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3:18=
:(2x)
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16 tháng 12 2021
\(\dfrac{5x}{2x-3}-\dfrac{x-18}{3-2x}-\dfrac{16x}{2x-3}=\dfrac{5x+x-18-16x}{2x-3}=\dfrac{-10x-18}{2x-3}\)
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
BA
0
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
SN
1
21 tháng 1
\(\dfrac{1}{2x-3}=\dfrac{2\left(x-3\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)\left(2x-3\right)}\)
\(\dfrac{2x-3}{2x^2-18}=\dfrac{2x-3}{2\left(x-3\right)\left(x+3\right)}=\dfrac{\left(2x-3\right)\cdot\left(2x-3\right)}{2\left(2x-3\right)\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(2x-3\right)^2}{2\left(2x-3\right)\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{2}{2x^2+3x-9}=\dfrac{2}{\left(x+3\right)\left(2x-3\right)}=\dfrac{2\cdot2\cdot\left(x-3\right)}{2\left(x-3\right)\cdot\left(x+3\right)\left(2x-3\right)}\)
\(=\dfrac{4x-12}{2\left(x-3\right)\left(x+3\right)\left(2x-3\right)}\)
20 tháng 10 2021
\(\left(2x-3\right)\left(x+1\right)+\left(4x^3-6x^2-6x\right):\left(-2x\right)=18\)
\(\Leftrightarrow2x^2+2x-3x-3-2x^2+3x+3=18\)
\(\Leftrightarrow x=9\)
CL
1
Y
29 tháng 6 2023
\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
29 tháng 6 2023
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
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 \(S=\left\{\dfrac{1}{2};\dfrac{-5}{2}\right\}\)
\(\frac{3}{18}=\frac{1}{2x}\)
=> \(3\cdot2x=18\cdot1\)
=> \(6x=18\)
=> \(x=18:6=3\)
Vậy x = 3
Ta có : \(\frac{3}{18}\)= \(\frac{1}{2x}\)
\(\Rightarrow\)3.2x = 1.18
\(\Rightarrow\)6x = 18
\(\Rightarrow\)x = 18 : 6
\(\Rightarrow\)x = 3