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

b) 2x² + 5x - 3 = 2x² + 6x - x - 3 = 2x(x + 3) - (x + 3) = (2x - 1)(x + 3)

c) x² - 4x - 5 = x² - 5x + x - 5 = x(x - 5) + (x - 5) = (x + 1)(x - 5)

d) x² + x - 12 =x² + 4x - 3x - 12 = x(x + 4) - 3(x + 4) = (x - 3)(x + 4)

\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2+\frac{1}{a^2}}=2\\ \)(do Bđt cosi)=> \(a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge6\\ \)

Dấu "=" xảy ra <=> a=b=c=1

=>B=3

Bất đẳng thức cosi mình chưa học

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

\(=x^3\left(x-4\right)-4x\left(x-4\right)\)

\(=\left(x^3-4x\right)\left(x-4\right)\)

\(=x\left(x^2-4\right)\left(x-4\right)\)

\(=\left(x-4\right)\left(x-2\right)x\left(x+2\right)\)

x chẵn nên x - 4; x - 2; x + 2 chẵn

Vậy \(\left(x-4\right)\left(x-2\right)x\left(x+2\right)\)là tích của 4 số chẵn liên tiếp

hay \(x^4-4x^3-4x^2+16x\)là tích của 4 số chẵn liên tiếp (đpcm)

A= x⁴ - 4x³ - 4x² + 16x

A= (x⁴ - 4x³) - (4x² - 16x)

A= x³ (x-4) - 4x( x -4)

A = (x³- 4x) (x-4)

A = x( x² - 4) (x-4)

A= x ( x-2) (x +2) (x-4)

A= (x-4) (x-2) x ( x+2) ( sắp xếp lại)

Vì x là số chẵn => A= (x-4) (x-2) x (x+2) là 4 số chẵn liên tiếp

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

\(\Rightarrow\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)

\(\Rightarrow\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)

\(\Rightarrow\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}\Rightarrow}\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\Rightarrow x:y:z=a:b:c\)

cảm ơn nhá

Ta có: 

\(1^4+\frac{1}{4}=\left(1^2-1+\frac{1}{2}\right)\left(1^2+1+\frac{1}{2}\right)=\frac{1}{2}.\left(2+\frac{1}{2}\right)\)

\(2^4+\frac{1}{4}=\left(2^2-2+\frac{1}{2}\right)\left(2^2+2+\frac{1}{2}\right)=\left(2+\frac{1}{2}\right).\left(6+\frac{1}{2}\right)\)

\(3^4+\frac{1}{4}=\left(3^2-3+\frac{1}{2}\right)\left(3^2+3+\frac{1}{2}\right)=\left(6+\frac{1}{2}\right).\left(12+\frac{1}{2}\right)\)

\(4^4+\frac{1}{4}=\left(4^2-4+\frac{1}{2}\right)\left(4^2+4+\frac{1}{2}\right)=\left(12+\frac{1}{2}\right).\left(20+\frac{1}{2}\right)\)

...

\(19^4+\frac{1}{4}=\left(19^2-19+\frac{1}{2}\right)\left(19^2+19+\frac{1}{2}\right)=\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)\)

\(20^4+\frac{1}{4}=\left(20^2-20+\frac{1}{2}\right)\left(20^2+20+\frac{1}{2}\right)=\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)\)

=> \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)

\(=\frac{\frac{1}{2}\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)\left(20+\frac{1}{2}\right)...\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)}\)

\(=\frac{\frac{1}{2}}{420+\frac{1}{2}}=\frac{1}{841}\)

Thời gian ôtô đi trên 1 / 3 quãng đường đầu là :

v₁ = s₁ / t₁ \(\Rightarrow\)t₁ = s₁ / v₁ = 1 / 3 ÷ 60 = 1 / 180 ( h )

Thời gian ôtô đi trên 2 / 3 quãng đường sau là :

v₂ = s₂ / t₂ \(\Rightarrow\)t₂ = s₂ / v₂ = 2 / 3 ÷ 50 = 1 / 75 ( h )

Vận tốc trung bình của ôtô là :

V_tb = s₁ + s₂ / t₁ + t₂

V_tb = ( 1 / 3 + 2 / 3 ) / ( 1 / 180 + 1 / 75 )

V_tb = 52,94 ( km / h )

Vậy : Vận tốc trung bình của ôtô là 52,94 km / h

a) 3x² - 4x + 1 + xy - y

= 4x² - x² - 4x + 1 + xy - y

= ( 4x² - 4x ) + ( xy - y ) - ( x² - 1 )

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

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

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

= ( x - 1 )( 3x + y - 1 )