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

a) 1/4(x-3)+2=1/5

1/4.(x-3) = 1/5-2

1/4.(x-3) = -9/5

x-3 = (-9/5):1/4

x-3 = -36/5

x = -36/5+3

x= -21/5

    3(x+3)-x(x+3)=0

     (x+3)(3-x)     =0

      x+3 =0 hoặc 3-x=0   =>x={-3;3}

(x-3)(x²+3x+9)+x(5-x²)=6x

x(x²+3x+9)-3(x²+3x+9)+x(5-x²)=6x

x³+3x²+9x-3x²-9x-27+5x-x³-6x=0

(x³-x³)+(3x²-3x²)+(9x-9x+5x-6x)=27

-x=27

x=-27

đéo bt

a) ( x+ 3 ) ( x - 3 ) = 3 ( x-3)

x+ 3 =3

x =0

a) x² - 9 = 3( x - 3 )

⇔ ( x - 3 )( x + 3 ) - 3( x - 3 ) = 0

⇔ ( x - 3 )( x + 3 - 3 ) = 0

⇔ ( x - 3 ).x = 0

⇔ x - 3 = 0 hoặc x = 0

⇔ x = 3 hoặc x = 0

b) 3( 3x² + 1 ) = 6 - 2( 3x + 2 )

⇔ 9x² + 3 = 6 - 6x - 4

⇔ 9x² + 6x + 3 - 6 + 4 = 0

⇔ 9x² + 6x + 1 = 0

⇔ ( 3x + 1 )² = 0

⇔ 3x + 1 = 0

⇔ x = -1/3

câu a chưa đủ đề em hấy

c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0

       (\(x\) - 2022).(\(x\) - 4) = 0

         \(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)

b, (\(x\) - 1)(\(x^2\) + \(x\) + 1) - \(x\)(\(x\) - 2)(\(x\) + 2)  = 7

     \(x^3\) - 1 - \(x\).(\(x^2\) - 4) = 7

      \(x^3\) - 1 - \(x^3\) + 4\(x\)  = 7

        (\(x^3\) - \(x^3\)) - 1 + 4\(x\) = 7

                       - 1 + 4\(x\) = 7

                               4\(x\) = 7 + 1

                               4\(x\) = 8

                                  \(x\) = 8:4

                                   \(x\) = 2

a, ĐK: \(x\ge0;x\ne9\)

\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

b, \(P>0\Leftrightarrow-\dfrac{3}{\sqrt{x}-3}>0\)

\(\Leftrightarrow\sqrt{x}-3>0\)

\(\Leftrightarrow x>9\)

c, \(P=-\dfrac{3}{\sqrt{x}-3}\in Z\)

\(\Leftrightarrow\sqrt{x}-3\inƯ_3=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2;4;6\right\}\)

\(\Leftrightarrow x\in\left\{0;4;16;36\right\}\)

a: Ta có: \(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

\(=\dfrac{-3}{\sqrt{x}-3}\)

b: Để P<0 thì \(\sqrt{x}-3< 0\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)