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Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{2x+2}{5x-3}=\frac{2x+12}{5x+18}=\frac{\left(2x+12\right)-\left(2x+2\right)}{\left(5x+18\right)-\left(5x-3\right)}=\frac{10}{21}\)
Có: \(\frac{2x+2}{5x-3}=\frac{10}{21}\)
<=> 21(2x+2) = 10(5x-3)
<=> 42x+42 = 50x-30
<=> 8x = 72
<=> x = 9
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2x+2}{5x-3}=\frac{2x+12}{5x+18}\)
\(\Rightarrow\left(2x+2\right)\left(5x+18\right)=\left(5x-3\right)\left(2x+12\right)\)
\(\Rightarrow2x\left(5x+18\right)+2\left(5x+18\right)=5x\left(2x+12\right)-3\left(2x+12\right)\)
\(\Rightarrow10x^2+36x+10x+36=10x^2+60x-6x-36\)
\(\Rightarrow46x+36=54x-36\)
\(\Rightarrow36+36=54x-46x\)
\(\Rightarrow72=8x\)
\(\Rightarrow x=9\)
tck cho to nha
2x(5x+18)+2(5x+18)=5x(2x+12)+3(3x-1)
10x^2+2x+10x+2=10x+-5x+9x-3
=> 10x^2+2x+10x-10x^2+5x-9x=-3-2
-2X=-5
x=5/2
x=5/2
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\(\frac{2x+2}{5x-3}=\frac{2x+12}{5x+18}\)
\(\Rightarrow\left(2x+2\right)\left(5x+18\right)=\left(2x+12\right)\left(5x-3\right)\)
\(\Rightarrow\left(2x+2\right)5x+\left(2x+2\right)18=\left(2x+12\right)5x-\left(2x+12\right)3\)
\(\Rightarrow10x^2+10x+36x+36=10x^2+60x-6x-36\)
\(\Rightarrow46x+36=54x-36\)
\(\Rightarrow54x-46x=36+36\)
\(\Rightarrow8x=72\)
\(\Rightarrow x=9\)
Vậy \(x=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\frac{2x+2}{5x-3}=\frac{2x+12}{5x+18}\)
=> ( 2x + 2 ) ( 5x + 18 ) = ( 2x + 12 ) ( 5x - 3 )
=> 2x ( 5x + 18 ) + 2 ( 5x + 18 ) = 2x ( 5x - 3 ) + 12 ( 5x - 3 )
=> 10 x 2 + 36x + 10x + 36 = 10 x 2 - 6x + 60 x - 36
=> 36x + 10x + 6x - 60x = - 36 - 36
=> - 8 x = - 72
=> x = 9
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\(a,\frac{-24}{x}+\frac{18}{x}=\frac{-24+18}{x}=\frac{-6}{x}\)
\(\Leftrightarrow x\inƯ(-6)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(b,\frac{2x-5}{x+1}=\frac{2x+2-7}{x+1}=\frac{2(x+1)-7}{x+1}=2-\frac{7}{x+1}\)
\(\Leftrightarrow7⋮x+1\Leftrightarrow x+1\inƯ(7)=\left\{\pm1;\pm7\right\}\)
Xét các trường hợp rồi tìm được x thôi :>
\(c,\frac{3x+2}{x-1}-\frac{x-5}{x-1}=\frac{3x+2-x-5}{x-1}=\frac{2x+7}{x-1}=\frac{2x-2+9}{x-1}=\frac{2(x-1)+9}{x-1}=2+\frac{9}{x-1}\)
\(\Leftrightarrow9⋮x-1\Leftrightarrow x-1\inƯ(9)=\left\{\pm1;\pm3;\pm9\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2;10;-8\right\}\)
d, TT
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a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
=> ( 2x+2).(5+18)= (2x+12). (5x-3)
=> (2x+2) . 23 = 2x. 5x - 2x.3 + 12.5x - 12.3
=> 2x. 23+ 2.23 = 10.x.x - 6x+ 60x - 36
=> 46x + 46 = 10x.x + 54 x -36
=> 46 + 36 = 10xx + 54x -46x
=> 82 = 10xx +8x
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