\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

giúp mik với

\(\left(2x-5\right)^3-\left(3x-4\right)^3+\left(x+1\right)^3=0\)

\(\Leftrightarrow\left(2x-5-3x+4\right)\left[\left(2x-5\right)^2+\left(2x-5\right)\left(3x-4\right)+\left(3x-4\right)^2\right]+\left(x+1\right)^3=0\)

\(\Leftrightarrow-\left(x+1\right)\left[\left(2x-5\right)^2+\left(2x-5\right)\left(3x-4\right)+\left(3x-4\right)^2\right]+\left(x+1\right)^3=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2-\left(2x-5\right)^2-\left(2x-5\right)\left(3x-4\right)-\left(3x-4\right)^2=0\left(1\right)\end{cases}}\)

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

\(\Leftrightarrow\left(3x-4\right)\left(6-x-2x+5-3x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-4=0\\-6x+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{-5}{2}\end{cases}}}\)

giúp mình nha mn

giúp mk với

Đặt a = 2x - 5; b = 3x - 4 => x + 1 = b - a

Khi đó ta có pt: a³ - b³ + (b - a)³ = 0

<=> (b - a)³ - (b³ - a³) = 0

<=> b³ - a³ - b³ + a³ - 3ab(b - a)= 0

<=> 3ab(b - a) = 0 <=> a = 0 hoặc b = 0 hoặc a= b

* Với a = 0, ta có: 2x - 5 = 0 <=> x = \(\frac{5}{2}\)

* Với b = 0, ta có: 3x - 4 = 0 <=> x = \(\frac{4}{3}\)

* Với a = b, ta có: 2x - 5 = 3x - 4 <=> x = -1

Vậy S = {-1;\(\frac{4}{3}\);\(\frac{5}{2}\)}

a) Trường hợp 1. Xét 4 - 5x = 5 - 6x.

Tìm được x = 1.

Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) đẻ được hỗ trợ tốt hơn. Viết như thế kia rất khó đọc => khả năng bị bỏ qua bài cao.

a: =>3x=3

=>x=1

b: =>12x-2(5x-1)=3(8-3x)

=>12x-10x+2=24-9x

=>2x+2=24-9x

=>11x=22

=>x=2

c: =>2x-3(2x+1)=x-6x

=>-5x=2x-6x-3=-4x-3

=>-x=-3

=>x=3

d: =>2x-5=0 hoặc x+3=0

=>x=5/2 hoặc x=-3

e: =>x+2=0

=>x=-2

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

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

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

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

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

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

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

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

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

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm