a: =>3x=6

=>x=2

b: =>\(\sqrt{2x+1}\left(\sqrt{2x-1}+1\right)=0\)

=>2x+1=0

=>x=-1/2

c: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

=>\(\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)\)

=>\(x+\sqrt{x}-6=x-1\)

=>căn x-6=-1

=>căn x=-1+6=5

=>x=25

a.\(\dfrac{1}{3}\) + x  = \(\dfrac{5}{6}\)

       x = \(\dfrac{5}{6}\) - \(\dfrac{1}{3}\)

      x = \(\dfrac{1}{2}\)

b. | x-1| - \(\dfrac{2}{5}\) = \(\dfrac{11}{10}\) 

   | x-1|        = \(\dfrac{11}{10}\) + \(\dfrac{2}{5}\)

  |x-1|        = \(\dfrac{3}{2}\)

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

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

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

c, \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = 1

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

             \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = \(\dfrac{2}{3}\)

                   \(\dfrac{x}{2}\) + 3 = 1

                   \(\dfrac{x}{2}\)       = 1 - 3

                    \(\dfrac{x}{2}\)    = -2

                     \(x\) = -4

d, \(\dfrac{x+2}{3}\) = \(\dfrac{27}{x+2}\)

(x+2)² = 27.3

(x+2) =9²

\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)

a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)

\(\Leftrightarrow18x+16=7\)

hay \(x=-\dfrac{1}{2}\)

c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)

hay x=0

\(a,\Rightarrow5x+3x^2-3x^2-x+2=6\\ \Rightarrow4x=4\Rightarrow x=1\\ b,\Rightarrow\left(2x+\dfrac{1}{2}-1+2x\right)\left(2x+\dfrac{1}{2}+1-2x\right)=2\\ \Rightarrow\dfrac{3}{2}\left(4x-\dfrac{1}{2}\right)=2\\ \Rightarrow6x-\dfrac{3}{4}=2\\ \Rightarrow6x=\dfrac{11}{4}\\ \Rightarrow x=\dfrac{11}{24}\\ c,\Rightarrow\left(x+3\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

a) A=(x-4)²+ |y-1|-6

Ta thấy:

(x-4)² ≥ 0 ∀ x

|y-1| ≥ 0 ∀ y

⇒ (x-4)²+ |y-1|  ≥ 0 ∀ x, y

⇒ (x-4)²+ |y-1|-6  ≥ -6 ∀ x, y

⇒ A ≥ -6 ∀ x, y

Dấu '=' xảy ra khi: \(\left[{}\begin{matrix}x-4=0\\y-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=4\\y=1\end{matrix}\right.\)

Vậy Min A = -6 tại x=4, y = 1

b) B= (x²-1)⁴+2.|2y-4|-3

Ta thấy:

(x²-1)⁴ ≥ 0 ∀ x

|2y-4| ≥ 0 ∀ y

⇒ 2|2y-4| ≥ 0 ∀ y

⇒ (x²-1)⁴+2.|2y-4| ≥ 0 ∀ x, y

⇒ (x²-1)⁴+2.|2y-4|-3 ≥ -3 ∀ x, y

Vậy Min B = -3 tại x=\(\pm\)1, y = 2

`a)[2x-3]/4-[4x-5]/3=[5-x]/6`

`<=>3(2x-3)-4(4x-5)=2(5-x)`

`<=>6x-9-16x+20=10-2x`

`<=>8x=1`

`<=>x=1/8`

(`b->c` mk đã làm r).

a: =>x+7/4=6:2/3=9

=>x=29/4

b: =>x:5/3=7/5

=>x=7/5*5/3=7/3

c:=>x+1/6=5/3

=>x=10/6-1/6=3/2

d: =>x+4/5=4/5+3/7+3/5

=>x=3/7+3/5=36/35

e: =>x/35=4/5-5/7=3/35

=>x=3

f: =>13/28+x=1/2

=>x=1/28

g: =>1/3-x=1/9

=>x=2/9

giúp mình vớiiii

x=1/6

x=1/4

a) x = 1/6

b) x = 1/4

HT

a) 2/3 + x = 5/6

X = 5/6 - 2/3

X = 1/6

b)  x : 5/6 = 3/10

X = 5/6 x 3/10

X = 17/15