Gọi độ dài quãng đường đi là x (km) với x>0

Thời gian đi là: \(\dfrac{x}{12}\) giờ

Độ dài quãng đường về là: \(x+4\) (km)

Thời gian về là: \(\dfrac{x+4}{20}\) giờ

Do tổng thời gian cả đi và về là 6h nên ta có pt:

\(\dfrac{x}{12}+\dfrac{x+4}{20}=6\)

\(\Leftrightarrow\dfrac{x}{12}+\dfrac{x}{20}=6-\dfrac{4}{20}\)

\(\Leftrightarrow\dfrac{2}{15}x=\dfrac{29}{5}\)

\(\Rightarrow x=43,5\left(km\right)\)

Mặc dù rất khâm phục với các nhà yêu nước,nhưng Nguyễn Tất Thành chọn hướng đi mới khác với các nhà yêu nước tiền bối.Vì những con đường đó có những hạn chế,sai lầm dẫn đến thất bài trong việc cứu nước. Nên Bác quyết định đi tìm con đường cứu nước mới hướng sang phương Tây

vì bác muốn chấm dứt mọi cuộc chiến liên quan đến việt nam 

  (bác muốn việt nam ko có chiến tranh)

    em chào cô Linh Trang, em là quân lớp 4a

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1 The boy said that they were picking up litter on the beach then

2 Before Lucas arrives at the restaurant, we won't order our meals

Q1: The boys said that they were picking up litter on the beach then.

Q2: Before Lucas arrives at the restaurant, we won't order our meals

1 During the flood, you should listen to the instructions from local authorities

2 Mars is one of the most promising planets for life in solar system

Q1: During the flood,you should listen to the instructions of the local authorities.

Q2: Mars is one of the most promising planets for life in the solar system.

ĐKXĐ: \(x\ne\left\{-1;0;1\right\}\)

a.

\(A=\dfrac{x^2+1}{x}+\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x-1\right)}+\dfrac{x^4-1-x\left(x^2-1\right)}{-x\left(x^2-1\right)}\)

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

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

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

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

b.

\(x^2+x=12\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2+4x-3x-12=0\)

\(\Leftrightarrow x\left(x+4\right)-3\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

Với \(x=3\Rightarrow A=\dfrac{\left(3+1\right)^2}{3}=\dfrac{16}{3}\)

Với \(x=-4\Rightarrow A=\dfrac{\left(-4+1\right)^2}{-4}=-\dfrac{9}{4}\)

c. Đề bài sai, \(A>4\) chỉ khi \(x>0\), còn khi \(x< 0\) thì \(A< -4\)

was nhé bn

Tom said that he was hungry