tìm x biết :2x^29=3x^39
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a) \(\dfrac{3x-4}{2x+5}=\dfrac{3x+7}{2x-20}\left(đk:x\ne-\dfrac{5}{2},x\ne10\right)\)
\(\Rightarrow\left(3x-4\right)\left(2x-20\right)=\left(3x+7\right)\left(2x+5\right)\)
\(\Rightarrow6x^2-68x+80=6x^2+29x+35\)
\(\Rightarrow97x=45\Rightarrow x=\dfrac{45}{97}\)
b) \(\dfrac{10x-5}{7x+2}=\dfrac{50x+10}{35x-29}\left(đk:x\ne-\dfrac{2}{7},x\ne\dfrac{29}{35}\right)\)
\(\Rightarrow\left(10x-5\right)\left(35x-29\right)=\left(50x+10\right)\left(7x+2\right)\)
\(\Rightarrow350x^2-465x+145=350x^2+170x+20\)
\(\Rightarrow635x=125\Rightarrow x=\dfrac{25}{127}\)
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a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)
\(\Rightarrow3x-6x^2+6x+14=29\)
\(\Rightarrow-6x^2+9x-15=0\)
\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)
\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)
Vậy \(S=\varnothing\)
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a. \(3\times2^{2x-1}=24\)
\(\Rightarrow2^{2x-1}=8=2^3\)
\(\Rightarrow2x-1=3\Rightarrow x=2\)
Tìm x biết:
5. ( x-1 ) - 7.( x-2 ) = 2x -39
Tìm x thuộc Z biết:
x - 3 - 14.( x-2 )= -3x -3
\(3x+7⋮x-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5 ( x - 1 ) - 7 ( x - 2 ) = 2x - 39
<=> 5x - 5 - 7x + 14 = 2x - 39
<=> 5x - 7x - 2x = -39 + 5 - 14
<=> -4x = -48
<=> x = 12
x - 3 - 14.( x-2 )= -3x -3\(\Rightarrow\chi-3-28-14\chi-28=-3\chi-3\)
\(\Rightarrow\chi-3-28+3=-3\chi-3\)
\(\Rightarrow\chi-28=11\chi\)
\(\Rightarrow\chi-11\chi=28\)
\(\Rightarrow10\chi=28\Rightarrow\chi=2,8\left(kot.m\chi\inℤ\right)\)
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a) Ta có : 2x2 + 3x = 0
<=> x(2x + 3) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x+3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{2}\end{cases}}\)
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75 + 68 : (5x + 7) = 29
=> 68 : (5x + 7) = -46
=> 5x + 7 = -3128
=> 5x = -3135
=> x = -627
vậy_
(x - 7)(2x - 8) = 0
=> x - 7 = 0 hoặc 2x - 8 = 0
=> x = 7 hoặc x = 4
vậy_
(3x - 5)10 = (3x - 5)9
=> (3x - 5)10 - (3x - 5)9 = 0
=> (3x - 5)9 .[(3x - 5) - 1] = 0
=> (3x - 5)9 = 0 hoặc (3x - 5) - 1 = 0
=> 3x - 5 = 0 hoặc 3x - 5 = 1
=> 3x = 5 hoặc 3x = 6
=> x = 5/3 hoặc x = 2
vậy_