e)

A = \(\frac{x+5}{x-2}\) = \(\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)

Muốn A nguyên thì:

=> \(\frac{7}{x-2}\) ∈ Z

=> 7 ⋮ x - 2

=> x - 2 ∈ Ư (7)

=> x - 2 ∈ { 1; 7; -1; -7 }

=> x ∈ { 3; 9; -5; 1 }

a) (5x - 1)(2x - 1/3) = 0

\(\Rightarrow5x-1=0\) hoặc \(2x-\frac{1}{3}=0\)

\(\Rightarrow\left[{}\begin{matrix}5x=0+1\\2x=0+\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1:5=\frac{1}{5}\\x=\frac{1}{3}:2=\frac{1}{3}.\frac{1}{2}=\frac{1}{6}\end{matrix}\right.\)

Vậy x = 1/5 hoặc x = 1/6

Lời giải:

a)

PT $\Leftrightarrow (x^2+4x-5)-(x^2-7x+10)=0$

$\Leftrightarrow 11x-15=0$

$\Leftrightarrow x=\frac{15}{11}$

b)

PT $\Leftrightarrow (x^2+4x-21)-(x^2+2x-8)=0$

$\Leftrightarrow 2x-13=0$

$x=\frac{13}{2}$

c)

PT $\Leftrightarrow (x^2-13x+42)-(x^2+4x)-(5x-5)=0$

$\Leftrightarrow -22x+47=0\Rightarrow x=\frac{47}{22}$

a: =>x+3>0

hay x>-3

b: =>4-x<0

hay x>4

c: =>x²-1=0 hoặc x+5=0

hay \(x\in\left\{1;-1;-5\right\}\)

a.(2x - 5)(3x + 4) - x(6x - 5) = 4

⇔ 6x² +8x -15x-20-6x²+5x=4

⇔-2x=24

⇔ x=-12

vậy x=12

b.(x - 2)² + x(x - 2) = 0

⇔(x-2)(x-2+x)=0

⇔(x-2) (2x-2)=0

⇔ (x-2)2(x-2)=0

⇔(x-2)².2=0

⇔(x-2)²=0

⇔x-2=0

⇔x=2

vậy x=2

c.(x³ + 4x² - x - 4) : (x + 4) = 0

⇔[(x³+4x²)-(x+4)] :(x+4)=0

⇔ [x²(x+4)-(x+4)] :(x+4)=0

⇔ (x+4)(x²-1):(x+4)=0

⇔(x-1)(x+1)=0

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

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

chẳng có đề bài biết làm ntn

bài 1: 5(x-2)=4(x-3) và m(x-2)-x(m-4)= 0

Xét 5(x-2)=4(x-3)

\(\Leftrightarrow\) \(5x-10-4x+12=0\)

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

\(\Leftrightarrow x=-2\)

Xét m(x-2)-x(m-4)= 0

\(\Leftrightarrow mx-2m-mx+4x=0\)

\(\Leftrightarrow4x-2m=0\left(1\right)\)

Thay x = -2 vào pt (1), ta có:

\(4\cdot\left(-2\right)-2m=0\)

\(\Leftrightarrow-8-2m=0\)

\(\Leftrightarrow-2m=8\)

\(\Leftrightarrow m=-4\)

Vậy m = -4 thì 2 pt 5(x-2)=4(x-3) và m(x-2)-x(m-4)= 0 tương đương

bài 2: 4(x-3)=3(x-5) và m(x-3)-x(m-9)=0

Xét 4(x-3)=3(x-5)

\(\Leftrightarrow4x-12-3x+15=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Xét m(x-3)-x(m-9)=0

\(\Leftrightarrow mx-3m-mx+9x=0\)

\(\Leftrightarrow9x-3m=0\left(2\right)\)

Thay x = -3 vào pt (2), ta có:

\(9\cdot\left(-3\right)-3m=0\)

\(\Leftrightarrow-27-3m=0\)

\(\Leftrightarrow-3m=27\)

\(\Leftrightarrow m=-9\)

Vậy m = -9 thì 2 pt 4(x-3)=3(x-5) và m(x-3)-x(m-9)=0 tương đương

em cảm ơn nhiều lắm ạ ^^

a: \(\Delta=\left(m-2\right)^2-4\left(m-5\right)\)

=m^2-4m+4-4m+20

=m^2-8m+24

=(m-4)^2+8>0

=>Phương trình luôn có nghiệm

b: \(\Delta=\left[-2\left(m+1\right)\right]^2-4\left(m-4\right)\)

=(2m+2)^2-4(m-4)

=4m^2+8m+4-4m+16

=4m^2+4m+20

=4m^2+4m+1+19

=(2m+1)^2+19>0

=>Phương trình luôn có nghiệm

1.

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

\(\Rightarrow \left[\begin{matrix} x=\frac{-1}{2}\\ x^2=-2< 0(\text{vô lý})\end{matrix}\right.\)

Vậy \(x=-\frac{1}{2}\)

2.\((x^2+4)(7x-3)=0\Rightarrow \left[\begin{matrix} x^2+4=0\\ 7x-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x^2=-4< 0(\text{vô lý})\\ x=\frac{3}{7}\end{matrix}\right.\)

Vậy \(x=\frac{3}{7}\)

3.

\((x-5)(3-2x)(3x+4)=0\)

\(\Rightarrow \left[\begin{matrix} x-5=0\\ 3-2x=0\\ 3x+4=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=5\\ x=\frac{3}{2}\\ x=-\frac{4}{3}\end{matrix}\right.\)

4.

\((x-2)(3x+5)=(2x-4)(x+1)\)

\(\Leftrightarrow (x-2)(3x+5)-(2x-4)(x+1)=0\)

\(\Leftrightarrow (x-2)(3x+5)-2(x-2)(x+1)=0\)

\(\Leftrightarrow (x-2)[(3x+5)-2(x+1)]=0\)

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

5.

\((2x+5)(x-4)=(x-5)(4-x)\)

\(\Leftrightarrow (2x+5)(x-4)-(x-5)(4-x)=0\)

\(\Leftrightarrow (2x+5)(x-4)+(x-5)(x-4)=0\)

\(\Leftrightarrow (x-4)[(2x+5)+(x-5)]=0\)

\(\Leftrightarrow (x-4).3x=0\)

\(\Rightarrow \left[\begin{matrix} x-4=0\\ 3x=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=4\\ x=0\end{matrix}\right.\)