áp dụng t/c của dãy tỉ số bằng nhau ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\frac{3}{6}=\frac{1}{2}\)

=> \(\frac{3x+2}{5x+7}=\frac{1}{2}\)=> 2(3x+2) = 5x + 7 => 6x + 4 = 5x + 7 => 6x - 5x = 7 - 4 => x = 3

Vậy x = 3

giải chi tiết hay đáp án thui?

bài 1

\(\frac{3x-y}{x+y}=\frac{3}{4}\)

\(\Rightarrow\left(3x-y\right).4=\left(x+y\right)3\)

\(\Rightarrow12x-4y=3x+3y\)

\(\Rightarrow9x=7y\)

\(\Rightarrow\frac{x}{y}=\frac{7}{9}\)

Vậy \(\frac{x}{y}=\frac{7}{9}\)

bài 2

\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+4}\)

\(\Rightarrow\) \(\left(3x+1\right)\left(5x+4\right)=\left(3x-1\right)\left(5x+7\right)\)

\(\Rightarrow15x^2+12x+10x+8=15x^2+21x-5x-7\)

\(\Rightarrow22x+8=16x-7\)

\(\Rightarrow22x+16x=-7-8\)

\(\Rightarrow6x=-15\)

\(\Rightarrow x=-2,5\)

Vậy x=-2,5

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

=> ( 5x + 1 ) . ( 3x + 2 ) = ( 3x - 1 ) . ( 5x + 7 )

=> 5x(3x + 2 ) + ( 3x + 2 ) = 3x(5x + 7 ) - ( 5x + 7 )

=> ( 15x² + 10x ) + ( 3x + 2 ) = ( 15x² + 21x ) - ( 5x + 7 )

=> ( 3x + 2 ) + ( 5x + 7 ) = ( 15x² + 21x ) - ( 15x² + 10x )

=> ( 3x + 5x ) + ( 2 + 7 ) = ( 15x² - 15x² ) + ( 21x - 10x )

=> 8x + 9 = 11x

=> 9 = 11x - 8x

=> 9 = 3x

=> x = 3

Vậy x = 3

~~Học tốt~~

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

=>       \(\frac{3x+2}{5x+7}\)=\(\frac{1}{2}\)=> ( 3x + 2 ) . 2 = 5x + 7 . 1 => 6x + 4 = 5x + 7 => x=3

Đặt \(B=\frac{2\sqrt{x}+3}{\sqrt{x}-1}=\frac{2\sqrt{x}-2+5}{\sqrt{x}-1}=\frac{2\left(\sqrt{x}-1\right)+5}{\sqrt{x}-1}=2+\frac{5}{\sqrt{x}-1}\)

\(\Rightarrow B\in Z\Leftrightarrow2+\frac{5}{\sqrt{x}-1}\in Z\Leftrightarrow\frac{5}{\sqrt{x}-1}\in Z\Leftrightarrow5⋮\sqrt{x}-1\Leftrightarrow\sqrt{x}-1\inƯ\left(5\right)\)

\(\Rightarrow\sqrt{x}-1\in\left\{-5;-1;1;5\right\}\)

Vì x dương\(\Rightarrow\sqrt{x}-1\ge0\)

\(\Rightarrow\sqrt{x}-1\in\left\{1;5\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{2;6\right\}\)

\(\Rightarrow x\in\left\{4;36\right\}\)

Vậy số phần tử của tập hợp A là 2

\(\frac{x-1}{x+5}=\frac{6}{7}\Leftrightarrow\frac{x-1}{6}=\frac{x+5}{7}\)

\(\Leftrightarrow\frac{7\left(x-1\right)}{42}=\frac{6\left(x+5\right)}{42}\)

\(\Leftrightarrow7\left(x-1\right)=6\left(x+5\right)\)

\(\Leftrightarrow7x-7=6x+30\)

\(\Leftrightarrow7x-6x=7+30\)

\(\Leftrightarrow x=37\)

Vậy nghiệm của phương trình là x = 37

\(a,\frac{3x+2}{5x+7}=\frac{3x-1}{5x-1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x-1\right)}=\frac{3}{8};\frac{3x+2}{5x+7}=\frac{3}{8}\Leftrightarrow24x+16=15x+21\Leftrightarrow9x=5\Leftrightarrow x=\frac{5}{9}\) \(b,\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow37.7-7x=3x+39\Leftrightarrow259-7x=3x+39\Leftrightarrow220-7x=3x\Leftrightarrow10x=220\Leftrightarrow x=22\) \(c,\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}=\frac{x+4}{2x+6}=\frac{\left(x+4\right)-\left(x+1\right)}{2x+6-\left(2x+1\right)}=\frac{3}{5};\frac{x+1}{2x+1}=\frac{3}{5}\Leftrightarrow5x+5=6x+3\Leftrightarrow x=2\) \(d,\frac{x-2}{x+2}=\frac{x+3}{x-4}=\frac{\left(x+3\right)-\left(x-2\right)}{\left(x-4\right)-\left(x+2\right)}=\frac{5}{-6};\frac{x-2}{x+2}=\frac{5}{-6}\Leftrightarrow6\left(2-x\right)=5x+10\Leftrightarrow2-6x=5x\Leftrightarrow x=\frac{2}{11}\) \(f,\frac{3x-5}{x}=\frac{9x}{3x+2}=\frac{9x-15}{3x}=\frac{9x-\left(9x-15\right)}{\left(3x+2\right)-3x}=\frac{15}{2};\frac{9x}{3x+2}=\frac{15}{2}\Leftrightarrow18x=45x+30\Leftrightarrow27x+30=0\Leftrightarrow x=\frac{-10}{9}\) \(e,\frac{x+2}{6}=\frac{5x-1}{5}\Leftrightarrow5\left(x+2\right)=6\left(5x-1\right)\Leftrightarrow5x+10=30x-6\Leftrightarrow10=25x-6\Leftrightarrow25x=16\Leftrightarrow x=\frac{16}{25}\)

a) Ta có: \(M=\frac{2x+5}{x+1}=\frac{2\left(x+1\right)+3}{x+1}=\frac{2x+2+3}{x+1}\)

Vì \(2x+2⋮\left(x+1\right)\Rightarrow3⋮\left(x+1\right)\)

Nên \(x+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow x=\left\{0;-2;2;-4\right\}\)

b) Tương tự