a) Ta có: \(36x^3-4x=0\)

\(\Leftrightarrow4x\left(9x^2-1\right)=0\)

\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)

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

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

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

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

d) Ta có: \(x^2-6x-16=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

e) Ta có: \(x^4-6x^2-7=0\)

\(\Leftrightarrow\left(x^2-7\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x\in\left\{\sqrt{7};-\sqrt{7}\right\}\)

A=-4x^2+4x-1+4

=-(4x^2-4x+1)+4

=-(2x-1)^2+4<=4

Dấu = xảy ra khi x=1/2

a.4h12p

b.x=12h15p-7h40p

x=4h25p

c.=6h38px3=19h54p

a: \(\left|3x-2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)

Lời giải:
$\frac{xy+3x-2y-6}{y+3}=3$

$\Rightarrow xy+3x-2y-6=3y+9$

$\Rightarrow xy+3x-5y-15=0$

$\Rightarrow x(y+3)-5(y+3)=0$

$\Rightarrow (y+3)(x-5)=0$

$\Rightarrow y+3=0$ hoặc $x-5=0$

Mà $y$ tự nhiên nên $y+3>0$. Do đó $x-5=0$

$\Rightarrow x=5$

Vậy $x=5$ và $y$ là số tự nhiên tùy ý.

\(1,3x-7=19\\ \Rightarrow3x=26\\ \Rightarrow x=\dfrac{26}{3}\\ 2,\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\\ 3,3x+\dfrac{2}{4}+1=5x-\dfrac{1}{3}\\ \Rightarrow5x-\dfrac{1}{3}-3x-\dfrac{2}{4}-1=0\\ \Rightarrow2x-\dfrac{11}{6}=0\\ \Rightarrow2x=\dfrac{11}{6}\\ \Rightarrow x=\dfrac{11}{12}\)

\(4,\dfrac{x}{15}+\dfrac{1}{2}-\dfrac{x}{50}=\dfrac{5}{6}\\ \Rightarrow\dfrac{x}{15}-\dfrac{x}{50}=\dfrac{5}{6}-\dfrac{1}{2}\\ \Rightarrow x\left(\dfrac{1}{15}-\dfrac{1}{50}\right)=\dfrac{1}{3}\\ \Rightarrow\dfrac{7}{150}x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{50}{7}\)

Tính:

\(\dfrac{10}{21}-1\cdot251+\dfrac{11}{21}+2\cdot251\\ 251\cdot\left(\dfrac{10}{21}+\dfrac{11}{21}\right)+\left(2-1\right)\\ 251\cdot1+1=252\)

Tìm x

c,\(\left(3x-1\right)^2:7=7\\ \left(3x-1\right)^2=7\cdot7=14\\ \left(3x-1\right)^2=\left(\pm7\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x-1=7\\3x-1=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=7+1=8\\3x=-7+1=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{6}{3}=-2\end{matrix}\right.\)

Vậy x={8/3;-2}

d; \(\left(7-2x\right)^3=\left(\dfrac{1}{2}\right)^2\cdot\left(-\dfrac{1}{2}\right)\\ \left(7-2x\right)^3=\left(-\dfrac{1}{2}\right)^3\\ 7-2x=-\dfrac{1}{2}\\ 2x=7--\dfrac{1}{2}=\dfrac{15}{2}\\ x=\dfrac{15}{2}:2=\dfrac{15}{4}\)

giúp với mn ơiii

A = \(\dfrac{10}{21}\) - 1.251 + \(\dfrac{11}{21}\) + 2.251

A = (\(\dfrac{10}{21}\) + \(\dfrac{11}{21}\)) - 1.251 + 2.251

A = \(\dfrac{21}{21}\) + 251

A = 1 + 251

A = 252

C, (3\(x\) - 1)²: 7 = 7

     (3\(x\) -1)²     = 7 x  7

     (3\(x\) -1)² = 49

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

    \(\left[{}\begin{matrix}3x=8\\3x=-6\end{matrix}\right.\)

   \(\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-2\end{matrix}\right.\)

Vậy \(x\) \(\in\){-2;\(\dfrac{8}{3}\)}

d, (7 - 2\(x\))³ = (\(\dfrac{1}{2}\))² \(\times\) (- \(\dfrac{1}{2}\))

    (7- 2\(x\))³ =  (- \(\dfrac{1}{2}\))³

    7 - 2\(x\)   = - \(\dfrac{1}{2}\)

       2\(x\)     = 7 + \(\dfrac{1}{2}\)

       2\(x\)     = \(\dfrac{15}{2}\)

          \(x\)     = \(\dfrac{15}{2}\): 2

         \(x\)     =   \(\dfrac{15}{4}\)

a:Ta có: \(x\left(x-1\right)+x=4\)

\(\Leftrightarrow x^2-x+x=4\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(3x\left(x-5\right)-2x+10=0\)

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

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

c: Ta có: \(5x^2-3x-2=0\)

\(\Leftrightarrow5x^2-5x+2x-2=0\)

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

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

d: Ta có: \(x^4-11x^2+18=0\)

\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)

\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)

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

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

a) x(x-1)+x=4

⇔x²=4⇔\(x=\pm2\)

b)3x(x-5)-2x+10=0

⇔3x(x-5)-2(x-5)=0

⇔(x-5)(3x-1)=0

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

c)5x²-3x-2=0

⇔ 5x(x-1)+2(x-1)=0

⇔ (x-1)(5x+2)=0

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

d)x⁴-11x²+18=0

⇔ x²(x²-2)-9(x²-2)=0

⇔ (x²-2)(x²-9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)

a) (3x - 1)³ + 17 = 710 : 5

(3x - 1)³ + 17 = 142

(3x - 1)³ = 142 - 17

(3x - 1)³ = 125

(3x - 1)³ = 5³

3x - 1 = 5

3x = 5 + 1

3x = 6

x = 6 : 3

x = 2