4x³ + x² = 4x

⇔ 4x³ + x² - 4x = 0

⇔ x( 4x² + x - 4 ) = 0

⇔ x = 0 hoặc 4x² + x - 4 = 0

+) 4x² + x - 4 = 0 (*)

Δ = b² - 4ac = 1² - 4.4.(-4) = 65

Δ > 0 nên (*) có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-1+\sqrt{65}}{8}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-1-\sqrt{65}}{8}\end{cases}}\)

Vậy \(S=\left\{0;\frac{-1\pm\sqrt{65}}{8}\right\}\)

Ca₃(PO4)₂

ko có CT nào là PO đâu

Ta có A = 3(x² + y²) - 2(x³ - y³)

= 3(x² + y²) - 2(x - y)(x² + xy + y²)

= 3(x² + y²) - 2(x² + xy + y²) Vì x - y = 1

= 3x² + 3y² - 2x² - 2xy - 2y²

= x² - 2xy + y²

= (x - y)²

= 1 (Vì x - y = 1)

Vậy A = 1 khi x - y = 1

Ta có: \(x^2-5x+1=0\)

\(\Leftrightarrow\left(x^2-5x+\frac{25}{4}\right)-\frac{21}{4}=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{21}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{2}=\frac{\sqrt{21}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{21}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=\frac{26+10\sqrt{21}}{4}\\x^2=\frac{26-10\sqrt{21}}{4}\end{cases}}\)

Thay vào ta được: \(\frac{x^2+1}{x^2}=1+\frac{1}{x^2}\) đến đây: \(\orbr{\begin{cases}=1+\frac{1}{\frac{26+10\sqrt{21}}{4}}\\=1+\frac{1}{\frac{26-10\sqrt{21}}{4}}\end{cases}}\)

\(\orbr{\begin{cases}=1+\frac{4}{26+10\sqrt{21}}=\frac{30+10\sqrt{21}}{26+10\sqrt{21}}\\=1+\frac{4}{26-10\sqrt{21}}=\frac{30-10\sqrt{21}}{26-10\sqrt{21}}\end{cases}}\)

Bài 1

1) 4x - x² - 4 = 0

⇔ -( x² - 4x + 4 ) = 0

⇔ -( x - 2 )² = 0

⇔ x - 2 = 0

⇔ x = 2

2) 4( x - 1 )² - ( 5 - 2x )² = 0

⇔ 2²( x - 1 )² - ( 5 - 2x )² = 0

⇔ ( 2x - 2 )² - ( 5 - 2x ) = 0

⇔ ( 2x - 2 - 5 + 2x )( 2x - 2 + 5 - 2x ) = 0

⇔ ( 4x - 7 ).3 = 0

⇔ 4x - 7 = 0

⇔ x = 7/4

3) 9( x - 2 )² - 4( 3 - x )² = 0

⇔ 3²( x - 2 )² - 2²( x - 3 )² = 0

⇔ ( 3x - 6 )² - ( 2x - 6 )² = 0

⇔ ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 ) = 0

⇔ x( 5x - 12 ) = 0

⇔ x = 0 hoặc 5x - 12 = 0

⇔ x = 0 hoặc x = 12/5

4) x² - 6x + 5 = 0

⇔ x² - 5x - x + 5 = 0

⇔ x( x - 5 ) - ( x - 5 ) = 0

⇔ ( x - 5 )( x - 1 ) = 0

⇔ x - 5 = 0 hoặc x - 1 = 0

⇔ x = 5 hoặc x = 1

Bài 2.

1) x² - z² + y² - 2xy

= ( x² - 2xy + y² ) - z²

= ( x - y )² - z²

= ( x - y - z )( x - y + z )

2) a³ - ay - a²x + xy

= ( a³ - a²x ) - ( ay - xy )

= a²( a - x ) - y( a - x )

= ( a - x )( a² - y )

3) 2xy + 3z + 6y + xz

= ( 2xy + 6y ) + ( xz + 3z )

= 2y( x + 3 ) + z( x + 3 )

= ( x + 3 )( 2y + z )

4) x² + 2xz + 2xy + 4yz

= ( x² + 2xy ) + ( 2xz + 4yz )

= x( x + 2y ) + 2z( x + 2y )

= ( x + 2y )( x + 2z )

5) ( x + y + z )³ - x³ - y³ - z³

= x³ + y³ + z³ + 3( x + y )( y + z )( x + z ) - x³ - y³ - z³

= 3( x + y )( y + z )( x + z )