Gọi hóa trị của Nitơ là x. Khi đó: \(N^xO_3^{II}\)

Theo quy tắc hóa trị ta có: x.1=3.II => \(x=\frac{3.II}{1}=VI\)( vô lý)

=> đề?

M+2019=2xy−yz−zx+2020M+2019=2xy−yz−zx+2020

=2xy−yz−zx+x2+y2+z2=2xy−yz−zx+x2+y2+z2

=(x+y−z2)2+3z24≥0=(x+y−z2)2+3z24≥0

⇒Mmin=0⇒Mmin=0 khi ⎧⎩⎨⎪⎪⎪⎪x+y−z2=03z24=0x2+y2+z2=2020{x+y−z2=03z24=0x2+y2+z2=2020

⇔⎧⎩⎨⎪⎪x+y=0z=0x2+y2=2020⇔{x+y=0z=0x2+y2=2020 ⇒⎧⎩⎨⎪⎪x=±1010−−−−√y=−xz=0

mình không hiểu ạ

a) 16( 2x - 3 ) + x²( 3x - 2 ) = 0 < xem lại đề bạn ơi -- >

b) x³ - 7x² = 7 - x

⇔ x²( x - 7 ) + x - 7 = 0

⇔ x²( x - 7 ) + ( x - 7 ) = 0

⇔ ( x - 7 )( x² + 1 ) = 0

⇔ x - 7 = 0 hoặc x² + 1 = 0

⇔ x = 7 ( x² + 1 ≥ 1 > 0 ∀ x )

a)16(2-3x)+x²(3x-2)=0

Với\(\left(x-3\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)\) và x = 1/3

\(\Rightarrow\left(\frac{1}{3}-3\right)\left(\frac{1}{3}+3\right)-\left(\frac{1}{2}+2\right)\left(\frac{1}{2}-1\right)=\left(\frac{1}{3}\right)^2-3^2-\left(\frac{5}{2}\cdot2-\frac{5}{2}\right)\)\(=\frac{1}{3^2}-\frac{3^2}{1}-\left(\frac{5}{1}-\frac{5}{2}\right)=\frac{1}{3^2}-\frac{3^4}{3^2}-\frac{5}{2}\)\(=\frac{-80}{3^2}-\frac{5}{2}=\frac{-160-45}{18}=\frac{-205}{18}\)

( x - 3 )( x + 3 ) - ( x + 2 )( x - 1 )

= x² - 9 - ( x² + x - 2 )

= x² - 9 - x² - x + 2

= -x - 7

Với x = 1/3 => Gtri bthuc = -1/3 - 7 = -22/3

Ta có : 3x³ + x² - 13x + 5

= 3x³ + 6x² - 5x² - 3x - 10x + 5

= ( 3x³ + 6x² - 3x ) - ( 5x² + 10x - 5 )

= 3x( x² + 2x - 1 ) - 5( x² + 2x - 1 )

= ( x² + 2x - 1 )( 3x - 5 )

=> ( 3x³ + x² - 13x + 5 ) : ( x² + 2x - 1 ) = 10x - 1

⇔ ( x² + 2x - 1 )( 3x - 5 ) : ( x² + 2x - 1 ) = 10x - 1

⇔ 3x - 5 = 10x - 1

⇔ 3x - 10x = -1 + 5

⇔ -7x = 4

⇔ x = -4/7

Cảm ơn nhé 

x\(^3\)+3x+2=0

x\(^3\)+2x+x+2=0

(x\(^3\)+x)+(2x+2)=0

x(x\(^2\)+1)+2(x+1)=0

x(x+1)(x-1)+2(x+1)=0

(x+1)[x(x-1)+2]=0

TH1:x+1=0

x=0-1

x=-1

TH2:x(x-1)+2=0

x(x-1)=0-2

x(x-1)=-2

x=1

a.
2x+16x^3+7x^2+x+33x^2+2x6x^3+3x^24x^2+x+34x^2+2x-x+3