Gọi vận tốc đi,là v₁  thời gian đi ; về lần lượt là t₁ ; t₂  (v₁ ; t₁ ; t₂ > 0)

=> vận tốc về v₁ - 5

Đổi 30 phút = 1/2 giờ

Ta có t₂ - t₁ = 1/2

<=> \(\frac{S}{v_1-5}-\frac{S}{v_1}=\frac{1}{2}\) 

<=> \(\frac{180}{v_1-5}-\frac{180}{v_1}=\frac{1}{2}\)

<=> \(\frac{1}{v_1-5}-\frac{1}{v_1}=\frac{1}{360}\)

 \(\Leftrightarrow\frac{5}{\left(v_1-5\right)v_1}=\frac{1}{360}\)

<=> (v₁ - 5).v₁ = 1800

<=> (v₁)² - 5.v₁ = 1800

<=> (v₁)² - 45.v₁ + 40v₁ - 1800 = 0

<=> v₁(v₁ - 45) + 40(v₁ - 45) = 0

<=> (v₁ + 40)(v₁ - 45) = 0

<=> \(\orbr{\begin{cases}v_1=-40\left(\text{loại}\right)\\v_1=45\left(\text{tm}\right)\end{cases}}\)

Vậy vận tốc lúc đi là 45 km/h

\(M:Cu\)

\(A:CuO\)

\(B:CuSO_4\)

\(C:CuSO_4\)

\(D:CuSO_4\)

\(E:CuSO_4\cdot5H_2O\)

PTHH : \(2Cu+O_2-t^o->2CuO\)

              \(CuO+H_2SO_4-->CuSO_4+H_2O\)

a) \(\left(2x+3\right)\left(x-3\right)+x\left(x-2\right)=3\left(x-2\right)^2\)

\(\Leftrightarrow2x^2-3x-9+x^2-2x=3\left(x^2-4x+4\right)\)

\(\Leftrightarrow3x^2-5x-9=3x^2-12x+12\)

\(\Leftrightarrow7x=21\Rightarrow x=3\)

b) \(\left(4x+7\right)\left(x-3\right)-x^2=3x\left(x+2\right)\)

\(\Leftrightarrow4x^2-5x-21-x^2=3x^2+6x\)

\(\Leftrightarrow11x=-21\Rightarrow x=-\frac{21}{11}\)

a,X=3

CBHT

ĐKXĐ : \(2014x+1\ne0;2015x+2\ne0;2016x+3\ne0;2017x+4\ne0\)

=> \(x\ne-\frac{1}{2014};x\ne-\frac{2}{2015};x\ne-\frac{3}{2016};x\ne-\frac{4}{2017}\)

Ta có \(\frac{1}{2014x+1}-\frac{1}{2015x+2}=\frac{1}{2016x+3}-\frac{1}{2017x+4}\)

<=> \(\frac{x+1}{\left(2014x+1\right)\left(2015x+2\right)}=\frac{x+1}{\left(2016x+3\right)\left(2017x+4\right)}\)

<=> \(\left(x+1\right)\left(\frac{1}{\left(2014x+1\right)\left(2015x+2\right)}-\frac{1}{\left(2016x+3\right)\left(2017x+4\right)}\right)=0\)

<=> \(\left(x+1\right)\left(\frac{8062x^2+8072x+10}{\left(2014x+1\right)\left(2015x+2\right)\left(2016x+3\right)\left(2017x+4\right)}\right)=0\)

<=> \(\left(x+1\right)\frac{\left(x+1\right)\left(8062x+10\right)}{\left(2014x+1\right)\left(2015x+2\right)\left(2016x+3\right)\left(2017x+4\right)}=0\)

<=> (x + 1)²(8062x + 10) = 0

<=> \(\orbr{\begin{cases}x+1=0\\8062x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\left(tm\right)\\x=-\frac{5}{4031}\left(tm\right)\end{cases}}\)

Vậy \(x\in\left\{-1;-\frac{5}{4031}\right\}\)là nghiệm phương trình

Có x = 3

<=> -4.3 - 10 = 5.3 + a

<=> -22 = 15 + a

<=> a = -37

Vậy a = -37

xin lỗi mình nhẩm sai :v 

\(\Leftrightarrow-12-10=15+a\Leftrightarrow-22=15+a\Leftrightarrow a=-37\)

Vậy a = -37 nếu x = 3

3y³ - 7y² - 7y + 3 = 0

<=> 3y³ + 3y² - 10y² - 10y + 3y + 3 = 0

<=> 3y²( y + 1 ) - 10y( y + 1 ) + 3( y + 1 ) = 0

<=> ( y + 1 )( 3y² - 10y + 3 ) = 0

<=> ( y + 1 )( 3y² - 9y - y + 3 ) = 0

<=> ( y + 1 )[ 3y( y - 3 ) - ( y - 3 ) ] = 0

<=> ( y + 1 )( y - 3 )( 3y - 1 ) = 0

<=> y = -1 hoặc y = 3 hoặc y = 1/3

Vậy ...

2y⁴ - 9y³ + 14y² - 9y + 2 = 0

<=> 2y⁴ - 4y³ - 5y³ + 10y² + 4y² - 8y - y + 2 = 0

<=> 2y³( y - 2 ) - 5y²( y - 2 ) + 4y( y - 2 ) - ( y - 2 ) = 0

<=> ( y - 2 )( 2y³ - 5y² + 4y - 1 ) = 0

<=> ( y - 2 )( 2y³ - 2y² - 3y² + 3y + y - 1 ) = 0

<=> ( y - 2 )[ 2y²( y - 1 ) - 3y( y - 1 ) + ( y - 1 ) ] = 0

<=> ( y - 2 )( y - 1 )( 2y² - 3y + 1 ) = 0

<=> ( y - 2 )( y - 1 )( 2y² - 2y - y + 1 ) = 0

<=> ( y - 2 )( y - 1 )[ 2y( y - 1 ) - ( y - 1 ) ] = 0

<=> ( y - 2 )( y - 1 )²( 2y - 1 ) = 0

<=> y = 2 hoặc y = 1 hoặc y = 1/2

Vậy ...