Hướng dẫn giải:

Để tìm được diện tích trồng trau đầu tiên ta phải thực hiện các bước sau:

1. Tìm chiều cao hình thang

Để tìm được h, ta sử dụng định lý Pytago để giải quyết:

- Đầu tiên dễ dàng nhận thấy 2d sẽ bằng b - a. => d = \(\dfrac{b-a}{2}\)

- Áp dụng ĐL Pytago ta có c² = h² + d² 

=> h = \(\sqrt{c^2-d^2}\)

2. Tìm diện tích cái giếng

Bán kính bằng = 1/4 độ dài được cao => \(r=h.\dfrac{1}{4}\)

Diện tích cái giếng bằng: S₁ = r x r x 3.14

3. Tìm diện tích hình thang

Sử dụng công thức hình thang bình thường để tính \(S_2=\dfrac{1}{2}\times\left(a+b\right)\times h\)

4. Diện tích trồng rau khi đó sẽ là S = S₂ - S₁

Code tham khảo:

\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+...+\dfrac{1}{2023\cdot4048}\)

\(=\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4046\cdot4048}\)

\(=\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4046}-\dfrac{1}{4048}\)

\(=\dfrac{1}{4}-\dfrac{1}{4048}=\dfrac{1012-1}{4048}=\dfrac{1011}{4048}\)

\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+\dfrac{1}{4\cdot10}+...+\dfrac{1}{2023\cdot4048}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2023\cdot2024}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{1012-1}{2024}\)

\(=\dfrac{1011}{4048}\)

Có đáp án luôn rồi!

@456 troll thật

a: x+(x+1)+(x+2)+...+(x+30)=496

=>(x+x+...+x)+(1+2+3+...+30)=496

=>\(31x+30\times\dfrac{31}{2}=496\)

=>\(31x+465=496\)

=>31x=31

=>x=1

b: \(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=1530\)

=>\(51x-\left(1+2+3+...+50\right)=1530\)

=>\(51x-\dfrac{50\times51}{2}=1530\)

=>\(51x-1275=1530\)

=>51x=1275+1530=2805

=>x=2805:51=55

a, \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=496\)

\(\Leftrightarrow31x+1+2+...+30=496\Leftrightarrow31x+\dfrac{\left(30+1\right).30}{2}=496\)

\(\Leftrightarrow31x+465=496\Leftrightarrow31x=31\Leftrightarrow x=1\)

b, \(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=1530\)

\(\Leftrightarrow51x+\dfrac{\left(-1-50\right).50}{2}=1530\Leftrightarrow51x-1275=1530\Leftrightarrow51x=2805\Leftrightarrow x=55\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\Leftrightarrow x=-2004\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

=>\(\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

=>\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

=>\(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

=>x+2004=0

=>x=-2004

b, \(275-5\left(2x-1\right)=200\Leftrightarrow5\left(2x-1\right)=75\Leftrightarrow2x-1=15\Leftrightarrow x=8\)

c, \(3x-38:2=206\Leftrightarrow3x-19=206\Leftrightarrow3x=225\Leftrightarrow x=75\)

d, \(2x+x+5x=400\Leftrightarrow8x=400\Leftrightarrow x=50\)

\(sin^210+sin^220+sin^245+sin^270+sin^280\)

\(=sin^210+sin^220+sin^245+cos^220+cos^210=1+1+sin^245=2+\dfrac{1}{2}=\dfrac{5}{2}\)

(x-15):12=79 dư 8

=>\(x-15=79\times12+8=956\)

=>x=956+15=971

(x - 15) : 12 = 79 dư 8

x - 15 = 79 x 12 + 8

x - 15 = 956

x = 956 + 15

x= 971 

\(\left\{47-\left[736:\left(5-3\right)^4\right]\right\}.2021\)

\(=\left\{47-\left[736:2^4\right]\right\}.2021\)

\(=\left\{47-\left[736:16\right]\right\}.2021\)

\(=\left\{47-46\right\}.2021\)

\(=1.2021\)

\(=2021\)

\(\left\{47-\left[736:\left(5-3\right)^4\right]\right\}\cdot2021\)

\(=\left\{47-736:16\right\}\cdot2021\)

\(=\left(47-46\right)\cdot2021=2021\)