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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)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=\left(x-3\right)^3-\left(x^3-3^3\right)+6\left(x^2+2x+1\right)\)
\(=\left(x-3\right)^3-\left[\left(x-3\right)^3+3.x.3.\left(x-3\right)\right]+6x^2+12x+6\)
\(=6x^2+12x+6-9x\left(x-3\right)=6x^2+12x+6-9x^2+27x\)
\(=39x-3x^2+6=3\left(13x-x^2+2\right).\)
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) \(\left(3x-2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=0\)
\(\Leftrightarrow6x^2+27x-4x-18-\left(6x^2+x+12x+2\right)=0\)
\(\Leftrightarrow6x^2+27x-4x-18-6x^2-13x-2=0\)
\(\Leftrightarrow10x-20=0\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
H
4 tháng 8 2019
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
V
1
PN
22 tháng 5 2020
Sửa đề thành rút gọn phân số nhé :
\(\frac{\left(\frac{x}{x+3+3}-\frac{x}{3.x^2+3x}+\frac{9}{x^2-9}\right):3}{x+3}\)
\(=\frac{\frac{x}{3\left(x+6\right)}-\frac{x}{9x^2+9x}+\frac{9}{\left(x^2-9\right).3}}{x+3}\)
\(=\frac{\frac{x}{3x+18}-9x-9+\frac{3}{x^2-9}}{x+3}\)
\(=\frac{\frac{x-9x\left(3x+18\right)}{3x+18}+\frac{3-9\left(x^2-9\right)}{x^2-9}}{x+3}\)
\(=\frac{\frac{x-27x^2-162x}{3x+18}+\frac{3-9x^2+81}{x^2-9}}{x+3}\)
\(=\frac{\frac{27x^2-161x}{3x+18}+\frac{-9x^2+84}{x^2-9}}{x+3}\)
\(=\frac{27x^2-161x}{\left(3x+18\right):\left(x+3\right)}+\frac{-9x^2+84}{\left(x^2-9\right):\left(x+3\right)}\)
Đến đây thì dễ r ha
24 tháng 7 2017
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
N
0
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 .