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a)
\(\dfrac{x-3}{5}+\dfrac{1-2x}{3}=6\\ < =>3x-9+5-10x=90\)
\(< =>3x-10x=90+9-5\\ < =>-7x=94\\ < =>x=-\dfrac{94}{7}\)
b)
\(\left(2x-3\right)\left(x^2+1\right)=0\\ < =>\left[{}\begin{matrix}2x-3=0\\x^2+1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x^2=-1\left(voli\right)\end{matrix}\right.\\ < =>x=\dfrac{3}{2}\)
c)
\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\left(x\ne-1;x\ne2\right)\)
suy ra: \(2\left(x-2\right)-x-1=3x-11\)
\(< =>2x-4-x-1-3x+11=0\)
\(< =>2x-x-3x=4+1-11\\ < =>-2x=-6\\ < =>x=3\left(tm\right)\)
a) \(\dfrac{x-3}{5}+\dfrac{1-2x}{3}=6\)
\(\Leftrightarrow3\left(x-3\right)+5\left(1-2x\right)=90\)
\(\Leftrightarrow-4-7x=90\)
\(\Leftrightarrow x=-\dfrac{94}{7}\)
b) \(\left(2x-3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x-3=0\) (Vì \(x^2+1>0\))
\(\Leftrightarrow x=\dfrac{3}{2}\)
c) \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\left(Đk:x\ne-1;x\ne2\right)\)
\(\Leftrightarrow2\left(x-2\right)-\left(x+1\right)=3x-11\)
\(\Leftrightarrow x-5=3x-11\)
\(\Leftrightarrow x=3\)
\(a,\dfrac{2}{2x+1}-\dfrac{3}{2x-1}=\dfrac{4}{4x^2-1}\\ \Leftrightarrow\dfrac{2\left(2x-1\right)-3\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4}{4x^2-1}\\ \Leftrightarrow\dfrac{4x-2-6x-3}{4x^2-1}=\dfrac{4}{4x^2-1}\\ \Leftrightarrow\dfrac{-2x-5}{4x^2-1}=\dfrac{4}{4x^2-1}\\ \Leftrightarrow\left(-2x-5\right)\left(4x^2-1\right)=4\left(4x^2-1\right)\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(-2x-5-4\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(-2x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\\-2x-9=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\\ Vậy......\)
\(b,\dfrac{2x}{x+1}+\dfrac{18}{x^2+2x-3}=\dfrac{2x-5}{x+3}\\ \Leftrightarrow\dfrac{2x}{x+1}+\dfrac{18}{x^2+3x-\left(x+3\right)}=\dfrac{2x-5}{x+3}\\ \Leftrightarrow\dfrac{2x\left(x^2+2x-3\right)+18\left(x+1\right)}{\left(x+1\right)\left(x-1\right)\left(x+3\right)}=\dfrac{2x-5}{x+3}\\ \Leftrightarrow\dfrac{2x^3+4x^2-6x+18x+18}{\left(x^2-1\right)\left(x+3\right)}=\dfrac{2x-5}{x+3}\\ \Leftrightarrow\dfrac{2x^3+4x^2+12x+18}{\left(x^2-1\right)\left(x+3\right)}=\dfrac{\left(2x-5\right)}{x+3}\\ \Leftrightarrow2\left(x^3+2x^2+6x+9\right)\left(x+3\right)=\left(2x-5\right)\left(x^2-1\right)\left(x+3\right)\\ \Leftrightarrow\left(x+3\right)\left(2x^3+4x^2+12x+18+2x^3-5x^2-2x+5\right)=0\\ \Leftrightarrow\left(x+3\right)\left(4x^3-x^2+10x+23\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\4x^3-x^2+10x+23=0\end{matrix}\right.\)
\(\dfrac{1}{x-1}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}\\ \Leftrightarrow\dfrac{x^2+x+1+2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4}{x^2+x+1}\\ \Leftrightarrow\dfrac{3x^2+x-4}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4}{x^2+x+1}\\ \Leftrightarrow\dfrac{3x^2+4x-3x-4}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4}{x^2+x+1}\\ \Leftrightarrow\dfrac{\left(3x+4\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4}{x^2+x+1}\\ \Leftrightarrow\left(3x+4\right)\left(x-1\right)\left(x^2+x+1\right)=4\left(x^2+x+1\right)\left(x-1\right)\\ \Leftrightarrow\left(x^2+x+1\right)\left(x-1\right)3x=0\\\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\\x-1=0\\3x=0\end{matrix}\right.\\ Vìx^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\3x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\\ Vậy.....\)
a) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\)
\(\Leftrightarrow\dfrac{4x+\left(2x-1\right)}{6}=\dfrac{24-2x}{6}\)
\(\Leftrightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow6x+2x=24+1\)
\(\Leftrightarrow8x=25\)
