1) Do x ∈ Z và 0 < x < 3

⇒ x ∈ {1; 2}

2) Do x ∈ Z và 0 < x ≤ 3

⇒ x ∈ {1; 2; 3}

3) Do x ∈ Z và -1 < x ≤ 4

⇒ x ∈ {0; 1; 2; 3; 4}

a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)

=>(x+5)(x-3)+8=x^2-1

=>x^2+2x-15+8=x^2-1

=>2x-7=-1

=>x=3(loại)

b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)

=>(x-4)(x+1)+x^2+3+5(x-1)=0

=>x^2-3x-4+x^2+3+5x-5=0

=>2x^2+2x-6=0

=>x^2+x-3=0

=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)

e: =>x^2-2x+1+2x+2=5x+5

=>x^2+3=5x+5

=>x^2-5x-2=0

=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)

g: (x-3)(x+4)*x=0

=>x=0 hoặc x-3=0 hoặc x+4=0

=>x=0;x=3;x=-4

a) (x - 3)(5 - 2x) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\5-2x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=\frac{5}{2}\end{matrix}\right.\)

b) (x + 5)(x - 1) - 2x(x - 1) = 0

<=> (x - 1)(x + 5 - 2x) = 0

<=> (x - 1)(5 - x) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\5-x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

c) 5(x + 3)(x - 2) - 3(x + 5)(x - 2) = 0

<=> (x - 2)[5(x + 3) - 3(x + 5)] = 0

<=> (x - 2)(5x + 3 - 3x - 15) = 0

<=> (x - 2)(2x - 12) = 0

<=> \(\left[{}\begin{matrix}x-2=0\\2x-12=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

d) (x - 6)(x + 1) - 2(x + 1) = 0

<=> (x + 1)(x - 6 - 2) = 0

<=> (x + 1)(x - 8) = 0

<=> \(\left[{}\begin{matrix}x+1=0\\x-8=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)

Câu e thì để mình nghĩ đã :)

#Học tốt!

Giúp luôn Đức Hải Nguyễn câu e:

e, (x - 1)² + 2(x - 1)(x + 2) + (x + 2)² = 0

\(\Leftrightarrow\) (x - 1 + x + 2)² = 0

\(\Leftrightarrow\) (2x + 1)² = 0

\(\Leftrightarrow\) 2x + 1 = 0

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

Vậy S = {\(\frac{-1}{2}\)}

Chúc bn học tốt!!

câu e nó là hàng đẳng thức đó (a+b)^2 với a là (x-1) B là x+2 ta có  (a+b)^2 = a^2+2.a.b+b^2

a) \(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)

\(\Rightarrow3x-1=0\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\)

\(\frac{-1}{2}x+5=0\Rightarrow\frac{-1}{2}x=-5\Rightarrow x=10\)

b) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)

\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)

\(\Rightarrow3x-5x-x=\frac{1}{5}+\frac{3}{2}+3\)

\(-3x=\frac{47}{10}\)

\(x=\frac{-47}{30}\)

c) \(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)

\(-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)

\(-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}+1-\frac{1}{3}\)

\(-7x=\frac{-1}{6}\)

\(x=\frac{1}{42}\)

d) \(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)

\(\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)

\(\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)

\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(3x=\frac{1}{6}\)

\(x=\frac{1}{18}\)

Học tốt nhé bn!
 

x = \(\frac{1}{18}\)nha

x=1/8

 HT!

a)  (x - 3)² - 5.(x - 2) + 5 = 0.

<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0

<=> x^2 - 11x + 24 = 0

<=> (x-3)(x-8)=0

<=> x = 3 hoặc x = 8

b) (2x - 1)² - 3.(x - 2).(x + 2) - 25 = 0.

<=> 4x^2 - 4x + 1 - 3x^2 + 12 - 25 = 0

<=> x² - 4x - 12 = 0

<=> (x+2)(x-6) = 0

<=> x = -2 hoặc x = 6

d)  x² - 4x + 5 = 0.

<=> (x - 2)² = -1 (vô lý)

Vậy phương trình vô nghiệm

1.

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

2.

\(\left(x^2-9\right)+2\left(x-3\right)=0\)

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

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

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

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

3.

\(\left(2x+1\right)^2+\left(x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x+1+x-1\right)=0\)

\(\Leftrightarrow\left(2x+1\right).3x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\2x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

4.

\(\left(x-1\right)\left(x+3\right)+\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-1+x+3\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

5.

\(\left(x+5\right)^2-16x^2=0\)

\(\Leftrightarrow\left(x+5+4x\right)\left(x+5-4x\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)

6.

\(x^3-2x^2-x+2=0\)

\(\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)

a: =>2/3-1/3x+1/2-x-1/2=5

=>-4/3x+2/3=5

=>-4/3x=13/3

=>x=-13/4

b: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x-\dfrac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)

c: =>1/3x+3/5x+3/5=0

=>14/15x=-3/5

=>x=-3/5:14/15=-3/5x15/14=-45/70=-9/14

d: =>x>8/2

e: =>x:1/45=1/2

=>x=1/90

g: =>1/2:x=-2/15

=>x=-1/2:2/15=-15/4

a)\(3x\left(x-1\right)+x-1=0\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\Leftrightarrow\hept{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)

\(S=\left\{1;\frac{1}{3}\right\}\)

b)\(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\Leftrightarrow\hept{\begin{cases}2-x=0\\x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=-3\end{cases}}}\)

\(S=\left\{2;-3\right\}\)