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

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

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

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

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 )

Bài 1:

a) \(3x\left(5x^2-2x+1\right)\)

\(=15x^3-6x^2+3x\)

b) \(\left(x^2-1\right)\left(x^2+2x\right)\)

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

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

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

Bài 2:

a) \(3x^2=2x\)

\(\Leftrightarrow3x^2-2x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)

b)\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-2x-1=12x-5\)

\(\Leftrightarrow14x=4\Leftrightarrow x=\frac{2}{7}\)

\(A=\left(2x-1\right)\left(4x+3\right)-\left(2x+5\right)^2-10x\)

\(=\left(8x^2+2x-3\right)-\left(4x^2+20x+25\right)-10x\)

\(=8x^2+2x-3-4x^2-20x-25-10x\)

\(=4x^2-28x-28\)

A=(2x-1)(4x+3)-(2x+5)²-10x

A=8x²+6x-4x-3-(4x²+10x+25)-10x

A=8x²+2x-3-4x²-10x-25-10x

A=4x²-18x-28