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\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
a) ĐKXĐ: \(x\ne1\)
Ta có: \(\frac{7x-3}{x-1}=\frac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-9-2x+2=0\)
\(\Leftrightarrow19x-7=0\)
\(\Leftrightarrow19x=7\)
hay \(x=\frac{7}{19}\)
Vậy: \(x=\frac{7}{19}\)
b) ĐKXĐ: \(x\ne-1\)
Ta có: \(\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=1+x\)
\(\Leftrightarrow12-28x-1-x=0\)
\(\Leftrightarrow11-29x=0\)
\(\Leftrightarrow29x=11\)
hay \(x=\frac{11}{29}\)
Vậy: \(x=\frac{11}{29}\)
c) ĐKXĐ: \(x\notin\left\{\frac{-2}{3};\frac{1}{3}\right\}\)
Ta có: \(\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)
\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow15x^2-8x+1-15x^2+11x+14=0\)
\(\Leftrightarrow3x+15=0\)
\(\Leftrightarrow3x=-15\)
hay x=-5
Vậy: x=-5
d) ĐKXĐ: \(x\notin\left\{1;\frac{-4}{3}\right\}\)
Ta có: \(\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\)
\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\)
\(\Leftrightarrow12x^2+37x+28=12x^2-7x-5\)
\(\Leftrightarrow12x^2+37x+28-12x^2+7x+5=0\)
\(\Leftrightarrow44x+33=0\)
\(\Leftrightarrow44x=-33\)
hay \(x=\frac{-3}{4}\)
Vậy: \(x=\frac{-3}{4}\)
a)
\(\frac{7x-3}{x-1}=\frac{2}{3}\\ \Leftrightarrow\frac{21x-9}{3\cdot\left(x-1\right)}-\frac{2x-2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow\frac{21x-9-2x+2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow\frac{19x-7}{3\cdot\left(x-1\right)}=0\\ \Rightarrow19x-7=0\\ \Rightarrow x=\frac{7}{19}\)
b)
\(\frac{2\cdot\left(3-7x\right)}{1+x}=\frac{1}{2}\\ \Leftrightarrow\frac{12-28x}{2\cdot\left(1+x\right)}-\frac{1+x}{2\cdot\left(1+x\right)}=0\\ \Leftrightarrow\frac{12-28x-1-x}{2\cdot\left(1+x\right)}=0\\ \Leftrightarrow\frac{11-29x}{2\cdot\left(1+x\right)}=0\\\Rightarrow11-29x=0\\ \Rightarrow x=\frac{11}{29}\)
c)
\(\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\\ \Leftrightarrow\frac{15x^2-8x+1}{\left(3x+2\right)\cdot\left(3x-1\right)}-\frac{15x^2-11x-14}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Leftrightarrow\frac{15x^2-8x+1-15x^2+11x+14}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Leftrightarrow\frac{3x+15}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Rightarrow3x+15=0\\ \Rightarrow x=-5\)
d)
\(\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\\ \Leftrightarrow\frac{12x^2+37x+28}{\left(x-1\right)\cdot\left(3x+4\right)}-\frac{12x^2-7x-5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow\frac{12x^2+37x+28-12x^2+7x+5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow\frac{44x+33}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow44x+33=0\\ \Rightarrow x=-\frac{3}{4}\)
1. \(\left(4x+7\right)\left(3x+4\right)=\left(12x-5\right)\left(x-1\right)\)
\(12x^2+16x+21x+28=12x^2-12x-5x+5\)
\(12x^2+37x+28-12x^2+17x-5=0\)
54x+23=0
54x=-23
x=-23/54
2. \(\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(15x^2-8x+1-15x^2+11x+14=0\)
3x+15=0
3x=-15
x=-5
1/ \(\frac{x-3}{3xy}\)+\(\frac{5x+3}{3xy}\)= \(\frac{6x}{3xy}\)=\(\frac{3}{y}\)
2/\(\frac{5x-7}{2x-3}\)+\(\frac{4-3x}{2x-3}\)=\(\frac{2x-3}{2x-3}\)=1
3/\(\frac{11x-7}{3-5x}\)-\(\frac{6x+4}{5x-3}\)=\(\frac{11x-7}{3-5x}\)+\(\frac{6x+4}{3-5x}\)=\(\frac{17x-3}{3-5x}\)
4/\(\frac{3}{2x+6}\)-\(\frac{x-6}{2x^2+6x}\)=\(\frac{3x}{x\left(2x+6\right)}\)-\(\frac{x-6}{x\left(2x+6\right)}\)=\(\frac{2x-6}{x\left(2x+6\right)}\)
5/\(\frac{1}{2x-10}\)+\(\frac{2x}{3x^2-15x}\)=\(\frac{1}{2\left(x-5\right)}\)+\(\frac{2x}{3x\left(x-5\right)}\)=\(\frac{3x}{6x \left(x-5\right)}\)+\(\frac{4x}{6x\left(x-5\right)}\)
=\(\frac{7x}{6x\left(x-5\right)}\)=\(\frac{7}{6\left(x-5\right)}\)
Giải:
a) ⇔⇔ 9x2 + 12x + 4 - 18x + 12 = 9x2 ⇔ 9x2 + 12x + 4 - 18x + 12 - 9x2 = 0
⇔ 16 + 6x = 0 ⇔ 2(8 + 3x) = 0 ⇔ 8 + 3x = 0 ⇔ x = \(\frac{-8}{3}\)
Vậy nghiệm của phương trình là x = \(\frac{-8}{3}\) .
b) \(\frac{3}{5x-1}+\frac{3}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\text{⇔ }\frac{-3}{1-5x}+\frac{-3}{5x-3}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)
⇔ \(\frac{9-15x}{\left(1-5x\right)\left(5x-3\right)}+\frac{15x-3}{\left(1-5x\right)\left(5x-3\right)}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\) ⇔ 9 - 15x + 15x - 3 = 4
⇔ 8 = 4 ( vô lí)
Vậy phương trình trên vô nghiệm.
Mình chỉ làm 2 câu a, b thôi nhé! Các bài tập này cách làm giống nhau, bạn tự hoàn thành những bài còn lại nhé!
Giải tiêu biểu câu a nhé.
a/ \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)
\(\Leftrightarrow19x+5=0\)
\(\Leftrightarrow x=-\frac{5}{19}\)
<=>(3x+2)(5x+1)=(3x-1)(5x+7)
<=>\(15x^2+10x+3x+2=15x^2-5x+21x-7\)
<=>-3x=-9
<=>x=3
tick hộ mình nhá