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3 tháng 5 2020
a) Ta có: \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)
\(\Leftrightarrow\frac{63\left(3x-11\right)}{693}-\frac{231x}{693}-\frac{99\left(3x-5\right)}{693}+\frac{77\left(5x-3\right)}{693}=0\)
\(\Leftrightarrow189x-693-231x-297x+495+385x-231=0\)
\(\Leftrightarrow46x-429=0\)
\(\Leftrightarrow46x=429\)
hay \(x=\frac{429}{46}\)
Vậy: \(x=\frac{429}{46}\)
b) Ta có: \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)
\(\Leftrightarrow\frac{9x-0,7}{4}-\frac{5x-1,5}{7}-\frac{7x-1,1}{6}+\frac{5\left(0,4-2x\right)}{5}=0\)
\(\Leftrightarrow105\left(9x-0,7\right)-60\left(5x-1,5\right)-70\left(7x-1,1\right)+420\left(0,4-2x\right)=0\)
\(\Leftrightarrow945x-\frac{147}{2}-300x+90-490x+77+168-840x=0\)
\(\Leftrightarrow-685x+261.5=0\)
\(\Leftrightarrow-685x=-261.5\)
hay \(x=\frac{523}{1370}\)
Vậy: \(x=\frac{523}{1370}\)
c) Ta có: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)
\(\Leftrightarrow\frac{14\left(5x-3\right)}{84}-\frac{21\left(7x-1\right)}{84}-\frac{24\left(2x-1\right)}{84}+\frac{420}{84}=0\)
\(\Leftrightarrow70x-42-147x+21-48x+24+420=0\)
\(\Leftrightarrow-125x+423=0\)
\(\Leftrightarrow-125x=-423\)
hay \(x=\frac{423}{125}\)
Vậy: \(x=\frac{423}{125}\)
d) Ta có: \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)
\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}-\frac{45x}{30}+\frac{20\left(x-7\right)}{30}=0\)
\(\Leftrightarrow435-12x-36-45x+20x-140=0\)
\(\Leftrightarrow-37x+259=0\)
\(\Leftrightarrow-37x=-259\)
hay \(x=7\)
Vậy: x=7
A
4 tháng 4 2020
a) \(\frac{9x-0,7}{4}\)\(-\)\(\frac{5x-1,5}{7}\)=\(\frac{12x-2,1}{3}\)
⇔\(\frac{21\left(9x-0,7\right)}{84}\)\(-\)\(\frac{12\left(5x-1,5\right)}{84}\)=\(\frac{28\left(12x-2,1\right)}{84}\)
⇒189x\(-\)14,7\(-\)60x+18=336x\(-\)58,8
⇔\(-\)207x=\(-\)62,1
⇔x=\(\frac{3}{10}\)
Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{3}{10}\)}
TN
0
16 tháng 3 2020
Bài 2:
a, \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(2x-1\right)-5\left(x+8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x-3\right)-\left(5x+40\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x-3-5x-40\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-43\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-43=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=43\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{1;43\right\}\)
b, \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow9x^2-1-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-\frac{1}{3};-2\right\}\)
c, \(\left(x+7\right)\left(3x-1\right)=49-x^2\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(49-x^2\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(7-x\right)\left(7+x\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1-7+x\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(4x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\4x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-7;2\right\}\)
d, \(x^3-5x^2+6x=0\)
\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow x\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{0;2;3\right\}\)
e, \(2x^3+3x^2-32x=48\)
\(\Leftrightarrow2x^3+3x^2-32x-48=0\)
\(\Leftrightarrow\left(2x^3-8x^2\right)+\left(11x^2-44x\right)+\left(12x-48\right)=0\)
\(\Leftrightarrow2x^2\left(x-4\right)+11x\left(x-4\right)+12\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x^2+11x+12\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(2x^2+8x\right)+\left(3x+12\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left[2x\left(x+4\right)+3\left(x+4\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{4;-4;3-\frac{3}{2}\right\}\)
DH
3 tháng 3 2020
\(b,\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}\) \(\left(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne-3\end{matrix}\right.\right)\)
\(\Leftrightarrow\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{\left(x-1\right)\left(x+3\right)}=1\)
\(\Leftrightarrow\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\left(3x^2+8x-3\right)-\left(2x^2+3x-5\right)+4=x^2+2x-3\)
\(\Leftrightarrow x^2+5x+6=x^2+2x-3\)
\(\Leftrightarrow9=-3x\)
\(\Leftrightarrow x=-3\left(ktmđk\right)\)
\(\Leftrightarrow Ptvn\)
T
1 tháng 5 2019
\(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{6}\)
\(\Leftrightarrow\frac{\left(9x-0,7\right)\cdot7}{4\cdot7}-\frac{\left(5x-1,5\right)\cdot4}{7\cdot4}=\frac{7x-1,1-2+10x}{6}\)
