\(\frac{2}{3x+5}\)\(\frac{2}{3x+5}\)
K
Khách
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KN
5 tháng 3 2020
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
KN
5 tháng 3 2020
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
19 tháng 2 2021
ĐKXĐ: \(x\notin\left\{2;5\right\}\)
Ta có: \(\dfrac{3x}{x-2}-\dfrac{2}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\dfrac{3x}{\left(x-2\right)\left(x-5\right)}=0\)
Suy ra: \(3x^2-15x-2x+4+3x=0\)
\(\Leftrightarrow3x^2-14x+4=0\)
\(\Delta=196-4\cdot3\cdot4=196-48=148\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là
\(\left\{{}\begin{matrix}x_1=\dfrac{14-\sqrt{148}}{6}=\dfrac{7-\sqrt{37}}{3}\left(nhận\right)\\x_2=\dfrac{14+\sqrt{148}}{6}=\dfrac{7+\sqrt{37}}{3}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{7-\sqrt{37}}{3};\dfrac{7+\sqrt{37}}{3}\right\}\)
19 tháng 2 2021
Lớp 8 chưa học delta nên mk sẽ trình bày theo cách khác nha!
Rút gọn pt trên ta được: 3x2 - 14x + 4 = 0 (Theo kết quả của Nguyễn Lê Phước Thịnh CTV)
\(\Leftrightarrow\) 3(x2 - \(\dfrac{14}{3}\)x + \(\dfrac{4}{3}\)) = 0
\(\Leftrightarrow\) x2 - 2.\(\dfrac{14}{6}\)x + \(\dfrac{196}{36}\) - \(\dfrac{37}{9}\) = 0
\(\Leftrightarrow\) (x - \(\dfrac{14}{6}\))2 - \(\left(\dfrac{\sqrt{37}}{3}\right)^2\) = 0
\(\Leftrightarrow\) (x - \(\dfrac{14}{6}\) - \(\dfrac{\sqrt{37}}{3}\))(x - \(\dfrac{14}{6}\) + \(\dfrac{\sqrt{37}}{3}\)) = 0
\(\Leftrightarrow\) (x - \(\dfrac{7}{3}\) - \(\dfrac{\sqrt{37}}{3}\))(x - \(\dfrac{7}{3}\) + \(\dfrac{\sqrt{37}}{3}\)) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{7+\sqrt{37}}{3}\\x=\dfrac{7-\sqrt{37}}{3}\end{matrix}\right.\) (TM)
Vậy ...
Chúc bn học tốt!
26 tháng 9 2016
\(\frac{1}{5}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)
Đặt \(a=\left|3x-\frac{1}{5}\right|\) ta đc:
\(\frac{1}{5}-\frac{5}{2}a=\frac{2}{3}a-\frac{2}{3}\)
\(\Leftrightarrow-\frac{5}{2}a=\frac{2}{3}a-\frac{13}{15}\)
\(\Leftrightarrow-\frac{5a}{2}=\frac{10a}{15}-\frac{13}{15}\)
\(\Leftrightarrow-\frac{5a}{2}=\frac{10a-13}{15}\Rightarrow-5a\cdot15=2\left(10a-13\right)\)
\(\Leftrightarrow-75a=20a-26\)
\(\Leftrightarrow-95a=-26\Leftrightarrow a=\frac{26}{95}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{26}{95}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{5}=\frac{26}{95}\\3x-\frac{1}{5}=-\frac{26}{95}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x=\frac{9}{19}\\3x=-\frac{7}{95}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{19}\\x=-\frac{7}{285}\end{array}\right.\)
IN
4
30 tháng 6 2019
\(\frac{3x+5}{2}+\frac{3x+5}{4}+\frac{3x+5}{6}=\frac{3x+5}{8}\)
\(\frac{3x+5}{2}+\frac{3x+5}{4}+\frac{3x+5}{6}-\frac{3x+5}{8}=0\)
\(\left(3x-5\right)\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}-\frac{1}{8}\right)=0\)
\(\Rightarrow x=-\frac{5}{3}\)
NP
2
Y
8 tháng 4 2019
+ Đặt \(t=x^2-3x+3\) thì pt đã cho trở thành :
\(\frac{1}{t}+\frac{2}{t+1}=\frac{6}{t+2}\)
\(\Leftrightarrow\frac{t+1+2t}{t\left(t+1\right)}=\frac{6}{t+2}\) \(\Leftrightarrow\frac{3t+1}{t^2+t}=\frac{6}{t+2}\)
\(\Leftrightarrow\left(3t+1\right)\left(t+2\right)=6\left(t^2+t\right)\)
\(\Leftrightarrow3t^2+7t+2=6t^2+6t\)
\(\Leftrightarrow3t^2-t-2=0\)
\(\Leftrightarrow3t^2-3t+2t-2=0\)
\(\Leftrightarrow\left(3t+2\right)\left(t-1\right)=0\)
\(\Leftrightarrow t-1=0\) ( do \(3t+2=3x^2-9x+11\)\(=3\left(x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}+\frac{17}{12}\right)=3\left[\left(x-\frac{3}{2}\right)^2+\frac{17}{12}\right]>0\forall x\))
\(\Leftrightarrow x^2-3x+3=1\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{3}{4}=1\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{2}=\frac{1}{2}\\x-\frac{3}{2}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\left(TM\right)\)
Vậy tập nghiệm của pt đã cho là \(S=\left\{1;2\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2019
\(\frac{1}{x^2-3x+3}-1+\frac{2}{x^2-3x+4}-1+2-\frac{6}{x^2-3x+5}=0\)
\(\Leftrightarrow\frac{-x^2+3x-2}{x^2-3x+3}+\frac{-x^2+3x-2}{x^2-3x+4}-\frac{2\left(-x^2+3x-2\right)}{x^2-3x+5}=0\)
\(\Leftrightarrow\left(-x^2+3x-2\right)\left(\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+3x-2=0\left(1\right)\\\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}=0\)
Do \(\left\{{}\begin{matrix}\frac{1}{x^2-3x+3}>\frac{1}{x^2-3x+5}\\\frac{1}{x^2-3x+4}>\frac{1}{x^2-3x+5}\end{matrix}\right.\) \(\forall x\Rightarrow\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}>0\)
\(\Rightarrow\left(2\right)\) vô nghiệm
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