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#

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

a) \(0,5x\left(2x-9\right)=1,5x\left(x-5\right)\Leftrightarrow x^2-4,5x=1,5x^2-7,5x\)

\(\Leftrightarrow x^2-1,5x^2=4,5x-7,5x\Leftrightarrow-0.5x=-3x\Leftrightarrow x=6\)

Vậy phương trình có tập nghiệm S = { 6 }

 b) \(5\left(x-1\right)-\left(2x-5\right)=16-x\Leftrightarrow5x-5-2x+5=16-x\)

\(\Leftrightarrow5x-2x+x=5-5+16\Leftrightarrow4x=16\Leftrightarrow x=4\)

Vậy phương trình có tập nghiệm S = { 4 }

d)\(-ĐKXĐ:\hept{\begin{cases}x+1\ne0\\x-2\ne0\\\left(x+1\right)\left(x-2\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+1\ne0\\x-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

Ta có:  \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\Leftrightarrow\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\left(x+2\right)}=\) \(\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow2\left(x-2\right)-\left(x+1\right)=3x-11\Leftrightarrow2x-4-x-1=3x-11\)

\(\Leftrightarrow2x-x+3x=4+1-11\Leftrightarrow4x=-6\Leftrightarrow x=-\frac{3}{2}\)

Vậy phương trình có tập nghiệm S = { -3/2 }

\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\\ \left(3x+2\right)\left(x^2-1\right)-\left(9x^2-4\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left[\left(x-1\right)-\left(3x-2\right)\right]=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\\ \left[{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)

\(b.x\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\\ x\left(x^2-9\right)-\left(x^3+8\right)=0\\ x^3-9x-x^3-8=0\\ -9x-8=0\\ -9x=8\\ x=\frac{-8}{9}\)

\(c.2x\left(x-3\right)+5\left(x-3\right)=0\\ \left(x-3\right)\left(2x+5\right)=0\\ \left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-5}{2}\end{matrix}\right.\)

\(d.\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\\ \left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\\ \left(3x-1\right)\left[\left(x^2+2\right)-\left(7x-10\right)\right]=0\\ \left(3x-1\right)\left(x^2+2-7x+10\right)=0\\ \left(3x-1\right)\left(x^2-7x+12\right)=0\\ \left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \left(3x-1\right)\left[\left(x^2-4x\right)+\left(-3x+12\right)\right]=0\\ \left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}3x-1=0\\x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\\x=3\end{matrix}\right.\)

\(e.\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\\ \left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\\ \left(x+2\right)\left[\left(3-4x\right)-\left(x+2\right)\right]=0\\ \left(x+2\right)\left(3-4x-x-2\right)=0\\ \left(x+2\right)\left(1-5x\right)=0\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{1}{5}\end{matrix}\right.\)

\(f.x\left(2x-7\right)-4x+14=0\\ x\left(2x-7\right)-2\left(2x-7\right)=0\\ \left(2x-7\right)\left(x-2\right)=0\\ \left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)

\(g.3x-15=2x\left(x-5\right)\\ 3\left(x-5\right)=2x\left(x-5\right)\\ 3\left(x-5\right)-2x\left(x-5\right)=0\\ \left(x-5\right)\left(3-2x\right)=0\\ \left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)

\(h.\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \left(2x+1\right)\left[\left(3x-2\right)-\left(5x-8\right)\right]=0\\ \left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \left(2x+1\right)\left(6-2x\right)=0\\ \left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=3\end{matrix}\right.\)

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

<=>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

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

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

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

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 = \(-\frac{11}{25}\)

a)\(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)

\(=>x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)

\(=>9x+7=17=>9x=10=>x=\frac{10}{9}\)

Bạn đăng 1 lần 4 câu làm mik nản quá!

b)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(=>x^3+8-x^3+2x=15=>2x+8=15=>2x=7=>x=\frac{7}{2}\)

\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)

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

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

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 \(S=\left\{1\right\}\)