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26 tháng 10 2018

0,5x(x – 3) = (x – 3)(1,5x – 1)

⇔ 0,5x(x – 3) – (x – 3)(1,5x – 1) = 0

⇔ (x – 3).[0,5x – (1,5x – 1)] = 0

⇔ (x – 3)(0,5x – 1,5x + 1) = 0

⇔ (x – 3)(1 – x) = 0

⇔ x – 3 = 0 hoặc 1 – x = 0

+ x – 3 = 0 ⇔ x = 3.

+ 1 – x = 0 ⇔ x = 1.

Vậy phương trình có tập nghiệm S = {1; 3}.

21 tháng 1 2022

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) \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow\left(3x-15\right)-2x\left(x-5\right)=0\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left(3-2x\right)\left(x-5\right)=0\)

\(\Rightarrow3-2x=\frac{3}{2}\Rightarrow x-5\Rightarrow x=5\)

\(\Rightarrow S=\left\{5;\frac{3}{2}\right\}\)

21 tháng 1 2022

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 :))

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#

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

Phải toán 8 ko zậy

0,5x( x - 3 )= (x- 3)(1,5x- 1)

=>0,5x(x -3) - (x -3)(1,5x -1)=0

=>(x -3)(0,5x - 1,5x + 1)= 0

=>(x -3)( 1 -x) =0 

 Từ đây tự làm nha

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=\(\frac{-1}{2}\)

19 tháng 1 2016

<=>x(2x-9)-3x(x-5)=0 

<=>(2x2-9)-(3x2-15x)=0

<=>[(2x2-9x-3x+15x)=0

<=>(-x2+6)=0

<=>-x2+6=0

<=>-x2=-6

<=>x2=6

vô lívì-x+6 lớn hơn hoặc bang82 với mọi x thuộc R

 

13 tháng 12 2017

<=>0,5x(x-3)-(x-3)(2,5x-4)=0

<=>(x-3)(0,5x-2,5x+4)=0

<=>(x-3)(-2x+4)=0

<=>(x-3)*2(4-x)=0

<=>x-3=0 hoặc 4-x=0

<=>x=3 hoặc x=4

a: \(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(\dfrac{1}{2}x\right)^2-\left(2x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(\dfrac{1}{2}x-2x+3\right)\left(\dfrac{1}{2}x+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(3-\dfrac{3}{2}x\right)\left(\dfrac{5}{2}x-3\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{6}{5}\right\}\)

b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{4}{3}\\\left(3x+4\right)^2-\left(2x\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{4}{3}\\\left(5x+4\right)\left(x+4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow x=-\dfrac{4}{5}\)

c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=12\\\left(5x-x+12\right)\left(5x+x-12\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=12\\\left(4x+12\right)\left(6x-12\right)=0\end{matrix}\right.\)

hay \(x\in\varnothing\)

d: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{10}{3}\\\left(2,5x-1,5x-5\right)\left(2,5x+1,5x+5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{10}{3}\\\left(x-5\right)\left(4x+5\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{5}{4};5\right\}\)

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: \(\Leftrightarrow\left|x-2\right|=0,5x-4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=8\\\left(0,5x-4-x+2\right)\left(0,5x-4+x-2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=8\\\left(-0,5x-2\right)\left(1,5x-6\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)