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)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

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

\(\left(x^2-5\right)\left(x^2+1\right)=0\)

<=> \(\hept{\begin{cases}x^2-5=0\\x^2+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x^2=5\\x^2=-1\end{cases}}\)

<=> \(\hept{\begin{cases}x=\sqrt{5};x=-\sqrt{5}\\x\in\varnothing\end{cases}}\)

câu còn lại tương tự nha

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!

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) tính thường

b) \(\left(x-1\right)\left(x+2\right)< 0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1>0\\x+2< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -2\end{cases}}\Leftrightarrow1< x< -2\left(ktm\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x-1< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 1\\x>-2\end{cases}}\Leftrightarrow-2< x< 1\left(tm\right)\)

vậy

c)\(\left(x+\frac{3}{5}\right)\left(x+1\right)< 0\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{5}< 0\\x+1>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< -\frac{3}{5}\\x>-1\end{cases}}\Leftrightarrow-1< x< -\frac{3}{5}\left(tm\right)\)

d) \(\left(x-\frac{1}{3}\right)\left(x+\frac{2}{5}\right)>0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>\frac{1}{3}\\x>-\frac{2}{5}\end{cases}}\Leftrightarrow x>\frac{1}{3}\left(tm\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< \frac{1}{3}\\x< -\frac{2}{5}\end{cases}}\Leftrightarrow x< \frac{-2}{5}\left(tm\right)\)

vậy ...

a) 5/2 - x + 4/5 = 2/3 + 4/7

<=> 33/10 - x = 26/21

<=> x = 433/210

b) ( x - 1 )( x + 2 ) < 0 ( cái " x " kia là nhân à :v )

Xét 2 trường hợp

1.\(\hept{\begin{cases}x-1>0\\x+2< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>1\\x< -2\end{cases}}\)( loại )

2. \(\hept{\begin{cases}x-1< 0\\x+2>0\end{cases}}\Rightarrow\hept{\begin{cases}x< 1\\x>-2\end{cases}}\Rightarrow-2< x< 1\)

Vậy -2 < x < 1

c) ( x + 3/5 )( x + 1 ) < 0

Xét hai trường hợp :

1. \(\hept{\begin{cases}x+\frac{3}{5}< 0\\x+1>0\end{cases}}\Rightarrow\hept{\begin{cases}x< -\frac{3}{5}\\x>-1\end{cases}}\Rightarrow-1< x< -\frac{3}{5}\)

2. \(\hept{\begin{cases}x+\frac{3}{5}>0\\x+1< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>-\frac{3}{5}\\x< -1\end{cases}}\)( loại )

Vậy -1 < x < -3/5

d) ( x - 1/3 )( x + 2/5 ) > 0

Xét hai trường hợp :

1.\(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\Rightarrow\hept{\begin{cases}x>\frac{1}{3}\\x>-\frac{2}{5}\end{cases}}\Rightarrow x>\frac{1}{3}\)

2.\(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< \frac{1}{3}\\x< -\frac{2}{5}\end{cases}\Rightarrow}x< -\frac{2}{5}\)

Vây \(\orbr{\begin{cases}x>\frac{1}{3}\\x< -\frac{2}{5}\end{cases}}\)

1, x(x - 5) - 4x + 20 = 0

=> x(x - 5) - 4(x - 5) = 0

=> (x - 4)(x - 5) = 0

=> x - 4 = 0 hoặc x - 5 = 0

=> x = 4 hoặc x = 5

=> x thuộc {4; 5}

2, 3(x + 1) + x(x + 1) 

= (3 + x)(x + 1)

3, 2x³ + x = 0

=> x(2x² + 1) = 0

=> x = 0 hoặc 2x² + 1 = 0

=> x = 0 hoặc 2x² = -1

=> x = 0 hoặc x² = -1/2 (vô lí vì x² > hoặc = 0 với mọi x)

=> x = 0

4, x³ - 16x = 0

=> x(x² - 16) = 0

=> x = 0 hoặc x² - 16 = 0

=> x = 0 hoặc x² = 16

=> x = 0 hoặc x = 4 hoặc x = -4

=> x thuộc {-4; 0; 4}

5, x² + 6x = -9

=> x² + 6x + 9 = 0

=> x² + 2.3.x + 3² = 0

=> (x + 3)² = 0

=> x + 3 = 0

=> x = -3

6, x⁴ - 2x³ + 10x² - 20x = 0

=> x²(x² + 10) - 2x(x² + 10) = 0

=> (x² + 2x)(x² + 10) = 0

=> x(x +2)(x² + 10) = 0

-TH1: x = 0

-TH2: x + 2 = 0 => x = -2

-TH3: x² + 10 = 0 => x² = -10 (vô lí vì x² > hoặc = 0 với mọi x)

=> x thuộc {0; -2}

7, (2x - 3)² = (x + 5)²

-TH1: 2x - 3 = x + 5

=> x = 8

- TH2: - 2x + 3 = x + 5

=> -3x = 2

=> x = \(\frac{-2}{3}\)

- TH3: 2x - 3 = - x - 5

=> 3x = -2

=> x = \(\frac{-2}{3}\)

- TH4: - 2x + 3 = - x - 5

=> -x = -8

=> x = 8`

=> x thuộc {\(\frac{-2}{3}\); 8}