a: Ta có: \(3x-\left(3x+2\right)=x+3\)

\(\Leftrightarrow x+3=-2\)

hay x=-5

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

\(\Leftrightarrow15x-3+8x-4=18x\)

\(\Leftrightarrow5x=7\)

hay \(x=\dfrac{7}{5}\)

\(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\\ \Leftrightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x\left(2+3\right)=\dfrac{3}{4}-\dfrac{1}{2}+6\\ \Leftrightarrow5x=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{25}{4}:5=\dfrac{5}{4}\\ ---\\ 4^{x+2}+4^x=272\\ \Leftrightarrow4^x\left(4^2+1\right)=272\\ \Leftrightarrow4^x.17=272\\ \Leftrightarrow4^x=\dfrac{272}{17}=16=4^2\\ Vậy:x=2\\ ----\\ \left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1,2-5x=0\\2,125+0,5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1,2\\0,5x=-2,125\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}=0,24\\x=\dfrac{-2,125}{0,5}=-4,25\end{matrix}\right.\)

a) \(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\)

\(\Rightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow5x-6=\dfrac{3}{4}-\dfrac{1}{2}\)

\(\Rightarrow5x-6=\dfrac{1}{4}\)

\(\Rightarrow5x=\dfrac{1}{4}+6\)

\(\Rightarrow5x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:5\)

\(\Rightarrow x=\dfrac{5}{4}\)

b) \(4^{x+2}+4^x=272\)

\(\Rightarrow4^x\cdot4^2+4^x\cdot1=272\)

\(\Rightarrow4^x\cdot\left(16+1\right)=272\)

\(\Rightarrow4^x\cdot17=272\)

\(\Rightarrow4^x=16\)

\(\Rightarrow4^x=4^2\)

\(\Rightarrow x=2\)

c) \(\left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}1,2-5x=0\\\dfrac{15}{8}+\dfrac{1}{2}x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=1,2\\\dfrac{1}{2}x=-\dfrac{15}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}\\x=-\dfrac{15}{8}:\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{25}\\x=-\dfrac{15}{4}\end{matrix}\right.\)

a/ (x-1)²-(4x+3)(2-x)=x²-2x+1-(8x-4x²+6-3x)

=x²-2x+1-8x+4x²-6+3x=5x²-7x-6

b/ (15x³y² - 6x²y³) : 3x²y² = 5x - 2y

c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

a, 3x - 7 = 0

<=> 3x = 7

<=> x = 7/3

b, 8 - 5x = 0

<=> -5x = -8

<=> x = 8/5

c, 3x - 2 = 5x + 8

<=> -2x = 10

<=> x = -5

e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)

`a ) 3x - 7 = 0`

`\(\Leftrightarrow \) 3x = 7`

`\(\Leftrightarrow \) x = 7/3`

Vậy `S = {-7/3}`

Câu 1:

a) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)

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

\(\Leftrightarrow12x-10x-4=21-9x\)

\(\Leftrightarrow11x=25\)

\(\Leftrightarrow x=\dfrac{25}{11}\)

b) \(\left(3x-1\right)\left(x-3\right)\left(7-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\Leftrightarrow x=\dfrac{1}{3}\\x-3=0\Leftrightarrow x=3\\7-2x=0\Leftrightarrow x=3,5\end{matrix}\right.\)

c) \(\left|3x\right|=4x+8\) (1)

Ta có: \(\left|3x\right|=3x\Leftrightarrow3x\ge0\Leftrightarrow x\ge0\)

\(\left|3x\right|=-3x\Leftrightarrow3x< 0\Leftrightarrow x< 0\)

Với \(x\ge0\), phương trình (1) có dạng:

\(3x=4x+8\Leftrightarrow-x=8\Leftrightarrow x=-8\)

(không thoả mãn điều kiện) \(\rightarrow\) loại

Với \(x< 0\), phương trình (1) có dạng:

\(-3x=4x+8\Leftrightarrow-7x=8\Leftrightarrow x=-\dfrac{8}{7}\)

(thoả mãn điều kiện) \(\rightarrow\) nhận

Vậy phương trình đã cho có 1 nghiệm \(x=-\dfrac{8}{7}\)

Câu 2:

\(2x\left(6x-1\right)\ge\left(3x-2\right)\left(4x+3\right)\)

\(\Leftrightarrow12x^2-2x\ge12x^2+9x-8x-6\)

\(\Leftrightarrow-3x\ge-6\)

\(\Leftrightarrow x\le2\)

Vậy bất phương trình đã cho có nghiệm \(x\le2\)

\(1.\)

\(a.\) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)

\(\Leftrightarrow\dfrac{24x}{24}-\dfrac{4\left(5x+2\right)}{24}=\dfrac{6\left(7-3x\right)}{24}\)

\(\Leftrightarrow24x-4\left(5x+2\right)=6\left(7-3x\right)\)

\(\Leftrightarrow24x-20x-8=42-18x\)

\(\Leftrightarrow24x-20x+18x=42+8\)

\(\Leftrightarrow22x=50\)

\(\Leftrightarrow x=\dfrac{50}{22}=\dfrac{25}{11}\)

Vậy : ...........

\(b.\) \(\left(3x-1\right)\left(x-3\right)\left(7-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\7-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy : ..............

\(c.\) \(\left|3x\right|=4x+8\) \(\left(1\right)\)

* Với \(3x< 0\Rightarrow x< 0\)

\(\left|3x\right|=-3x\)

Khi đó : \(\left(1\right)\Rightarrow-3x=4x+8\)

\(\Rightarrow-3x-4x=8\)

\(\Rightarrow-7x=8\)

\(\Rightarrow x=-\dfrac{8}{7}\) ( Thoả mãn điều kiện )

* Với \(3x\ge0\Rightarrow x\ge0\)

\(\left|3x\right|=3x\)

Khi đó : \(\left(1\right)\Rightarrow3x=4x+8\)

\(\Rightarrow3x-4x=8\)

\(\Rightarrow-x=8\)

\(\Rightarrow x=-8\) ( Không thoả mãn điều kiện )

Vậy : ..............

\(2.\)

\(2x\left(6x-1\right)\ge\left(3x-2\right)\left(4x+3\right)\)

\(\Leftrightarrow12x^2-2x\ge\left(12x^2+9x-8x-6\right)\)

\(\Leftrightarrow12x^2-2x-12x^2-9x+8x\ge-6\)

\(\Leftrightarrow-3x\ge-6\)

\(\Leftrightarrow x\le2\)

Vậy : ..............

a: =>x^2+4x-4x+1=0

=>x^2+1=0

=>Loại

b: =>2x-6+4=2x+2

=>-2=2(loại)

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

=>6-1=1

=>5=1(loại)

d =>x+3=0

=>x=-3(loại)

e: =>x^2-3x^2+3x-3x-2=0

=>-2x^2-2=0

=>x^2+1=0

=>Loại

a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)

a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1

=>1,7x=6,7

hay x=67/17

b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)

=>150x+120-45x-75=96x+216-40x+360

=>105x+45=56x+576

=>49x=531

hay x=531/49