Giải:

Gọi vận tốc của xe 1, xe 2 là \(v_1,v_2\)

Ta có: xe 1 đi từ A đến B hết 6 giờ

xe 2 đi từ A đến B hết 4 giờ 30 phút \(=4,5\)h

\(6v_1=4,5v_2\Rightarrow12v_1=9v_2\Rightarrow\dfrac{v_1}{9}=\dfrac{v_2}{12}\)và \(v_2-v_1=20\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{v_1}{9}=\dfrac{v_2}{12}=\dfrac{v_2-v_1}{12-9}=\dfrac{20}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}v_1=60\\v_2=80\end{matrix}\right.\)

Vậy vận tốc xe 1 là 60 km/h

vận tốc xe 2 là 80 km/h

C...á...i ...g...ì ?

Nhiều thế ư ?

\(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

Thay x - y = 7

\(\Rightarrow A=49+14+37=100\)

Vậy A = 100 khi x - y = 7

a, \(A=4x^2-4x+2017\)

\(=4x^2-4x+1+2016\)

\(=\left(2x-1\right)^2+2016\ge2016\)

Dấu " = " khi \(\left(2x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(MIN_A=2016\) khi \(x=\dfrac{1}{2}\)

b, \(B=-x^2+5x-2018\)

\(=-\left(x^2-5x+2018\right)\)

\(=-\left(x^2-\dfrac{5}{2}x2+\dfrac{25}{4}+\dfrac{8047}{4}\right)\)

\(=-\left[\left(x-\dfrac{5}{2}\right)^2+\dfrac{8047}{4}\right]\)

\(=-\left(x-\dfrac{5}{2}\right)^2-\dfrac{8047}{4}\le\dfrac{-8047}{4}\)

Dấu " = " khi \(\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)

Vậy \(MAX_B=\dfrac{-8047}{4}\) khi \(x=\dfrac{5}{2}\)

a, \(A=x^3-30x^2-31x+1\)

\(=x^3-31x^2+x^2-31x+1\)

\(=x^2\left(x-31\right)+x\left(x-31\right)+1\)

\(=\left(x^2+x\right)\left(x-31\right)+1\)

Thay x = 31 \(\Rightarrow A=1\)

Vậy A = 1 khi x = 31

b, tách ra làm tương tự phần a

1,sai đề

2, \(\dfrac{x^2}{xy+x}+\dfrac{y}{y^2-1}-\dfrac{x}{x\left(y-1\right)}\)

\(=\dfrac{x^2}{x\left(y+1\right)}+\dfrac{y}{\left(y-1\right)\left(y+1\right)}-\dfrac{x}{x\left(y-1\right)}\)

\(=\dfrac{x}{y+1}+\dfrac{y}{\left(y-1\right)\left(y+1\right)}-\dfrac{1}{y-1}\)

\(=\dfrac{x\left(y-1\right)-y-1}{\left(y+1\right)\left(y-1\right)}+\dfrac{y}{\left(y-1\right)\left(y+1\right)}\)

\(=\dfrac{xy-x-y-1+y}{y^2-1}=\dfrac{xy-x-1}{y^2-1}\)

3, \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Vậy x = 2 hoặc x = -5

1, Ta có: \(\dfrac{2727}{2323}=\dfrac{27.101}{23.101}=\dfrac{27}{23}=\dfrac{27.1010101}{23.1010101}=\dfrac{27272727}{23232323}\)

2, \(3^{n+2}+2^{n+3}+3^n+2^{n+1}\)

\(=3^n.3^2+3^n+2^n.2^3+2^n.2\)

\(=3^n\left(3^2+1\right)+2^n\left(2^3+2\right)\)

\(=3^n.10+2^n.10=\left(3^n+2^n\right).10⋮10\forall n\in N\)

Vậy...

\(-1-\dfrac{1}{3}-\dfrac{1}{6}-...-\dfrac{1}{1225}\)

\(=\dfrac{-1}{2}\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{2450}\right)\)

\(=\dfrac{-1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)

\(=\dfrac{-1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(=\dfrac{-1}{2}\left(1-\dfrac{1}{50}\right)\)

\(=\dfrac{-1}{2}.\dfrac{49}{50}=\dfrac{-49}{100}\)

Vậy...

a, \(xy-x-y=2\)

\(\Leftrightarrow x\left(y-1\right)-y+1=3\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)=3\)

Ta có bảng sau:

bạn xét các trường hợp ra nhé!

b, \(xy-3x+5y=22\)

\(\Leftrightarrow x\left(y-3\right)+5y-15=7\)

\(\Leftrightarrow x\left(y-3\right)+5\left(y-3\right)=7\)

\(\Leftrightarrow\left(x+5\right)\left(y-3\right)=7\)

Ta có bảng sau:

.............

c, d tương tự

a, \(3a^2b^2-6a^2b^3+3a^2b^2\)

\(=6a^2b^2-6a^2b^3=6a^2b^2\left(1-b\right)\)

b, \(a^{n+1}-2a^{n-1}=a^2.a^{n-1}-2a^{n-1}=a^{n-1}\left(a^2-2\right)\)

c, \(3a^2b\left(a+b-2\right)-4ac^2-4bc^2+8c^2\)

\(=3a^2b\left(a+b-2\right)-4c^2\left(a+b-2\right)\)

\(=\left(3a^2b-4c^2\right)\left(a+b-2\right)\)

c, \(5a^n\left(a^2-ab+1\right)-2a^2b^n+2ab^{n+1}-2b^n\)

\(=5a^n\left(a^2-ab+1\right)-2a^2b^n+2ab^n.b-2b^n\)

\(=5a^n\left(a^2-ab+1\right)-2b^n\left(a^2-ab+1\right)\)

\(=\left(5a^n-2b^n\right)\left(a^2-ab+1\right)\)