Giải các phương trình :
a) x(2x - 9) = 3x(x - 5)
b) 0,5x(x - 3) = (x - 3)(1,5x - 1)
c) 3x - 15 = 2x(x - 5)
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DL
22 tháng 2 2019
a, 0,5x.(2x - 9) = 1,5x.(x - 5)
<=> x2 - 4,5x = 1,5x2 - 7,5x
<=> 0,5x2 + 3x = 0
<=> 0,5x.( x + 6 ) = 0
<=> x = 0 hoặc x + 6 = 0
<=> x = 0 hoặc x = -6
Vậy....
#Đức Lộc#
S
22 tháng 2 2019
Làm thử nha :v
a) 0,5x.(2x - 9) = 1,5x.(x - 5)
<=> 0,5x.(2x - 9) - 1,5x.(x - 5) = 1,5x.(x - 5) - 1,5x.(x - 5)
<=> 0,5x.(2x - 9) - 1,5x.(x - 5) = 0
<=> -x(0,5x - 3) = 0
=> x = 0 hoặc 6
b) 5(x - 1) - (2x - 5) = 16 - x
<=> 3x = 16 - x
<=> 3x + x = 16
<=> 4x = 16
<=> x = 16 : 4
=> x = 4
TN
5 tháng 3 2020
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\\ \left(3x+2\right)\left(x^2-1\right)-\left(9x^2-4\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left[\left(x-1\right)-\left(3x-2\right)\right]=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\\ \left[{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
\(b.x\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\\ x\left(x^2-9\right)-\left(x^3+8\right)=0\\ x^3-9x-x^3-8=0\\ -9x-8=0\\ -9x=8\\ x=\frac{-8}{9}\)
\(c.2x\left(x-3\right)+5\left(x-3\right)=0\\ \left(x-3\right)\left(2x+5\right)=0\\ \left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-5}{2}\end{matrix}\right.\)
\(d.\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\\ \left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\\ \left(3x-1\right)\left[\left(x^2+2\right)-\left(7x-10\right)\right]=0\\ \left(3x-1\right)\left(x^2+2-7x+10\right)=0\\ \left(3x-1\right)\left(x^2-7x+12\right)=0\\ \left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \left(3x-1\right)\left[\left(x^2-4x\right)+\left(-3x+12\right)\right]=0\\ \left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}3x-1=0\\x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\\x=3\end{matrix}\right.\)
TN
5 tháng 3 2020
\(e.\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\\ \left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\\ \left(x+2\right)\left[\left(3-4x\right)-\left(x+2\right)\right]=0\\ \left(x+2\right)\left(3-4x-x-2\right)=0\\ \left(x+2\right)\left(1-5x\right)=0\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{1}{5}\end{matrix}\right.\)
\(f.x\left(2x-7\right)-4x+14=0\\ x\left(2x-7\right)-2\left(2x-7\right)=0\\ \left(2x-7\right)\left(x-2\right)=0\\ \left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)
\(g.3x-15=2x\left(x-5\right)\\ 3\left(x-5\right)=2x\left(x-5\right)\\ 3\left(x-5\right)-2x\left(x-5\right)=0\\ \left(x-5\right)\left(3-2x\right)=0\\ \left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
\(h.\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \left(2x+1\right)\left[\left(3x-2\right)-\left(5x-8\right)\right]=0\\ \left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \left(2x+1\right)\left(6-2x\right)=0\\ \left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=3\end{matrix}\right.\)
VT
2
H
19 tháng 1 2016
<=>0,5x(x-3)-(x-3)(1,5x-1)=0
<=>(x-3)[0,5x-(1,5x-1)]=0
<=>(x-3)[0,5-1,5x+1]=0
<=>(x-3)(2x+1)=0
<=>x-3=0
<=>2x+1=0
<=>x=3
<=>2x=-1
<=>x=3
<=>x=\(\frac{-1}{2}\)
HP
0
17 tháng 5 2022
a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x\right)\left(3x-2+2x\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-2\right)\left(5x-2\right)=0\end{matrix}\right.\)
hay x=2
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{10}{3}\\\left(-3,5x-1,5x-5\right)\left(-3,5x+1,5x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{10}{3}\\\left(-5x-5\right)\left(-2x+5\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-1;\dfrac{5}{2}\right\}\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(3x-1-x-15\right)\left(3x-1+x+15\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(2x-16\right)\left(4x+14\right)=0\end{matrix}\right.\Leftrightarrow x=8\)
