Gọi độ dài quãng đường là s

Thời gian đi \(\frac{1}{3}\) quãng đường đầu là: \(t_1=\frac{s}{3v_1}=\frac{s}{3.18}=\frac{s}{54}h\)

Gọi thời gian đi \(\frac{2}{3}\) quãng đường còn lại là \(t_2\)

Quãng đường đi được trong \(\frac{1}{2}\) thời gian này là: \(s_2=v_2.\frac{t_2}{2}=24.\frac{t_2}{2}=12t_2\left(km\right)\)

Quãng đường cuối cùng là: \(s_3=v_3.\frac{t_2}{2}km\)

Có

 \(s_2+s_3=\frac{2s}{3}\)

\(\rightarrow12t_2+\frac{v_3}{2}.t_2=\frac{2s}{3}\)

\(\rightarrow\left(12+\frac{v_3}{2}\right)t_2=\frac{2s}{3}\)

\(\rightarrow t_2=\frac{2s}{12+\frac{v_3}{2}}h\)

Vận tốc trung bình trên cả quãng đường là: \(v_{tb}=\frac{s}{t_1+t_2}=\frac{s}{\frac{s}{54}+\frac{2s}{12+\frac{v_3}{2}}}\)

Theo đề cho, có \(\frac{s}{\frac{s}{54}+\frac{2s}{3\left(12+\frac{v_3}{2}\right)}}=27\)

Đến đây bạn tự làm nốt nhé

1,did not eat/comed

2,told/met

3,were be/started

4 ,will leave/bought/needed

5,burned/was arriving

6,does/was

Put the verbs in brackets in past pefect or past simple.

1. He (not eat)___didn't eat__until his parent (come)___came__home

2. She (tell)__told____me she never (meet)___met___him.

3. They (be)__we____out for an hour when it (start)___started_____to rain.

4. They (leave)___left_____the shop as soon as they (buy)__bought_____every thing they (need)___needed____.

5. The house (burn)___had burned____to the ground buy the time the firemen (arrive)____had arrived___.

6. As he (do)___had done___a lot of work that day he (be)____had been___very tired.

Phân tích đa thức thành nhân tử

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

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

Tìm x

a) x( x - 3 ) + 7x - 21 = 0

<=> x( x - 3 ) + 7( x - 3 ) = 0

<=> ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

b) ( x - 2 )² + x( 3 - x ) = 6

<=> x² - 4x + 4 + 3x - x² = 6

<=> -x + 4 = 6

<=> -x = 2

<=> x = -2

\(A=\frac{x-2}{x}\)và \(B=\frac{x}{x-2}-\frac{2x}{x^2-4}\)( x ≠ 0 ; x ≠ ±3 )

a) Tại x = 23 ( tmđk ) => \(A=\frac{23-2}{23}=\frac{21}{23}\)

b) P = A.B

\(=\frac{x-2}{x}\times\left(\frac{x}{x-2}-\frac{2x}{x^2-4}\right)\)

\(=\frac{x-2}{x}\times\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=\frac{x-2}{x}\times\frac{x^2+2x-2x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{1}{x}\times\frac{x^2}{x+2}=\frac{x}{x+2}\)

Để P = 4 => \(\frac{x}{x+2}=4\)

=> 4( x + 2 ) = x

=> 4x + 8 - x = 0

=> 3x + 8 = 0

=> x = -8/3 ( tmđk )

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

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

                                                                     \(=\frac{-2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{-1}{x}\)

\(\frac{2x-1}{2x}+\frac{2x}{1-2x}+\frac{1}{4x^2-2x}\)

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

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

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

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