26000

2 và 3/5 ha = 26000 m2

\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{10}=\frac{y}{15}\\\frac{z}{4}=\frac{y}{5}\Leftrightarrow\frac{z}{12}=\frac{y}{15}\end{cases}\Leftrightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12};x+y-z=-39}\)

Tính chất dãy tỉ số bằng nhau:

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y-z}{10+15-12}=-\frac{39}{13}=-3\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=-3\Leftrightarrow x=-30\\\frac{y}{15}=-3\Leftrightarrow y=-45\\\frac{z}{12}=-3\Leftrightarrow z=-36\end{cases}}\)

Bằng - 45

mình trả lời rồi nha bạn

2/5ha = 4000m2

=4(2,86+3,14)-6,01x5+3

=24-30,05+3

=27-30,05

=-3,05

\(=4\left(2,86+3,14\right)-6,01.5+3\)

\(=4.6-30,05+3\)

\(=27-30,05=-3,05\)

3,05 nha

1 We should recycle plastic bags and jars

2 How can I get to the nearest bus stop?

3 We are going rafting so you should bring a towel

4 Because of feeling tired, I went to bed early

5 It is not difficult to recycle paper

Câu t3 bạn viết lại cho mình đề dc k, mình k rõ á 

B - B -  A-  D - A - B - A - A - C - D

6B

7B

8A

9D

10A

11B

12A

13A

14A

15D

`(x^2 - 5x + 7)^2 - (2x - 5)^2 = 0`

`<=> x^4 + 25^2 + 49 - 10x^3 - 70x + 14x^2 - (4x^2 - 20x + 25) = 0`

`<=> x^4 - 10x^3 + 39x^2 - 70x + 49 - 4x^2 + 20x - 25 = 0`

`<=> x^4 - 10x^3 + 35x^2 - 50x + 24 = 0`

`<=> x^4 - 4x^3 - 6x^3 + 24x^2 + 11x^2 - 44x - 6x + 24 = 0`

`<=> (x - 4)(x^3 - 6x^2 + 11x - 6) = 0`

`<=> (x - 4)(x^3 - 3x^2 - 3x^2 + 9x + 2x - 6) = 0`

`<=> (x - 4)(x - 3)(x^2 - 3x + 2) = 0`

`<=> (x - 4)(x - 3)(x - 2)(x - 1) = 0`

`<=> x ∈ {4,3,2,1}`

Vậy `S = {4; 3; 2; 1}`

\(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)

\(\Leftrightarrow\left(x^2-5x+7\right)^2=\left(2x-5\right)^2\)

\(\Leftrightarrow\left|x^2-5x+7\right|=\left|2x-5\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+7=-2x+5\\x^2-5x+7=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=0\\x^2-7x+12=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\\\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\end{matrix}\right.\)

\((x^2-5x+7)^2-(2x-5)^2=0 \)

\(⇔(x^2-5x+7+2x-5)(x^2​-5x+7-2x+5)=0 \)

\(⇔(x^2-3x+2)(x^2-7x+12)=0 \)

\(⇔(x^2-2x-x+2)(x^2-3x-4x+12)=0 \)

\(⇔[x(x-2)-(x-2)][x(x-3)-4(x-3)]=0 \)

\(⇔(x-1)(x-2)(x-3)(x-4)=0\)

Vậy\(S=\) \(\left\{1;2;3;4\right\}\)

B