a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)

\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)

\(\Leftrightarrow15x^2-15x^2+3x+10x+5x-21x=-7-2\)

\(\Leftrightarrow-3x=-9\)

\(\Leftrightarrow x=3\)

Vậy x = 3

b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)

                            \(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)

                             \(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)

                              \(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)

                             \(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)

Vậy x = 2

bài này sử dụng tích chéo nha bạn

z cau lamđi

\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)

\(\Rightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)

\(\Rightarrow15x^2+10x+3x+2=15x^2-5x+21x-7\)

\(3x=9\)

\(x=3\)

Ta có:  (3x+2)(5x+1)=(5x+7)(3x-1)   (ĐKXĐ: x khác  -7/5, x khác -1/5)

=> 15x²+13x+2=15x²+16x-7

=> 3x-9=0  =>x=3 (chọn)

a ) \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)

\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=3x\left(5x+7\right)-\left(5x+7\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)

\(\Leftrightarrow15x^2+13x+2=15x^2+16x-7\)

\(\Leftrightarrow13x+2=16x-7\)

\(\Leftrightarrow13x-16x=-7-2\)

\(\Leftrightarrow-3x=-9\)

\(\Rightarrow x=3\)

b ) tương tự

@Nguyễn Việt Lâm em sắp ktra, anh giúp em bài này với ạ ....

Akai Haruma giúp em giải phương trình trên được ko ạ ^_^

@Nguyễn Việt Lâm anh giải bải này đc ko ạ .

\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)

\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)

\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)

\(\Rightarrow-16x-272=120x+312\)

\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)

0.4736842105

\(\frac{3x+2}{5x+7}=\frac{5x-1}{5x+1}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x-1\right)\left(5x+7\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=25x^2+35x-5x-7\)

\(\Leftrightarrow25x^2+30x-7-15x^2-13x-2=0\)

\(\Leftrightarrow10x^2+17x-9=0\)

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