\(\Leftrightarrow x=\dfrac{25}{8}\)
Vậy phương trình có một nghiệm là x = \(\dfrac{25}{8}\)
b) \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)
\(\Leftrightarrow\dfrac{6\left(x-1\right)+3\left(x-1\right)}{12}=\dfrac{12-8\left(x-1\right)}{12}\)
\(\Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=12-8\left(x-1\right)\)
\(\Leftrightarrow9\left(x-1\right)+8\left(x-1\right)=12\)
\(\Leftrightarrow17\left(x-1\right)=12\)
\(\Leftrightarrow17x-17=12\)
\(17x=12+17\)
\(\Leftrightarrow17x=29\)
\(\Leftrightarrow x=\dfrac{29}{17}\)
Vậy phương trình có một nghiệm là x = \(\dfrac{29}{17}\)
c) \(\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow\dfrac{2-x}{2001}-\dfrac{1-x}{2002}-\dfrac{\left(-x\right)}{2003}=1\)
\(\Leftrightarrow\dfrac{2-x}{2001}+1-\dfrac{1-x}{2002}-1-\dfrac{\left(-x\right)}{2003}-1=1+1-1-1\)
\(\Leftrightarrow\dfrac{2-x}{2001}+\dfrac{2001}{2001}-\dfrac{1-x}{2002}-\dfrac{2002}{2002}-\dfrac{\left(-x\right)}{2003}-\dfrac{2003}{2003}=0\)
\(\Leftrightarrow\dfrac{2003-x}{2001}-\dfrac{2003-x}{2002}-\dfrac{2003-x}{2003}=0\)
\(\Leftrightarrow\left(2003-x\right)\left(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow2003-x=0\)
\(\Leftrightarrow-x=-2003\)
\(\Leftrightarrow x=2003\)
Vậy phương trình có một nghiệm là x = 2003
a) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\)
\(\Leftrightarrow\dfrac{4x}{6}+\dfrac{2x-1}{6}=\dfrac{24}{6}-\dfrac{2x}{6}\)
\(\Leftrightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow4x+2x+2x=1+24\)
\(\Leftrightarrow8x=25\)
\(\Leftrightarrow x=\dfrac{25}{8}\)
Vậy S={\(\dfrac{25}{8}\)}
b) \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)
\(\Leftrightarrow\dfrac{6\left(x-1\right)}{12}+\dfrac{3\left(x-1\right)}{12}=\dfrac{12}{12}-\dfrac{8\left(x-1\right)}{12}\)
\(\Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=12-8\left(x-1\right)\)
\(\Leftrightarrow6x-6+3x-3=12-8x+8\)
\(\Leftrightarrow6x+3x+8x=6+3+12+8\)
\(\Leftrightarrow17x=29\)
\(\Leftrightarrow x=\dfrac{29}{17}\)
Vậy S={\(\dfrac{29}{17}\)}
ĐKXĐ : \(x\ne\left\{2;3;4;5;6\right\}\)
\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x+2\right).\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right).\left(x+6\right)=32\)
\(\Leftrightarrow x^2+8x-20=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(\text{loại}\right)\\x=-10\end{matrix}\right.\Leftrightarrow x=-10\)
Vậy tập nghiệm phương trình S = {10}
a) x - 5 = 7 - x
<=> 2x = 12
<=> x = 6
Vậy tập nghiệm phương trình S = {6}
b) 3x - 15 = 2x(x - 5)
<=> 3(x - 5) = 2x(x - 5)
<=> (2x - 3)(x - 5) = 0
<=> \(\left[{}\begin{matrix}2x-3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=5\end{matrix}\right.\)
Tập nghiệm phương trình \(S=\left\{\dfrac{3}{2};5\right\}\)
a) \(\left(2x-1\right)^2-\left(3x+5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1-3x-5\right)=0\\ \text{}\Leftrightarrow\left(2x-1\right)\left(-x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};-6\right\}\)
b) \(\dfrac{x+5}{4}-\dfrac{2x-3}{3}=\dfrac{2x-1}{12}\)
\(\Leftrightarrow3\left(x+5\right)-4\left(2x-3\right)=2x-1\\ \Leftrightarrow3x+15-8x+12=2x-1\\ \Leftrightarrow-5x+27=2x-1\\ \Leftrightarrow-5x-2x=-1-27\\ \Leftrightarrow-7x=-28\\ \Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
\(c)\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{-4}{1-x^2}\)(ĐKXĐ: \(x\ne\pm1\))
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}-\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \dfrac{\left(x+1\right)^2-\left(x-1\right)^2-4}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4x-4}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4}{x+1}=0\)
\(\Leftrightarrow4=0\)(vô lý)
Vậy .....