d: \(\Leftrightarrow\left|x-2\right|=0,5x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=8\\\left(0,5x-4-x+2\right)\left(0,5x-4+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=8\\\left(-0,5x-2\right)\left(1,5x-6\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
24 tháng 7 2019
a) (x - 1)3 + (2 - x)(4 + 2x + x2) + 3x(x + 2) = 12
<=> x3 - 2x2 + x - x2 + 2x - 1 + 8 + 4x + 2x2 - 4x - 2x2 + 3x2 + 6x = 17
<=> 9x + 7 = 17
<=> 9x = 17 - 7
<=> 9x = 10
<=> x = \(\frac{10}{9}\)
b) (x + 2)(x2 - 2x + 4) - x(x2 - 2) = 15
<=> x3 - 2x2 + 4x + 2x2 - 4x + 8 - x3 + 2x = 15
<=> 2x + 8 = 15
<=> 2x = 15 - 8
<=> 2x = 7
<=> x = \(\frac{7}{2}\)
c) (x - 3)3 - (x - 3)(x2 + 3x + 9) + 9(x2 + 1)2 = 15
<=> x3 + 45x - 18 - x3 - 3x2 - 9x + 3x2 + 9x + 27 = 15
<=> 45x + 9 = 15
<=> 45x = 15 - 9
<=> 45x = 6
<=> x = \(\frac{6}{45}\)
d) x(x - 5)(x + 5) - (x + 2)(x2 - 2x + 4) = 3
<=> x3 - 25x - x3 + 2x2 - 4x - 8 = 3
<=> -25x - 8 = 3
<=> -25x = 3 + 8
<=> -25x = 11
<=> x = \(-\frac{11}{25}\)
21 tháng 1 2018
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{3;-\dfrac{5}{2}\right\}\)
\(b,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\Leftrightarrow-\left(3x-2\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(-x-11-2+5x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(4x-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{\dfrac{2}{3};\dfrac{13}{4}\right\}\)
\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-\dfrac{1}{2};3\right\}\)
\(d,\left(x-1\right)\left(2x-1\right)=x\left(1-x\right)\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1+x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{1;\dfrac{1}{3}\right\}\)
\(e,0,5x\left(x-3\right)=\left(x-3\right)\left(1,5x-1\right)\)
\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(1,5x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(0,5x-1,5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{1;3\right\}\)
\(f,\left(x+2\right)\left(3-4x\right)=x^2+4x=4\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-x^2-4x-4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-2;\dfrac{1}{5}\right\}\)
\(g,\left(2x^2+1\right)\left(4x-3\right)=\left(x-12\right)\left(2x^2+1\right)\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(x-12\right)\left(2x^2+1\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\forall x\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\\x=-3\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-3\right\}\)
\(h,2x\left(x-1\right)=x^2-1\)
\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy nghiệm của pt là \(S=\left\{1\right\}\)
a) \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x.\left(2x-9\right)-x.3\left(x-5\right)=0\)
\(\Leftrightarrow x.\left[\left(2x-9\right)-3\left(x-5\right)\right]=0\)
\(\Leftrightarrow x.\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x.\left(6-x\right)=0\)
\(\Leftrightarrow S=\left\{0;6\right\}\)
b) \(0,5x\left(x-3\right)=\left(x-3\right)\left(1,5x-1\right)\)
\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(1,5x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left[0,5x-\left(1,5x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(0,5x-1,5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(1-x\right)=0\)
\(+x-3=0\Rightarrow x=3\)
\(+1-x=0\Rightarrow x=1\)
\(\Rightarrow S=\left\{1;3\right\}\)
c) \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow\left(3x-15\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(3-2x\right)\left(x-5\right)=0\)
\(\Rightarrow3-2x=\frac{3}{2}\Rightarrow x-5\Rightarrow x=5\)
\(\Rightarrow S=\left\{5;\frac{3}{2}\right\}\)
a)
\(x\left(2\times-9\right)=3\times\left(\times-5\right)\)
\(\text{⇔}x.\left(2\times-9\right)-x.3\left(x-5\right)=0\)
\(\text{⇔}x.[\left(2\times-9\right)-3\left(x-5\right)]=0\)
\(\text{⇔}x.\left(2x-9-3x+15\right)=0\)
\(\text{⇔}x.\left(6-x\right)=0\)
\(\text{⇔}x=0\) hoặc \(6-x=0+6-x=0\)
\(\text{⇔}x=6\)
Vậy tập nghiệm của phương trình là \(S=\left\{0;6\right\}\) BIẾT MỖI CÂU A :))