\(\Leftrightarrow\frac{63x-4,9-20x+6}{28}=\frac{7x-1,1-2+10x}{6}\)
\(\Leftrightarrow\left(63x-4,9-20x+6\right)\cdot6=28\left(7x-1,1-2+10x\right)\)
\(\Leftrightarrow378x-120x+6,6=196x-86,8+280x\)
\(\Leftrightarrow378x-120x-196x-280x=-86,8-6,6\)
\(\Leftrightarrow-218x=-93,4\)
\(\Leftrightarrow x=\frac{467}{1090}\)
NV
18 tháng 4 2017
\(1.\frac{7x-3}{x-1}=\frac{2}{3}\) ( \(x\ne1\))
\(\Leftrightarrow\frac{3\left(7x-1\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)
\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\frac{7}{19}\)
\(2.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)
\(\Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x-1\right)\left(3x+2\right)}\)
\(\Rightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow\left(15x^2-15x^2\right)+\left(-8x+11x\right)=-14-1\)
\(\Leftrightarrow3x=-15\)
\(\Leftrightarrow x=-5\)
\(3.\frac{1-x}{x+1}+3=\frac{2x+3}{3x-1}\)
\(\Leftrightarrow\frac{\left(1-x\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}+\frac{3\left(x+1\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}=\frac{\left(2x+3\right)\left(x+1\right)}{\left(3x-1\right)\left(0+1\right)}\)
\(\Rightarrow\left(1-x\right)\left(3x-1\right)+3\left(x+1\right)\left(3x-1\right)=\left(2x+3\right)\left(x+1\right)\)
\(\Leftrightarrow3x-1-3x^2+x+3\left(3x^2-x+3x-1\right)=2x^2+2x+3x+3\)
\(\Leftrightarrow3x-1-3x^2+x+9x^2-3x+9x-3=2x^2+2x+3x+3\)
\(\Leftrightarrow6x^2+10x-4=2x^2+5x+3\)
\(\Leftrightarrow\left(6x^2-2x^2\right)+\left(10x-5x\right)=7\)
\(\Leftrightarrow4x^2+5x-7=0\)
\(\Leftrightarrow\left(2x\right)^2+4x.\frac{5}{4}+\frac{16}{25}+\frac{191}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{5}{4}\right)^2-\frac{191}{25}=0\)
\(\left(2x+\frac{5}{4}\right)^2>0\)
\(\Rightarrow\left(2x+\frac{5}{4}\right)^2+\frac{191}{25}>0\)
=> PT vô nghiệm
\(4.\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{x^2-4}+\frac{\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{2\left(3x-2\right)+1}{x^2-4}\)
\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3\left(3x-2\right)+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)
\(\Leftrightarrow3x^2-25x-6=3x^2-2x+1\)
\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-25x+2x\right)+\left(-6-1\right)=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=\frac{-7}{23}\)
\(5.\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2}{9x^2-4}-\frac{6\left(3x-2\right)}{9x^2-4}=\frac{9x^2}{9x^2-4}\)
\(\Rightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(12x-18x\right)+\left(4+12\right)=0\)
\(\Leftrightarrow-6x+16=0\)
\(\Leftrightarrow-6x=-16\)
\(\Leftrightarrow x=\frac{16}{6}\)
\(6.1+\frac{1}{x+2}=\frac{12}{8-x^3}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}+\frac{1\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}=\frac{12\left(x+2\right)}{\left(x+2\right)\left(8-x^3\right)}\)
\(\Rightarrow\left(x+2\right)\left(8-x^3\right)+1\left(8-x^3\right)=12\left(x+2\right)\)
\(\Leftrightarrow8x+x^4+16+2x^3+8-x^3=12x+24\)
\(\Leftrightarrow x^4+\left(2x^3-x^3\right)+\left(8x-12x\right)+\left(16-24\right)=0\)
\(\Leftrightarrow x^4+x^3-4x-8=0\)
\(\Leftrightarrow\left(x^4-4x\right)+\left(x^3-8\right)=0\)
Đến đấy mk tắc r xl bạn nhé
\(\frac{5x-1}{3}+\frac{7x-1,1}{3}-\frac{1,5-5x}{7}=\frac{9x-0,7}{4}\)
⇔ \(\frac{5x-1+7x-1,1}{3}-\frac{1,5-5x}{7}-\frac{9x-0,7}{4}=0\)
⇔ \(\frac{12x-2,1}{3}-\frac{1,5-5x}{7}-\frac{9x-0,7}{4}=0\)
⇔ \(\frac{28\left(12x-2,1\right)}{84}-\frac{12\left(1,5-5x\right)}{84}-\frac{21\left(9x-0,7\right)}{84}=0\)
⇔ \(\frac{336x-58,8}{84}-\frac{18-60x}{84}-\frac{189x-14,7}{84}=0\)
⇔ \(\frac{336x-58,8-18+60x-189x+14,7}{84}=0\)
⇔ \(\frac{207x-62,1}{84}=0\)
⇔ 207x - 62, 1 = 0
⇔ 207x = 62, 1
⇔ x = 0, 3
\(\frac{5x-1}{3}+\frac{7x-1.1}{3}-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)
\(\Leftrightarrow\left(\frac{5x-1}{3}+\frac{7x-1.1}{3}\right)-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)
\(\Leftrightarrow\left(\frac{5x-1+7x-1.1}{3}\right)-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)
\(\Leftrightarrow\frac{12x-2.1}{3}-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)
\(\Leftrightarrow\frac{28\left(12x-2.1\right)}{84}-\frac{12\left(1.5-5x\right)}{84}-\frac{21\left(8x-0,7\right)}{84}=0\)
\(\Leftrightarrow\frac{336x-58.8-18+60x-189x+14.7}{84}=0\)
\(\Leftrightarrow336x-58.8-18+60x-189x+14.7=0\)
\(\Leftrightarrow207x-62.1=0\)
\(\Leftrightarrow207x=62.1\)
\(\Leftrightarrow x=\frac{62.1}{207}=\frac{3}{10}=0.3\)