\right)\)
\(=-3x^4+3x^2=3\left(x^2-x^4\right)=3\left(x-x^2\right)\left(x+x^2\right)=\left(3x-3x^2\right)\left(x+x^2\right).\)
DL
23 tháng 7 2019
b)\(\left(x^4-3x^2+9\right)\left(x^2+3-\left(3+x^2\right)\right)^3=\left(x^4-3x^2+9\right).0^3=0\)
c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=\left(x-3\right)^3-\left(x^3-3^3\right)+6\left(x^2+2x+1\right)\)
\(=\left(x-3\right)^3-\left[\left(x-3\right)^3+3.x.3.\left(x-3\right)\right]+6x^2+12x+6\)
\(=6x^2+12x+6-9x\left(x-3\right)=6x^2+12x+6-9x^2+27x\)
\(=39x-3x^2+6=3\left(13x-x^2+2\right).\)
AH
Akai Haruma
Giáo viên
30 tháng 11 2023
Lời giải:
a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$
$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$
$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$
$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$
$\Leftrightarrow -x+2=0$
$\Leftrightarrow x=2$
b.
$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$
$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$
$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$
$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$
$\Leftrightarrow -x+10=0\Leftrightarrow x=10$
AH
Akai Haruma
Giáo viên
30 tháng 11 2023
c.
$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$
$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$
$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$
$\Leftrightarrow 3x-28=25$
$\Leftrightarrow x=\frac{53}{3}$
d.
$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$
$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$
$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$
$\Leftrgihtarrow 24x=22$
$\Leftrightarrow x=\frac{11}{12}$
HT
21 tháng 7 2017
1)
a) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow\left(6x-3\right)\left(3x-1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-6x-9x+3-18x^2 +29x-3=0\)
\(\Leftrightarrow14x=0\)
\(\Rightarrow x=0\)
b) \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+10\)
\(\Leftrightarrow10x-16-12x+15=12x-16+10\)
\(\Leftrightarrow10x-12x+15=12x+10\)
\(\Leftrightarrow-2x+15=12x+10\)
\(\Leftrightarrow-2x-12x=10-15\)
\(\Leftrightarrow-14x=-5\)
\(\Rightarrow x=\dfrac{5}{14}\)
c) \(\left(3x-2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=0\)
\(\Leftrightarrow6x^2+27x-4x-18-\left(6x^2+x+12x+2\right)=0\)
\(\Leftrightarrow6x^2+27x-4x-18-6x^2-13x-2=0\)
\(\Leftrightarrow10x-20=0\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
NL
1
TH
31 tháng 12 2020
(\(3+\dfrac{x}{3-x}+\dfrac{2x}{3+x}-\dfrac{4x^2-3x-9}{x^2-9}\) ):\(\left(\dfrac{2}{3-x}-\dfrac{x-1}{3x-x^2}\right)\)\(=\left(\dfrac{3x^2-27}{\left(x-3\right)\left(x+3\right)}+\dfrac{-x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2-3x-9}{\left(x-3\right)\left(x+3\right)}\right)\)\(:\left(\dfrac{2x}{x\left(3-x\right)}-\dfrac{x-1}{x\left(3-x\right)}\right)\)
\(=\dfrac{3x^2-27-x^2-3x+2x^2-6x-4x^2+3x+9}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\)
\(=\dfrac{-6x-18}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) \(=\dfrac{-6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\)
\(=\dfrac{6}{3-x}.\dfrac{x\left(x-3\right)}{x+1}\) \(=\dfrac{6x}{x+1}\)
LT
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 12 2023
Em ơi em đăng câu hỏi thì dùng công thức toán học để viết chữ viết thế này ai hiểu đúng được em?
13 tháng 8 2021
a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)
\(\Leftrightarrow39x=-34\)
hay \(x=-\dfrac{34}{39}\)
b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x=28\)
hay x=7
c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow x=2\)
bn có chép thiếu hay sai j k?
Mik thấy đề bài k đc hợp lí cho lắm
Đề là rút gọn chăng ?
\(\frac{x}{x+3}+\frac{3-x}{x+3}.\frac{x^2+3x+9}{x^2-9}\)
\(\frac{x}{x+3}+\frac{\left(3-x\right)\left(x^2+3x+9\right)}{\left(x+3\right)^2\left(x-3\right)}\)
\(\frac{x\left(x+3\right)^2\left(x-3\right)}{\left(x+3\right)^3\left(x-3\right)}+\frac{\left(3-x\right)\left(x^2+3x+9\right)\left(x+3\right)}{\left(x+3\right)^3\left(x-3\right)}\)
\(x\left(x+3\right)^2\left(x-3\right)+\left(3-x\right)\left(x^2+3x+9\right)\left(x+3\right)\)
Cậu làm tiếp .