`@` `\text {Ans}`

`\downarrow`

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{1000}\right)\)

`=`\(\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}-\dfrac{1}{4}\right)...\left(\dfrac{1000}{1000}-\dfrac{1}{1000}\right)\)

`=`\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{999}{1000}\)

`=`\(\dfrac{1}{1000}\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.......\dfrac{99}{100}=\dfrac{1}{100}\)

                                                BG

Đổi 2/5=4/10

Sau khi bán 10 con gà trống thì số phần bị giảm đi là:

4/10-3/10=1/10

Số gà trống bán đi bằng số phần số gà trống ban đầu là:

1/10:4/10=1/4

Số gà trống bạn đầu là:

10:1/4=40(gà)

Số gà mái ban là:

40:2/5=100(gà)

Lúc đầu nhà An có số con là:

40+100=140(con)

Đ/S:140 con

sao bài này ko có câu lập luận

x - 56/5 = -3x + 4

x + 3x = 4 + 56/5

4x = 76/5

x = 76/5 : 4

x = 19/5

`@` `\text {Ans}`

`\downarrow`

\(x-\dfrac{56}{5}=-3x+4\)

`\Rightarrow `\(x-\dfrac{56}{5}+3x-4=0\)

`\Rightarrow `\(\left(x+3x\right)+\left(-\dfrac{56}{5}-4\right)=0\)

`\Rightarrow `\(4x-\dfrac{76}{5}=0\)

`\Rightarrow `\(4x=\dfrac{76}{5}\)

`\Rightarrow `\(x=\dfrac{76}{5}\div4\)

`\Rightarrow `\(x=\dfrac{19}{5}\)

Vậy, `x=`\(\dfrac{19}{5}\)

`@` `\text {Ans}`

`\downarrow`

`1.`

`23.7 + 34.6 = 161 + 204 = 365`

`2.`

`4.7 - 1.2`

`= 28 - 2`

`= 26`

`3.`

`34.7 + 45.5`

`= 238 + 225`

`= 463`

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

\(\dfrac{4}{2}A=\dfrac{4}{2}\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)

\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+..\left(\dfrac{1}{32}-\dfrac{1}{32}\right)+\left(1-\dfrac{1}{64}\right)\)

\(A=1-\dfrac{1}{64}\)

\(A=\dfrac{63}{64}\)

\(\dfrac{127}{128}\)

2 x 25 x 36 x 50

= (2 x 50) x( 25 x 36)

= 100 x 900

= 90000

nhanh giúp mình vội quá

Đặt: \(S=2+2^2+2^3+...+2^{60}\)

\(S=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)

\(S=3\cdot\left(2+2^3+...+5^{59}\right)\)

Vậy S chia hết cho 3

________________________

\(S=2+2^2+2^3+...+2^{60}\)

\(S=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)

\(S=2\left(1+2^2\right)+2^2\left(1+2^2\right)+...+2^{58}\left(1+2^2\right)\)

\(S=5\left(2+2^2+....+2^{58}\right)\)

Vậy S chia hết cho 5 

___________________________

\(S=2+2^2+2^3+...+2^{60}\)

\(S=2\left(1+2+2^2\right)+2^2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(S=7\cdot\left(2+2^2+...+2^{58}\right)\)

Vậy  S chia hết cho 7

\(n=a^2+b^2\)

\(\Rightarrow n^2=\left(a^2+b^2\right)^2-4a^2b^2+4a^2b^2=\)

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

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

\(=\left[\left(a-b\right)\left(a+b\right)\right]^2+\left(2ab\right)^2=\)

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

\(A=\left(x^2+x+1\right)^2-4\left(x+2\right)^2+15\)

\(\Rightarrow A=\left(x^2+x+1\right)^2-\left[2\left(x+2\right)\right]^2+15\)

\(\Rightarrow A=\left(x^2+x+1+2x+2\right)\left(x^2+x+1-2x-2\right)+15\)

\(\Rightarrow A=\left(x^2+3x+3\right)\left(x^2-x-1\right)+15\)

\(\Rightarrow A=\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}+3\right)\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1\right)+15\)

\(\Rightarrow A=\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\right]+15\left(1\right)\)

Ta có : \(\left\{{}\begin{matrix}\left(x+\dfrac{3}{2}\right)^2\ge0,\forall x\\\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\\\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4},\forall x\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\left[\left(-\dfrac{3}{2}-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\right]+15\left(x=-\dfrac{3}{2}\right)\\A\ge\left[\left(\dfrac{1}{2}+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right].\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\left[4-\dfrac{5}{4}\right]+15\left(x=-\dfrac{3}{2}\right)\\A\ge\left[4+\dfrac{3}{4}\right].\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\dfrac{9}{4}+15\left(x=-\dfrac{3}{2}\right)\\A\ge\dfrac{19}{4}.\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{27}{16}+15\left(x=-\dfrac{3}{2}\right)\\A\ge-\dfrac{95}{16}+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{267}{16}\left(x=-\dfrac{3}{2}\right)\\A\ge\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow A\ge\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\)

\(\Rightarrow GTNN\left(A\right)=\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\)

Câu 1:

Tổng độ dài 2 đáy:

25 + 15= 40(cm)

Chiều cao hình thang:

(360 x 2): 40 = 18(cm)

Đ.số: 18cm

Câu 2:

Đổi: 2dm= 0,2m

Thể tích bể cá:

1,5 x 0,8 x 0,2= 0,24 (m³)

Đ.số: 0,24m³

Câu 3:

Người đó đi bộ 5 vòng hết:

6 phút x 5  + 43 giây x 5

= 30 phút + 215 giây

= 30 phút + 3 phút + 35 giây

= 33 phút 35 giây

Đ.số: 33 phút 35 giây