a: \(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

2x+1.3y=12x => 2x+1.3y=4x.3x

=> 2x+1.3y=22x.3x => x + 1 = 2x và y = x

=> x = 1 và y = x = 1

 Vậy x=y=1

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x ( 12x - 4 ) - 9x( 4x - 3 ) = 30`

`=> 3x (12x-4) - 3*3x (4x - 3) = 30`

`=> 3x [12x - 4 - 3(4x-3)] = 30`

`=> 3x (12x - 4 - 12x + 9) = 30`

`=> 3x (-4+9)=30`

`=> 3x*5=30`

`=> 3x=6`

`=> x=2`

Vậy, `x=2`

`b)`

`x( 5 - 2x) + 2x( x - 1)`

`=> x(5-2x) + 2x^2 - 2x=15`

`=> 5x - 2x^2 + 2x^2 - 2x =15`

`=> 3x = 15`

`=> x=5`

Vậy, `x=5.`

a: =>36x^2-12x-36x^2+27x=30

=>15x=30

=>x=2

b: =>5x-2x^2+2x^2-2x=15

=>3x=15

=>x=5

a)x.(5-2x)-2x.(1-x)=15
   x [ 5 - 2x -2.(1-x) ] = 15
   x ( 5 - 2x -2 + 2x ) =15
   x . 3 =15
   x = 5
b)(3x+2)²+(1+3x).(1-3x)=2
   9x²+12x+4+1-9x²=2
   12x + 5 = 2
    12x = -3
        x = -1/4

a)\(\Leftrightarrow\)\(5x-2x^2-2x+2x^2=15\)

\(\Leftrightarrow\)\(3x=15\)

\(\Leftrightarrow\)\(x=5\)

b)\(\Leftrightarrow\)\(9x^2+12x+4+1-9x^2-2=0\)

\(\Leftrightarrow\)\(12x+3=0\)

\(\Leftrightarrow\)\(x=-0,25\)

\(a,\sqrt{9x^2}=2x+1\\ \Leftrightarrow\left[{}\begin{matrix}3x=2x+1,\forall x\ge0\\-3x=2x+1,\forall x< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1,\forall x\ge0\left(N\right)\\x=-1,\forall x< 0\left(N\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-1,\forall x+3\ge0\\x+3=1-3x,\forall x+3< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2,\forall x\ge-3\left(N\right)\\x=-\dfrac{1}{2},\forall x< -3\left(L\right)\end{matrix}\right.\Leftrightarrow x=2\)

\(c,\sqrt{x^2-2x+4}=2x-3\left(x\in R\right)\\ \Leftrightarrow x^2-2x+4=\left(2x-3\right)^2\\ \Leftrightarrow x^2-2x+4=4x^2-12x+9\\ \Leftrightarrow3x^2-10x+5=0\\ \Delta=100-4\cdot3\cdot5=40\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10-\sqrt{40}}{6}\\x=\dfrac{10+\sqrt{40}}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{10}}{3}\\x=\dfrac{5+\sqrt{10}}{3}\end{matrix}\right.\)

\(a.\sqrt{9x^2}=2x+1\)

<=> \(\sqrt{9}x=2x+1\)

<=> 3x = 2x + 1

<=> 3x - 2x = 1

<=> x = 1

a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)

\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)

\(\Leftrightarrow3x=3\)

hay x=1

Vậy: S={1}

b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)

\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)

\(\Leftrightarrow6x=-20\)

hay \(x=-\dfrac{10}{3}\)

c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)

\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)

\(\Leftrightarrow17x=17\)

hay x=1

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

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

a: Ta có: \(2x^3-18x=0\)

\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)

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

b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)

\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)

\(\Leftrightarrow-13x=13\)

hay x=-1

c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)

\(\Leftrightarrow3x=12\)

hay x=4

a) 2x³-18x=0

⇔ 2x(x²-9)=0

⇔ 2x(x-3)(x+3)=0

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

b)(3x-1)(2x+1)-6x(x+2)=11

⇔ 6x²+x-1-6x²-12x=11

⇔ -11x=12

\(\Leftrightarrow x=-\dfrac{12}{11}\)

c) (x-1)³-(x+2).(x²-2x+4)=3.(1-x²)

⇔ x³-3x²+3x-1-x³-8-3+3x²=0

⇔ 3x=12

⇔   x=4

c. (x - 1)³ - (x + 2)(x² - 2x + 4) = 3(1 - x²)

<=> (x³ - 3x² + 3x - 1) - (x³ - 2x² + 4x + 2x² - 4x + 8) = 3 - 3x²

<=> x³ - 3x² + 3x - 1 - x³ + 2x² - 4x - 2x² + 4x - 8 = 3 - 3x²

<=> x³ - x³ - 3x² + 2x² - 2x² + 3x² + 3x - 4x + 4x = 3 + 1 + 8

<=> 3x = 12

<=> x = 4

a.ĐKXĐ: \(x\ge0\)
 \(\sqrt{2x}< \dfrac{1}{3}\) \(\Leftrightarrow2x< \dfrac{1}{3}\Leftrightarrow6x< 1\Leftrightarrow x< \dfrac{1}{6}\)

b. ĐKXĐ: \(x\ge\dfrac{1}{6}\)

\(\sqrt{-3x+\dfrac{1}{2}}\ge5\Leftrightarrow-3x+\dfrac{1}{2}\ge25\Leftrightarrow x=-\dfrac{49}{6}\) 

c. ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{-2x+1}>7\) \(\Leftrightarrow-2x+1>49\Leftrightarrow x=-24\)

d. ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}\le\dfrac{3}{2}\Leftrightarrow2x-1\le\dfrac{9}{4}\Leftrightarrow x=\dfrac{13}{8}\)

a: Ta có: \(\sqrt{2x}< \dfrac{1}{3}\)

\(\Leftrightarrow2x< \dfrac{1}{9}\)

\(\Leftrightarrow x< \dfrac{1}{18}\)

Kết hợp ĐKXĐ, ta được: \(0\le x< \dfrac{1}{18}\)

b: Ta có: \(\sqrt{-3x+\dfrac{1}{2}}\ge5\)

\(\Leftrightarrow-3x+\dfrac{1}{2}\ge25\)

\(\Leftrightarrow-3x\ge\dfrac{49}{2}\)

hay \(x\le-\dfrac{49}{6}\)

c: Ta có: \(\sqrt{-2x+1}>7\)

\(\Leftrightarrow-2x+1>49\)

\(\Leftrightarrow-2x>48\)

hay x<-24

a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)