\(d)\dfrac{1}{x+1}+\dfrac{2x-1}{x^3+1}=\dfrac{2}{x^2-x+1}\)(ĐKXĐ: \(x\ne-1\))
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2}{x^2-x+1}=0\\ \Leftrightarrow\dfrac{x^2-x+1+2x-1-2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x+2x-2x-2}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x-2}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow x^2+x-2x-2=0\\ \Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)
Vậy ....
a) 4x - 5=0
\(\Leftrightarrow\)4x =5
\(\Leftrightarrow\) x =\(\dfrac{5}{4}\)
b) 2x + 1 = 7 - x
\(\Leftrightarrow\)2x + x = 7-1
\(\Leftrightarrow\)3x = 6
\(\Leftrightarrow\)x = 2
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
a)\(\frac{3+2x}{2+x}-1=\frac{2-x}{2+x}\) (x khác -2)
\(\Leftrightarrow\frac{3+2x}{2+x}-\frac{2-x}{2+x}=1\)
\(\Leftrightarrow\frac{1+3x}{2+x}=1\)
\(\Leftrightarrow1+3x=2+x\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
b) \(\frac{5-2x}{3}+\frac{x^2-1}{3}x-1=\frac{\left(x-2\right)\left(1-3x\right)}{9x-3}\) (x khác 1/3)
\(\Leftrightarrow\frac{x^3-3x+5}{3}+\frac{\left(x-2\right)\left(3x-1\right)}{3\left(3x-1\right)}=1\)
\(\Leftrightarrow\frac{x^2-2x+3}{3}=1\)
\(\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\frac{1}{\left(3-2x\right)^2}-\frac{4}{\left(3+2x\right)^2}=\frac{3}{9-4x^2}\) (x khác +- 3/2)
\(\Leftrightarrow\frac{\left(3+2x\right)^2}{\left(3+2x\right)^2\left(3-2x\right)^2}-\frac{4\left(3-2x\right)^2}{\left(3+2x\right)^2\left(3-2x\right)^2}=\frac{9}{\left(3+2x\right)^2\left(3-2x\right)^2}\)
\(\Leftrightarrow9+12x+4x^2-4\left(9-12x+4x^2\right)-9=0\)
\(\Leftrightarrow-12x^2+60x-36=0\)
\(\Leftrightarrow-12\left(x^2-5x+3\right)=0\Leftrightarrow x^2-5x+3=0\)
\(\Rightarrow\Delta=b^2-4ac=25-12=13>0\)
\(x_1=\frac{-b+\sqrt{\Delta}}{2ac}=\frac{5+\sqrt{13}}{6}\)
\(x_2=\frac{5-\sqrt{13}}{6}\)
d) \(\frac{1}{x^2+2x+1}=\frac{4}{x+2x^2+x^3}=\frac{5}{2x+2x^2}\)
\(\Leftrightarrow\frac{x^2+2x+1}{1}=\frac{x+2x^2+x^3}{4}=\frac{2x+2x^2}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{x^2+2x+1}{1}=\frac{x+2x^2+x^3}{4}=\frac{2x+2x^2}{5}=\frac{x^2+2x+1-\left(x+2x^2+x^3\right)+2x+2x^2}{1-4+5}\)
(dấu bằng thứ nhất của câu d là dấu cộng à???)
`1+(x-2)/(1-x)+(2x^2-5)/(x^3-1)=4/(x^2+x+1)(x ne 1)`
`<=>(x^3-1)/(x^3-1)-((x-2)(x^2+x+1))/(x^3-1)+(2x^2-5)/(x^3-1)=(4(x-1))/(x^3-1)`
`<=>x^3-1-(x-2)(x^2+x+1)+2x^2-5=4(x-1)`
`<=>x^3-1-(x^3-x^2-x-2)+2x^2-5=4x-4`
`<=>x^3-1-x^3+x^2+x+2+2x^2-5-4x+4=0`
`<=>3x^2-3x+2=0`
`<=>x^2-2/3 x+2/3=0`
`<=>x^2-2.x. 1/3+1/9+5/9=0`
`<=>(x-1/3)^2=-5/9` vô lý
Vậy phương trình vô nghiệm.
ĐKXĐ: \(x\ne1\)
Ta có: \(1+\dfrac{x-2}{1-x}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}\)
\(\Leftrightarrow\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
Suy ra: \(x^3-1-\left(x^3+x^2+x-2x^2-2x-2\right)+2x^2-5=4x-4\)
\(\Leftrightarrow x^3-1-x^3+x^2+x+2+2x^2-5-4x+4=0\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
mà 3>0
nên x(x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}