2x + 2x-3 =18
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16 tháng 12 2021
\(\dfrac{5x}{2x-3}-\dfrac{x-18}{3-2x}-\dfrac{16x}{2x-3}=\dfrac{5x+x-18-16x}{2x-3}=\dfrac{-10x-18}{2x-3}\)
BA
0
SN
1
21 tháng 1
\(\dfrac{1}{2x-3}=\dfrac{2\left(x-3\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)\left(2x-3\right)}\)
\(\dfrac{2x-3}{2x^2-18}=\dfrac{2x-3}{2\left(x-3\right)\left(x+3\right)}=\dfrac{\left(2x-3\right)\cdot\left(2x-3\right)}{2\left(2x-3\right)\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(2x-3\right)^2}{2\left(2x-3\right)\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{2}{2x^2+3x-9}=\dfrac{2}{\left(x+3\right)\left(2x-3\right)}=\dfrac{2\cdot2\cdot\left(x-3\right)}{2\left(x-3\right)\cdot\left(x+3\right)\left(2x-3\right)}\)
\(=\dfrac{4x-12}{2\left(x-3\right)\left(x+3\right)\left(2x-3\right)}\)
0
26 tháng 3 2018
Thay x = 1 vào phương trình 2(2x+1)+18=3(x+2)(2x+k)2(2x+1)+18=3(x+2)(2x+k), ta có:
2(2.1+1)+18=3(1+2)(2.1+k)
⇔2(2+1)+18=3.3(2+k)⇔2.3+18=9(2+k)
⇔6+18=18+9k⇔24−18=9k⇔6=9k
⇔k=69=232(2.1+1)+18=3(1+2)(2.1+k)
⇔2(2+1)+18=3.3(2+k)
⇔2.3+18=9(2+k)
⇔6+18=18+9k
⇔24−18=9k⇔6=9k
⇔k=\(\frac{6}{9}\)=\(\frac{2}{3}\)
Vậy khi thì phương trình có nghiệm x = 1
CL
1
1 tháng 9 2016
Đặt x2 + 2x = a ta có
\(\frac{1}{a-3}\)+ \(\frac{18}{a+2}\)= \(\frac{18}{a+1}\)
<=> a2 - 15a + 56 = 0
<=> a = (7;8)
Thế vô tìm được nghiệm
20 tháng 10 2021
\(\left(2x-3\right)\left(x+1\right)+\left(4x^3-6x^2-6x\right):\left(-2x\right)=18\)
\(\Leftrightarrow2x^2+2x-3x-3-2x^2+3x+3=18\)
\(\Leftrightarrow x=9\)
H
25 tháng 2 2022
(2x+1)(x+1)2(2x+3)-18=0
\(\Leftrightarrow\)(2x+1)(x+1)2(2x+3)=18
\(\Leftrightarrow\left(2x+2+1\right)\left(2x+2-1\right)\left(x+1\right)^2=18\)
\(\Leftrightarrow\left(\left(2x+2\right)^2-1\right)\left(x+1\right)^2=18\)
\(\Leftrightarrow4\left(x+1\right)^4-\left(x+1\right)^2-18=0\)
Đặt \(t=\left(x+1\right)^2\left(t\ge0\right)\)
\(\Leftrightarrow4t^2-t-18=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\left(tm\right)\\t=-2\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left(x+1\right)^2-\dfrac{9}{4}=0\)
\(\Leftrightarrow\left(x+1-\dfrac{2}{3}\right)\left(x+1+\dfrac{2}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
25 tháng 2 2022
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)-18=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+3\right)\left(x^2+2x+1\right)-18=0\)
\(\Leftrightarrow\left(4x^2+6x+2x+3\right)\left(x^2+2x+1\right)-18=0\)
\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18=0\)
\(\Leftrightarrow\left(4x^2+8x+3\right).4.\left(x^2+2x+1\right)-4.18=0\)
\(\Leftrightarrow\left(4x^2+8x+3\right)\left(4x^2+8x+4\right)-72=0\)
-Đặt \(t=4x^2+8x+3\)
PT\(\Leftrightarrow t\left(t+1\right)-72=0\)
\(\Leftrightarrow t^2+t-72=0\)
\(\Leftrightarrow t^2-8t+9t-72=0\)
\(\Leftrightarrow t\left(t-8\right)+9\left(t-8\right)=0\)
\(\Leftrightarrow\left(t-8\right)\left(t+9\right)=0\)
\(\Leftrightarrow t-8=0\) hay \(t+9=0\)
\(\Leftrightarrow4x^2+8x+3-8=0\) hay \(4x^2+8x+3+9=0\)
\(\Leftrightarrow4x^2+8x-5=0\) hay \(4x^2+8x+12=0\)
\(\Leftrightarrow4x^2-2x+10x-5=0\) hay \(\left(2x\right)^2+2.2x.2+4+8=0\)
\(\Leftrightarrow2x\left(2x-1\right)+5\left(2x-1\right)=0\) hay \(\left(2x+2\right)^2+8=0\) (phương trình vô nghiệm vì \(\left(2x+2\right)^2+8\ge8\))
\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow2x-1=0\) hay \(2x+5=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\) hay \(x=\dfrac{-5}{2}\)
-Vậy \(S=\left\{\dfrac{1}{2};\dfrac{-5}{2}\right\}\)
CL
1
NL
1
NN
22 tháng 11 2017
a) \(\dfrac{16+x}{x^2-2x}+\dfrac{18}{2x-x^2}\)
\(=\dfrac{16+x}{x^2-2x}-\dfrac{18}{x^2-2x}\)
\(=\dfrac{16+x-18}{x^2-2x}\)
\(=\dfrac{x-2}{x\left(x-2\right)}\)
\(=\dfrac{1}{x}\)
\(2^x+2^{x-3}=18\)
\(2^x+2^x:2^3=18\)
\(2^x+2^x.\frac{1}{2^3}=18\)
\(2^x.\left(1+\frac{1}{2^3}\right)=18\)
\(2^x.\frac{9}{8}=18\)
\(2^x=18:\frac{9}{8}\)
\(2^x=16=2^4\)
Vậy \(x=4\)
2x + 2x-3 = 18
<=> 2x + 2x : 23 = 18
<=> 2x + 2x . \(\frac{1}{2^3}\)= 18
<=> 2x(1 + \(\frac{1}{2^3}\)) = 18
<=> 2x = 16
<=> 2x = 24
<=> x = 4