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Bạn đưa quá nhiều bài 1 lúc nên người ta giải được cũng chẳng ai muốn giải đâu, vì nhìn vào đã thấy ngộp rồi. Kinh nghiệm là muốn được giải quyết nhanh thì chỉ đăng 2-3 bài 1 lúc thôi
Bài 1:
a/ \(11-\left(2x+3\right)=3\left(x-4\right)\)
\(\Leftrightarrow11-2x-3=3x-12\)
\(\Leftrightarrow5x=20\)
\(\Rightarrow x=4\)
b/ \(5\left(2x-3\right)-4\left(5x-7\right)=19-2x\)
\(\Leftrightarrow10x-15-20x+28=19-2x\)
\(\Leftrightarrow8x=-6\)
\(\Rightarrow x=-\frac{3}{4}\)
c/
\(\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)
\(\Leftrightarrow2x-3\left(2x+1\right)=x-6x\)
\(\Leftrightarrow x=3\)
d/
\(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow79x=158\)
\(\Rightarrow x=2\)
e/
\(\frac{2-6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)
\(\Leftrightarrow4\left(2-6x\right)-2\left(2+3x\right)=140-5\left(6x+3\right)\)
\(\Leftrightarrow0=-121\) (vô lý)
Vậy pt vô nghiệm
f/
\(\frac{3x+2}{2}-\frac{3x+1}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow3\left(3x+2\right)-\left(3x+1\right)=12x+10\)
\(\Leftrightarrow6x=-5\)
\(\Rightarrow x=-\frac{5}{6}\)
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
a/ \(\left|\frac{3x-6}{1-2x}\right|=x-2\) \(\left(x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{3x-6}{1-2x}=x-2\\\frac{3x-6}{1-2x}=2-x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x-6=\left(x-2\right)\left(1-2x\right)\\3x-6=\left(2-x\right)\left(1-2x\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x-6=x+4x-2-2x^2\\3x-6=-x-4x+2+2x^2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2x^2+2x+4=0\\2x^2-8x+8=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\\x=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
KL: .............
b/ Tương tự
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
Bài 1:
a/ \(x\ne1;2\)
\(\frac{x-2}{\left(x-1\right)\left(x-2\right)}-\frac{7\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x-2-7x+7+1=0\)
\(\Leftrightarrow-6x+6=0\)
\(\Rightarrow x=1\) (loại)
Vậy pt vô nghiệm
b/ \(x\ne\frac{3}{2}\)
\(\frac{2x+3}{2x-3}-\frac{3}{2\left(2x-3\right)}-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{10\left(2x+3\right)}{10\left(2x-3\right)}-\frac{15}{10\left(2x-3\right)}-\frac{4\left(2x-3\right)}{10\left(2x-3\right)}=0\)
\(\Leftrightarrow20x+30-15-8x+12=0\)
\(\Leftrightarrow12x+27=0\)
\(\Rightarrow x=-\frac{9}{4}\)
c/ \(x\ne\pm1\)
\(\frac{x+1}{x-1}-\frac{4}{x+1}+\frac{3-x^2}{x^2-1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}+\frac{3-x^2}{x^2-1}=0\)
\(\Leftrightarrow x^2+2x+1-4x+4+3-x^2=0\)
\(\Leftrightarrow-2x+8=0\)
\(\Rightarrow x=4\)
Bài 1:
d/\(x\ne\pm3\)
\(\frac{x-1}{x+3}-\frac{x}{x-3}+\frac{7x-3}{x^2-9}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{x^2-9}-\frac{x\left(x+3\right)}{x^2-9}+\frac{7x-3}{x^2-9}=0\)
\(\Leftrightarrow x^2-4x+3-x^2-3x+7x-3=0\)
\(\Rightarrow0=0\)
Vậy pt có vô số nghiệm \(x\ne\pm3\)
e/ \(x\ne\pm1\)
\(\frac{1}{x+1}+\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x^2-1\right)\left(x-1\right)}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Leftrightarrow x^2-2x+1+2+3x-3=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\left(l\right)\end{matrix}\right.\)
Bài 1 :
a, Ta có : \(3x-1=2x+4\)
=> \(3x-2x=4+1\)
=> \(x=5\)
Vậy phương trình có tập nghiệm \(S=\left\{5\right\}\)
b, Ta có : \(5x-2=0\)
=> \(5x=2\)
=> \(x=\frac{2}{5}\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{2}{5}\right\}\)
c, Ta có : \(7x-4=3x+12\)
=> \(7x-3x=12+4\)
=> \(4x=16\)
=> \(x=4\)
Vậy phương trình có tập nghiệm \(S=\left\{4\right\}\)
d, Ta có : \(\frac{x-1}{2}+\frac{3x+2}{4}=\frac{x-7}{12}\)
=> \(\frac{6\left(x-1\right)}{12}+\frac{3\left(3x+2\right)}{12}=\frac{x-7}{12}\)
=> \(6\left(x-1\right)+3\left(3x+2\right)=x-7\)
=> \(6x-6+9x+6=x-7\)
=> \(6x+9x-x=6-7-6\)
=> \(14x=-7\)
=> \(x=-\frac{1}{2}\)
Vậy phương trình có tập nghiệm \(S=\left\{-\frac{1}{2}\right\}\)
Bài 2 :
a, ĐKXĐ : \(\left\{{}\begin{matrix}x^2-2x+1\ne0\\x-1\ne0\end{matrix}\right.\)
=> \(x-1\ne0\)
=> \(x\ne1\)
- Ta có : \(\left(\frac{x+1}{x^2-2x+1}+\frac{1}{x-1}\right):\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\left(\frac{x+1}{\left(x-1\right)^2}+\frac{x-1}{\left(x-1\right)^2}\right):\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\left(\frac{2x}{\left(x-1\right)^2}\right):\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\left(\frac{2x}{\left(x-1\right)^2}\right)\left(\frac{x-1}{x}\right)-\frac{2}{x-1}\)
= \(\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\frac{x-2}{x-1}\)
a,<=>\(\frac{20\left(1-2x\right)+6x}{12}\)=\(\frac{9\left(x-5\right)-24}{12}\)
=> 20-40x+6x = 9x-45-24
<=> -40x+6x-9x = -20-45-24
<=> -43x = -89
<=> x = \(\frac{89}{43}\)
c,ĐKXĐ :x\(\ne\pm1\)
<=>\(\frac{3\left(x+1\right)}{x^2+1}\) = -\(\frac{3x+2}{x^2+1}\) - \(\frac{4\left(x-1\right)}{x^2+1}\)
=> 3x+1 = -3x-2-4x+4
<=>3x+3x+4x = -1-2+4
<=> 10x = 1
<=> x =\(\frac{1}{10}\)(TMĐK)
\(1a,\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(x=-\frac{1}{12}\)
Vậy ................
\(b,\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow\frac{5\left(7x-1\right)}{30}+\frac{30.2x}{30}=\frac{6\left(16-x\right)}{30}\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
Vậy ....................
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