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Question

1. giải phương trình

a, $\frac{7x-3}{x-1}=\frac{2}{3}$

b, $\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}$

c,$\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}$

d,$\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}$

Nguyễn Lê Phước Thịnh · Accepted Answer

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: $x=\frac{-3}{4}$Câu trả lời: 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: $x=\frac{-3}{4}$

Hoàng Ngọc Anh · Answer

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)

$\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\
\Leftrightarrow\frac{12x^2+37x+28}{\left(x-1\right)\cdot\left(3x+4\right)}-\frac{12x^2-7x-5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\
\Leftrightarrow\frac{12x^2+37x+28-12x^2+7x+5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\
\Leftrightarrow\frac{44x+33}{\left(x-1\right)\cdot\left(3x+4\right)}=0\
\Leftrightarrow44x+33=0\
\Rightarrow x=-\frac{3}{4}$Câu trả lời: 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)

$\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\
\Leftrightarrow\frac{12x^2+37x+28}{\left(x-1\right)\cdot\left(3x+4\right)}-\frac{12x^2-7x-5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\
\Leftrightarrow\frac{12x^2+37x+28-12x^2+7x+5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\
\Leftrightarrow\frac{44x+33}{\left(x-1\right)\cdot\left(3x+4\right)}=0\
\Leftrightarrow44x+33=0\
\Rightarrow x=-\frac{3}{4}$

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