\(\frac{x}{3}=\frac{2x}{9}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
13 tháng 7 2017
Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)
\(\left(1\right)\Leftrightarrow\frac{x}{x-3}-\frac{2x^2+9}{\left(x-3\right)\left(2x+3\right)}=\frac{1}{2x+3}\)
ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)
\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)
\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)
Thỏa mãn ĐK
Các trường hợp khác làm tương tự
PN
22 tháng 5 2021
\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)
\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)
\(< =>4x-12-4x+2=10x+10+5\)
\(< =>10x=-10-10-5=-25\)
\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)
PN
22 tháng 5 2021
\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)
\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)
\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)
NH
2 tháng 3 2023
\(\dfrac{x+3}{x-3}-\dfrac{x}{x+3}=\dfrac{2x^2+9}{x^2-9}\left(x\ne-3;x\ne3\right)\\ < =>\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{2x^2+9}{\left(x-3\right)\left(x+3\right)}\)
suy ra
`x^2 +6x+9-x^2 +3x=2x^2 +9`
`<=> 2x^2 - x^2 +x^2 - 6x -3x +9 -9=0`
`<=> 2x^2 -9x=0`
`<=> x(2x-9)=0`
\(< =>\left[{}\begin{matrix}x=0\\2x-9=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
PA
30 tháng 11 2016
\(MTC:\left(x-3\right)^2\left(x^2+3x+9\right)\)
\(\frac{x}{x^3-27}=\frac{x}{\left(x-3\right)\left(x^2+3x+9\right)}=\frac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\frac{2x}{x^2-6x+9}=\frac{2x}{\left(x-3\right)^2}=\frac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\frac{1}{x^2+3x+9}=\frac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(MTC:2\left(x-1\right)\left(x+1\right)\)
\(\frac{x-1}{2x+2}=\frac{x-1}{2\left(x+1\right)}=\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(\frac{x+1}{2x-2}=\frac{x+1}{2\left(x-1\right)}=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(\frac{1}{1-x^2}=-\frac{1}{\left(x-1\right)\left(x+1\right)}=-\frac{2}{2\left(x-1\right)\left(x+1\right)}\)
\(MTC:2\left(x+1\right)\left(x^2-x+1\right)\)
\(\frac{1}{x^3+1}=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{2}{2\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{3}{2x+2}=\frac{3}{2\left(x+1\right)}=\frac{3\left(x^2-x+1\right)}{2\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2}{x^2-x+1}=\frac{4\left(x+1\right)}{2\left(x+1\right)\left(x^2-x+1\right)}\)
1 tháng 6 2020
1) Ta có: x-4=2x+4
\(\Leftrightarrow x-4-2x-4=0\)
\(\Leftrightarrow-x-8=0\)
\(\Leftrightarrow-x=8\)
hay x=-8
Vậy: S={8}
2) Ta có: \(\frac{2x-1}{2}-\frac{x}{3}=x-\frac{x}{6}\)
\(\Leftrightarrow\frac{3\left(2x-1\right)}{6}-\frac{2x}{6}=\frac{6x}{6}-\frac{x}{6}\)
\(\Leftrightarrow3\left(2x-1\right)-2x-6x+x=0\)
\(\Leftrightarrow6x-3-2x-6x+x=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: S={-3}
3) ĐKXĐ: \(x\notin\left\{\frac{-1}{2};3\right\}\)
Ta có: \(\frac{x+3}{2x+1}-\frac{x}{x-3}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-3\right)}{\left(2x+1\right)\left(x-3\right)}-\frac{x\left(2x+1\right)}{\left(x-3\right)\left(2x+1\right)}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)
Suy ra: \(x^2-9-\left(2x^2+x\right)-3x^2-x-9=0\)
\(\Leftrightarrow-2x^2-x-18-2x^2-x=0\)
\(\Leftrightarrow-4x^2-2x-18=0\)
\(\Leftrightarrow-4\left(x^2+\frac{1}{2}x+\frac{4}{5}\right)=0\)
\(\Leftrightarrow x^2+\frac{1}{2}x+\frac{4}{5}=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{1}{4}+\frac{1}{16}+\frac{59}{80}=0\)
\(\Leftrightarrow\left(x+\frac{1}{4}\right)^2+\frac{59}{80}=0\)(vô lý)
Vậy: S=\(\varnothing\)
4) Ta có: \(\frac{2x}{3}+\frac{2x-1}{6}=4-\frac{x}{3}\)
\(\Leftrightarrow\frac{4x}{6}+\frac{2x-1}{6}=\frac{24}{6}-\frac{2x}{6}\)
\(\Leftrightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow6x-1-24+2x=0\)
\(\Leftrightarrow8x-25=0\)
\(\Leftrightarrow8x=25\)
hay \(x=\frac{25}{8}\)
Vậy: \(S=\left\{\frac{25}{8}\right\}\)
8 tháng 3 2020
a) \(\frac{1}{x+3}+\frac{x}{x^2-6x+9}\left(x\ne\pm3\right)\)
\(=\frac{1}{x+3}+\frac{x}{\left(x-3\right)^2}=\frac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)^2}+\frac{x^2+3x}{\left(x+3\right)\left(x-3\right)^2}\)
\(=\frac{x^2-6x+9-x^2+3x}{\left(x-3\right)^2\left(x+3\right)}=\frac{-3x+9}{\left(x-3\right)^2\left(x+3\right)}=\frac{-3\left(x-3\right)}{\left(x-3\right)^2\left(x+3\right)}=\frac{-3}{\left(x-3\right)\left(x+3\right)}\)
8 tháng 3 2020
anhdun_•Ŧ๏áйツɦọς• giải a r nha , tớ giải b+c cho
\(b,\frac{2x}{x^2-9}-\frac{x-1}{x+3}\)
\(\frac{2x}{x^2-3^2}-\frac{x-1}{x+3}\)
\(\frac{2x}{\left(x+3\right)\left(x-3\right)}-\frac{x-1}{x+3}\)
\(\frac{2x-\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(\frac{2x-x^2+3x+x-3}{\left(x+3\right)\left(x-3\right)}\)
\(\frac{\left(2x+3x+x\right)-x^2-3}{\left(x+3\right)\left(x-3\right)}\)
\(\frac{6x-x^2-3}{\left(x+3\right)\left(x-3\right)}\)